Schatz

Find Pivot Index

2022-11-16T00:00:00-06:00

Problem Statement

Given an array of integers nums, calculate the pivot index of this array.

The pivot index is the index where the sum of all the numbers strictly to the left of the index is equal to the sum of all the numbers strictly to the index’s right.

If the index is on the left edge of the array, then the left sum is 0 because there are no elements to the left. This also applies to the right edge of the array.

Return the leftmost pivot index. If no such index exists, return -1.

Solution

class Solution:
    def pivotIndex(self, nums: List[int]) -> int: 
        aa = 0
        bb = sum(nums)
        
        for i,j in enumerate(nums):
            if aa == (bb-aa-j):
                return i
            aa += j
        return -1

Count distinct elements in every window

2021-10-16T00:00:00-05:00

Problem Statement

Given an array of integers and a number K. Find the count of distinct elements in every window of size K in the array.

Edge Case

Check with window size 1 and N

Solution

class Solution
{
    ArrayList<Integer> countDistinct(int A[], int n, int k)
    {
        // code here 
        HashMap<Integer,Integer> map = new HashMap<>();
        int i=0;
        for(;i<k-1;i++){
            map.put(A[i],map.getOrDefault(A[i],0)+1);
        }
        
        ArrayList<Integer>  res  = new ArrayList<Integer>();
        
        for(;i<n;i++){
            map.put(A[i],map.getOrDefault(A[i],0)+1);
            
            res.add(map.size());
            
            map.put(A[i-k+1],map.get(A[i-k+1])-1);
            
            if(map.get(A[i-k+1])<1){
                map.remove(A[i-k+1]);
            }
        }
        return res;
    }
}

All Unique Permutations of an array

2021-10-15T00:00:00-05:00

Problem Statement

Given an array arr[] of length n. Find all possible unique permutations of the array.

Solution

from itertools import permutations
class Solution:
    def uniquePerms(self, arr, n):
        # code here 
        result = []
        per = set(permutations(arr))
        for i in per:
            result.append(i)
        return sorted(result)

Swap all odd and even bits

2021-10-14T00:00:00-05:00

Problem Statement

Given an unsigned integer N. The task is to swap all odd bits with even bits. For example, if the given number is 23 (00010111), it should be converted to 43(00101011). Here, every even position bit is swapped with adjacent bit on the right side(even position bits are highlighted in the binary representation of 23), and every odd position bit is swapped with an adjacent on the left side.

Appris

Refeesfhed about data
Edge Case
Check with numbers other greater than 10⁹ like 2³¹ and 2³²

Notes

Refer Java Docs for Integer Parse Method
Refer Java Docs of String Builder

Solution

class Solution{
    //Function to swap odd and even bits.
    public static int swapBits(int n) 
    {
	    // Your code
	    String s = Integer.toBinaryString(n);
	    StringBuilder sb = new StringBuilder(s);
	    if(s.length()%2==1 && sb.length()<31){
	        sb.insert(0,"0");
	    }
	    
	    for(int i=0;i<s.length();i+=2){
	        char temp = sb.charAt(i);
	        sb.setCharAt(i,sb.charAt(i+1));
	        sb.setCharAt(i+1,temp);
	    }
	    String e = sb.toString();
	    
	    int res = Integer.parseInt(e, 2);
	    return res;
	}
    
}

Valid Pair Sum

2021-10-10T00:00:00-05:00

Problem Statement

Given an array of size N, find the number of distinct pairs {i, j} (i != j) in the array such that the sum of a[i] and a[j] is greater than 0.

Solution

Code:

class Solution:
    def ValidPair(self, a, n): 
        # Sorting the given array 
        a = sorted(a) 
  
        # Variable to store the count of pairs 
        ans = 0
  
        # Loop to iterate through the array 
        for i in range(n): 
  
            # Ignore if the value is negative 
            if (a[i] <= 0): 
                continue
  
            # Finding the index using lower_bound 
            j = lower_bound(a,-a[i] + 1) 
  
            # Finding the number of pairs between 
            # two indices i and j 
            ans += i - j 
        return ans

Maximum difference of zeros and ones in binary string

2021-10-04T00:00:00-05:00

Problem Statement

Given a binary string S consisting of 0s and 1s. The task is to find the maximum difference of the number of 0s and the number of 1s (number of 0s – number of 1s) in the substrings of a string.

Note: In the case of all 1s, the answer will be -1.

Appris

Application of Kadan Algorithim

Edge Case

Solution

class Solution {
    int maxSubstring(String S) {
        // code here
        int val = 0;
        int res = 0;
        for(int i=0;i<S.length();i++){
            if(S.charAt(i)=='0'){
                val++;
            }else{
                val--;
            }
            res = Math.max(val,res);
            if(val<0){
                val=0;
            }
        }
        if(val ==0 && res ==0){
            return -1;
        }
        return res;
    }
}

Check For Bst

2021-10-03T00:00:00-05:00

Problem Statement

Given an undirected graph with V vertices and E edges, check whether it contains any cycle or not.

Solution

class Solution:
    
    #Function to detect cycle in an undirected graph.
    #def isCycle(self, V, adj):
	
	def isCycle(self, V, adj):
	    #Code here
		vis = [0]*V
		
		def dfs(cur, parent):
		    vis[cur] = 1
		    
		    for i in adj[cur]:
		        if not vis[i]:
		            if dfs(i, cur): return True
		        elif i != parent:
		            return True
		            
		    return False
		
		for i in range(V):
		    if not vis[i]:
		        if dfs(i, -1): return True
		        
		return False

Smallest range in K lists

2021-10-02T00:00:00-05:00

Problem Statement

Given K sorted lists of integers, KSortedArray[] of size N each. The task is to find the smallest range that includes at least one element from each of the K lists. If more than one such range’s are found, return the first such range found.

Note

This a officail solution given by leetcode editorial
Solution

```java

class Solution { class Element implements Comparable { int val; int ind; int row; Element (int val, int ind, int row) { this.val= val; this.ind= ind; this.row= row; } public int compareTo(Element e){ return this.val- e.val; } } int[] findSmallestRange(int[][] KSortedArray,int n,int k) { int length=Integer.MAX_VALUE, low= Integer.MAX_VALUE, high= Integer.MIN_VALUE, max= Integer.MIN_VALUE; PriorityQueue pq= new PriorityQueue<>(); for(int i=0; i<k; i++){ max= Math.max(max, KSortedArray[i][0]); pq.add(new Element(KSortedArray[i][0], 0, i)); } length= max- pq.peek().val; low= pq.peek().val; high= max; // System.out.println(length + " " + low + " " + high); boolean flag= true; while(flag){ // System.out.println(length + " " + low + " " + high); Element x= pq.poll(); int row= x.row; int ind= x.ind; if(!(ind+1<n)){ flag= false; break; } else { Element ne= new Element(KSortedArray[row][ind+1], ind+1, row); pq.add(ne); max= Math.max(max, KSortedArray[row][ind+1]); if(length> max- pq.peek().val){ length= max- pq.peek().val; low= pq.peek().val; high= max; } } } int arr[]= {low, high}; return arr; } } ```

Check for BST

2021-10-01T00:00:00-05:00

Problem Statement

Given the root of a binary tree. Check whether it is a BST or not. Note: We are considering that BSTs can not contain duplicate Nodes. A BST is defined as follows:

The left subtree of a node contains only nodes with keys less than the node’s key.
The right subtree of a node contains only nodes with keys greater than the node’s key.
Both the left and right subtrees must also be binary search trees.

Appris

Inorder Traversal of Binary Search Tree results sorted array.

Edge Case

3 2 5 1 4

Understanding Edge Case

Note

This is the official solution provided by GeeksForGeeks Editorial

Solution

Efficient Solution

public class Solution
{
    Node prev;
    //Function to check whether a Binary Tree is BST or not.
    boolean isBST(Node root)
    {
        // code here.
        if (root != null)
        {
            if (!isBST(root.left))
                return false;
            
            // allows only distinct values node
            if (prev != null && root.data <= prev.data )
                return false;
            prev = root;
            return isBST(root.right);
        }
        return true;
    }
}

🔴🔴🔴 Edge Case Failing Code

public class Solution{ 
 Node prev;
 //Function to check whether a Binary Tree is BST or not.
 boolean isBST(Node root){
     // code here.
     if (root != null){
         /* False if left is > than node */
         if (node.left != null && node.left.data > node.data)
             return false;
            
         /* False if right is < than node */
         if (node.right != null && node.right.data < node.data) 
             return false; 
            
         // allows only distinct values node
         if (prev != null && root.data <= prev.data )
             return false;
            
         prev = root;
         return isBST(root.right);
     }
     return true;
 }
}

Vertical Traversal of Binary Tree

2021-09-30T00:00:00-05:00

Problem Statement

Given a Binary Tree, find the vertical traversal of it starting from the leftmost level to the rightmost level. If there are multiple nodes passing through a vertical line, then they should be printed as they appear in level order traversal of the tree.

Note

Check for traversal type asked.

Solution

//User function Template for Java
class MetaNode{
    Node head;
    int pos;
    MetaNode(Node root,int position){
        head = root;
        pos = position;
    }
}


class Solution
{
    //Function to find the vertical order traversal of Binary Tree.
    static ArrayList <Integer> verticalOrder(Node root){
        // add your code here
        Map<Integer, List <Integer>>  res = new TreeMap<>();
        ArrayDeque<MetaNode> q = new ArrayDeque<>();
        
        MetaNode head = new MetaNode(root,0);
        q.add(head);
        
        while(q.isEmpty()!=true){
            int cnt = q.size();
            for(int i=0;i<cnt;i++){
                MetaNode cur = q.pop();
                if (res.containsKey(cur.pos)!=true){
                    res.put(cur.pos,new LinkedList <Integer>());
                }
                res.get(cur.pos).add(cur.head.data);
                if(cur.head.left!=null){
                    q.addLast(new MetaNode(cur.head.left,cur.pos-1));
                }
                if(cur.head.right!=null){
                    q.addLast(new MetaNode(cur.head.right,cur.pos+1));
                }
            }
        }
        
        ArrayList<Integer> soln =  new ArrayList<Integer>();
        for(List<Integer> temp:res.values()){
            for(Integer val:temp){
                soln.add(val);
            }
        }
        return soln;
    }
}

Restricted Pacman

2021-09-29T00:00:00-05:00

Problem Statement

In the game of Restricted Pacman, an infinite linear path is given. Pacman has to start at position 0 and eat as many candies as possible. In one move he can only jump a distance of either M or N. If M and N are coprime numbers, find how many candies will be left on the board after the game is over.
Note: The result is always finite as after a point X every index in the infinite path can be visited.

Edge Case

Check with upper limits.
M and N are coprime, to check for prime factors before testing with code.

Note

M and N are coprimes, so there is no chance of double counting a specific number.

Trivia

Find the largest index that can’t be obtained using any combination of M & N using Frobenius Number which defines X = (M * N) – M – N.

Solution

O(1) if you are familar with math behind Coin Problem

class Solution{
 static int candies(int m, int n){
     // Your code goes here
     return ((m-1)*(n-1)/2);
 }
}

O(n) Efficient Iterative DP Solution

class Solution{
 static int candies(int m, int n){
     // Your code goes here
     int cnt=0;
     int X = (n*m)-(n+m)+1;
     boolean[] flags = new boolean[X];
     flags[0]=true;
     for(int i=1;i<X;i++){
         if(i-n >= 0 && flags[i-n]==true){
             flags[i]=true;
         }
         if(i-m >= 0 && flags[i-m]==true){
             flags[i]=true;
         }
         if(!flags[i]){
             cnt++;
         }
     }
     return cnt;
 }
}

Brute Force Recusrive DP Solution

class Solution{
 static int N,M;
 static int candies(int m, int n){
     // Your code goes here
     N = n;
     M = m;
     boolean[] flags = new boolean[n*m];
     rectoggle(flags,0);
        
     return cntflag(flags);
 }
    
 static void rectoggle(boolean[] flags,int i){
     if(i<flags.length){
         flags[i]=true;
         rectoggle(flags,i+M);
         rectoggle(flags,i+N);
     }
     return;
 }
    
 static int cntflag(boolean[] flags){
     int cnt = 0;
     for(boolean flag:flags){
         if(!flag){
             cnt++;
         }
     }
     return cnt;
 }
}

Sort a stack

2021-09-28T00:00:00-05:00

Problem Statement

Given a stack, the task is to sort it such that the top of the stack has the greatest element.

Solution

class GfG{
	public Stack<Integer> sort(Stack<Integer> s){
		//add code here.
		if (s.isEmpty() != true){
		    Integer temp = s.pop();
		    s = sort(s);
		    sortInsert(s,temp);
		}
		return s;
	}
	
	public void sortInsert(Stack<Integer> s, Integer val){
	    if(s.isEmpty() == true || s.peek() <= val){
	        s.push(val);
	    }else{
	        Integer temp = s.pop();
	        sortInsert(s,val);
	        s.push(temp);
	    }
	}
}

The Preparations

2021-09-27T00:00:00-05:00

Problem Statement

Chef has an exam which will start exactly M minutes from now. However, instead of preparing for his exam, Chef started watching Season-1 of Superchef. Season-1 has N episodes, and the duration of each episode is K minutes.

Will Chef be able to finish watching Season-1 strictly before the exam starts?

Solution

# cook your dish here
T = int(input())
for _ in range(T):
    M,N,K = list(map(int,input().split(" ")))
    
    if M>(N*K):
        print("YES")
    else:
        print("NO")

Maximum Light Up

2021-09-27T00:00:00-05:00

Problem Statement

Chef has allocated himself a budget of P rupees to buy Diwali crackers. There are only 3 types of crackers available in the market and each type of cracker can be bought any number of times.

Fuljhari, where each costs a rupees Anar, where each costs b rupees Chakri, where each costs c rupees The crackers have the following interesting properties:-

A Fuljhari can be lit on its own To light an Anar, you will need x Fuljharis To light a Chakri, you will need y Fuljharis What’s the maximum total of Anars and Chakris that Chef can light up with his given budget.

Solution

# cook your dish here
T = int(input())
for _ in range(T):
    P,a,b,c,x,y = list(map(int,input().split(" ")))
    b = b+a*x;
    c = c+a*y;
    
    cnt = 0
    if(P//b > P//c):
        cnt+= P//b
        rem = P - (P//b * b)
        cnt += rem//c
    else:
        cnt+= P//c
        rem = P - (P//c * c)
        cnt += rem//b
    
    print(cnt)

Magical Doors

2021-09-27T00:00:00-05:00

Problem Statement

Chef wants to cross a hallway of N doors. These N doors are represented as a string. Each door is initially either open or close, represented by 1 or 0 respectively. Chef is required to go through all the doors one by one in the order that they appear, starting from the leftmost door and moving only rightwards at each step.

To make the journey easier, Chef has a magic wand, using which Chef can flip the status of all the doors at once. Determine the minimum number of times Chef has to use this wand to cross the hallway of doors.

For example, the doors are 10011. Chef will start from the left and enter the first door. After passing through that door, the magic wand is waved. This flips the string to 01100. Now Chef passes through the next two doors in one go. Again, just before reaching the 4th door, the magic wand is waved. Now that the string is in its original state, Chef passes through the last two doors as well. The minimum number of times the magic wand needed to be used here was 2.

Solution

# cook your dish here
T = int(input())
for _ in range(T):
    S = input()
    cnt = 0
    
    if(S[0]=="1"):
        i = 1;
        while(i<len(S)):
            while(i<len(S) and S[i]==S[i-1]):
                i+=1;
            if(i<len(S)):
                cnt+=1
                i+=1;
    else:
        cnt+=1
        i = 1;
        while(i<len(S)):
            while(i<len(S) and S[i]==S[i-1]):
                i+=1;
            if(i<len(S)):
                cnt+=1
                i+=1;
    print(cnt)

Finding middle element in a linked list

2021-09-27T00:00:00-05:00

Problem Statement

Given a singly linked list of N nodes. The task is to find the middle of the linked list. For example, if the linked list is 1-> 2->3->4->5, then the middle node of the list is 3. If there are two middle nodes(in case, when N is even), print the second middle element. For example, if the linked list given is 1->2->3->4->5->6, then the middle node of the list is 4.

Edge Case

Check with Single Element

Solution

/* Node of a linked list
 class Node {
   int data;
    Node next;
    Node(int d)  { data = d;  next = null; }
}
*/

class Solution
{
    int getMiddle(Node head){
         // Your code here.
        if (head == null ){
            return 0;
        }
        if(head.next ==null){
            return head.data;
        }
        Node fast = head;
        Node slow = head;
        while(fast!=null && fast.next!=null){
            fast = fast.next.next;
            slow = slow.next;
        }
        return slow.data;
    }
}

Chef in Vaccination Queue

2021-09-27T00:00:00-05:00

Problem Statement

There are N people in the vaccination queue, Chef is standing on the Pth position from the front of the queue. It takes X minutes to vaccinate a child and Y minutes to vaccinate an elderly person. Assume Chef is an elderly person.

You are given a binary array A1,A2,…,AN of length N, where Ai=1 denotes there is an elderly person standing on the ith position of the queue, Ai=0 denotes there is a child standing on the ith position of the queue. You are also given the three integers P,X,Y. Find the number of minutes after which Chef’s vaccination will be completed.

Solution

# cook your dish here2
T = int(input())
for _ in range(T):
    N,P,X,Y = list(map(int,input().split(" ")))
    A = input().split(" ")
    val = 0
    for i in range(0,P):
        if(A[i]=="0"):
           val+=X
        else:
            val+=Y
    print(val)

Count Occurences of Anagrams

2021-09-26T00:00:00-05:00

Problem Statement

Given a word pat and a text txt. Return the count of the occurences of anagrams of the word in the text.

Note

Check for substrings not subsequence or subset. Read Here

Solution

class Solution {

    int search(String pat, String txt) {
        // code here
        if(pat.length()>txt.length()){
            return 0;
        }
        int a = (int)'a';
        int idx = 0;
        int pl = pat.length();
        int tl = txt.length();
        
        int[] pat_cnt = new int[26];
        int[] txt_cnt = new int[26];
        
        for(;idx<pl;idx++){
            pat_cnt[((int)pat.charAt(idx))-a]++;
            txt_cnt[((int)txt.charAt(idx))-a]++;
        }
        
        int res = 0;
        if(Arrays.equals(pat_cnt,txt_cnt)){
            res++;
        }
        
        for(;idx<tl;idx++){
            txt_cnt[((int)txt.charAt(idx-pl))-a]--;
            txt_cnt[((int)txt.charAt(idx))-a]++;
            
            if(Arrays.equals(pat_cnt,txt_cnt)){
                res++;
            }
        }
        
        return res;
    }
}

Largest subarray with 0 sum

2021-09-25T00:00:00-05:00

Problem Statement

Given an array having both positive and negative integers. The task is to compute the length of the largest subarray with sum 0.

Appris

Check whether the prefix sum has appeared earlier, cause this shows the sum of values between current element and previous is equivalent to zero

Solution

class GfG
{
    int maxLen(int arr[], int n){
        // Your code here
        Map<Integer, Integer> map = new HashMap<>();
        int sum = 0;
        map.put(sum,0);
        sum+=arr[0];
        
        int res = 0;
        map.put(sum,1);
        
        for(int i=1;i<n;i++){
            sum+=arr[i];
            Integer val = map.get(sum);
            if(val!=null){
                res = Math.max(res,i-val+1);
            }else{
                map.put(sum,i+1);
            }
        }
        
        return res;
    }
}

Check Tree Traversal

2021-09-24T00:00:00-05:00

Problem Statement

Given Preorder, Inorder and Postorder traversals of some tree of size N. The task is to check if they are all of the same tree or not.

Note

This Soln is By Ayush Modi https://auth.geeksforgeeks.org/user/ritumodi5/profile

Solution

class Tree{
    int val;
    Tree left, right;
    Tree(int val){
        this.val = val;
        left = null; right = null;
    }
}

class Solution{
    static int idx;
    static boolean isPossible;
    static boolean checktree(int preorder[], int inorder[], int postorder[], int N){
        idx = 0;
        isPossible = true;
        Map<Integer, Integer> hmap = new HashMap<>();
        for(int i = 0; i < inorder.length; i++) hmap.put(inorder[i], i);
        
        Tree root = buildTree(inorder, preorder, hmap, 0, N-1);
        if(!isPossible) return false;
        List<Integer> post = new ArrayList<>();
        buildPost(root, post);

        return Arrays.equals(post.stream().mapToInt(i->i).toArray(), postorder);
    }
    
    private static void buildPost(Tree root, List<Integer> post){
        if(root == null) return;
        buildPost(root.left, post);
        buildPost(root.right, post);
        post.add(root.val);
    }
    
    private static Tree buildTree(int[] inorder, int[] preorder, Map<Integer, Integer> hmap, int start, int end){
        if(start > end) return null;
        if(!isPossible) return null;
        int val = preorder[idx++];
        Tree root = new Tree(val);
        if(!hmap.containsKey(val)){
            isPossible = false;
            return null;
        }
        int pos = hmap.get(val);
        if(pos < start || pos > end){
            isPossible = false;
            return null;
        }
        root.left = buildTree(inorder, preorder, hmap, start, pos-1);
        root.right = buildTree(inorder, preorder, hmap, pos+1, end);
        return root;
    }
    
}

Count total set bits

2021-09-23T00:00:00-05:00

Problem Statement

You are given a number N. Find the total count of set bits for all numbers from 1 to N(both inclusive).

Solution

class Solution{
    
    public static int log2(int N){
  
        // calculate log2 N indirectly
        // using log() method
        int result = (int)Math.floor(Math.log(N) / Math.log(2));
        return result;
    }
    
    //Function to return sum of count of set bits in the integers from 1 to n.
    public static int countSetBits(int n){
        // Your code here
        if((n&(n-1)) == 0){
            //if n is power of 2 directly return result
            return 1+ log2(n)*(n/2);
       }else{
           // y is 2th power  before n
           int y =(int) Math.pow(2,log2(n));
           return (n-y) + countSetBits(y)+ countSetBits(n-y);
       }
    }
}

Repeated String Match

2021-09-22T00:00:00-05:00

Problem Statement

Given two strings A and B, find the minimum number of times A has to be repeated such that B becomes a substring of the repeated A. If B cannot be a substring of A no matter how many times it is repeated, return -1.

Appris

String questions, Python is goto.

Solution

class Solution:
    def repeatedStringMatch(self, A, B):
        # code here
        cnt = 1
        S = A;
        while(len(B) > len(S)):
            S+=A
            cnt+=1
        if B in S:
            return cnt
        if B in S+A:
            return cnt+1
        return -1;

Dam of Candies

2021-09-21T00:00:00-05:00

Problem Statement

Geek wants to make a special space for candies on his bookshelf. Currently, it has N books, each of whose height is represented by the array height[] and has unit width. Help him select 2 books such that he can store maximum candies between them by removing all the other books from between the selected books. The task is to find out the area between two books that can hold the maximum candies without changing the original position of selected books

Solution

 
class Solution:
    def maxCandy(self, height, n): 
        # Your code goes here
        maximum=0;
        f=0
        l=n-1
        while(f<l):
            if height[f]<height[l]:
                maximum=max(maximum,(l-f-1)*height[f])
                f+=1
            else:
                maximum=max(maximum,(l-f-1)*height[l])
                l-=1
        return maximum

Factorials of large numbers

2021-09-20T00:00:00-05:00

Problem Statement

Given an integer N, find its factorial.

Appris

Python is better for these types of problems, datatype overflow will not be an issue.
Integers in Python 3 are limited only by available memory. Python 2 had two integer types - int and long. They are unified from Python 3.
BigInteger is an 0ption in Java

Edge Case

Factorial problems, look for corner cases of the input range. Look for overflow and run-time errors.

```
1
```
```
2
```
```
50
```

Solution

Java

class Solution {
    static void arrayMultiplication(ArrayList<Integer> res, int factor){
        int carry = 0;
        for(int i=res.size()-1;i>=0;i--){
            int temp = res.get(i)*factor + carry;
            res.set(i, temp%10);
            carry = temp/10;
        }
        while(carry>0){
            res.add(0,carry%10);
            carry = carry/10;
        }
    }
    
    static ArrayList<Integer> factorial(int N){
        //code here
        ArrayList<Integer> res = new ArrayList<Integer>();
        if(N==1 || N==2){
            res.add(N);
            return res;
        }
        
        res.add(2);
        for(int i=3;i<=N;i++){
            arrayMultiplication(res,i);
        }
        
        return res;
    }
}

Python

class Solution:
    def factorial(self, N):
        #code here
        temp = 1
        for i in range(1,N+1):
            temp*=i;
        
        res = []
        for i in str(temp):
            res.append(i)
        return res;

Median of 2 Sorted Arrays of Different Sizes

2021-09-19T00:00:00-05:00

Problem Statement

Given two sorted arrays array1 and array2 of size m and n respectively. Find the median of the two sorted arrays.

Appris

In Python, it is important have integer decimal, as integer itself
```
5.0 -> 5
5.1 -> 5.1 
```
In Java, Divide by double to convert the return into double, rather than type conversion.

Note

The naive approach would be to merge the two arrays and return the median element(s).
But for the naive method, there is a lot of overhead. We only need the midst element(s). So we could do what we do in merge sort, iterate over the two arrays and have a counter to reach the middle element(s).
This can be achieved in serval ways, one of which is to create an auxiliary array of size (m + n) /2 +1. (We only need the middle element(s)). Therefore this space is more than enough.
But I believe this can be further optimized, by only storing one extra element, in the case of an even number of elements.May reduce the space complexity but not the complexity, cause a check to be done several times

Edge Case

Look for Odd and Even number of elements. Check explicitly for even number of total elements

4
1 2 3 4
6
5 6 7 8 9 10

Solution

Java

class GFG 
{ 
    static double medianOfArrays(int n, int m, int a[], int b[]) 
    {
        // Your Code Here
        int mid = ((n+m)/2)+1;
        int[] arr = new int[mid];
        
        int i=0,j=0,k=0;
        
        while(k<mid && i<n && j<m ){
            if(a[i]<=b[j]){
                arr[k++]=a[i++];
            }else{
                arr[k++]=b[j++];
            }
        }
        
        while(k<mid && i<n){
            arr[k++]=a[i++];
        }
        
        while(k<mid && j<m){
            arr[k++]=b[j++];
        }
        
        if((n+m)%2==1){
            return arr[mid-1];
        }else{
            return (arr[mid-1] + arr[mid-2])/2.0;
        }

    }
}

Python

class Solution:
    def MedianOfArrays(self, array1, array2):
        #code here
        """
        Naive approach
        """
        a = array1+array2
        a.sort();
        
        n = len(a);
        if(n%2==0):
            val = (a[n//2]+a[n//2-1])/2
            if int(val)!=val:
                return val
            else:
                return int(val);
        else:
            return a[n//2];

Expression Add Operators

2021-09-19T00:00:00-05:00

Problem Statement

Given a string num that contains only digits and an integer target, return all possibilities to add the binary operators ‘+’, ‘-‘, or ‘*’ between the digits of num so that the resultant expression evaluates to the target value.

Edge Case

10-5, this should give a hint of an option of no operation is also an operation
```
Input: num = "105", target = 5
Output: ["1*0+5","10-5"]
```

Appris

String Builder preety much works like python list.

Note

This the officail solution given by leetcode editorial

Solution

class Solution {

  public ArrayList<String> answer;
  public String digits;
  public long target;

  public void recurse(
      int index, long previousOperand, long currentOperand, long value, ArrayList<String> ops) {
    String nums = this.digits;

    // Done processing all the digits in num
    if (index == nums.length()) {

      // If the final value == target expected AND
      // no operand is left unprocessed
      if (value == this.target && currentOperand == 0) {
        StringBuilder sb = new StringBuilder();
        ops.subList(1, ops.size()).forEach(v -> sb.append(v));
        this.answer.add(sb.toString());
      }
      return;
    }

    // Extending the current operand by one digit
    currentOperand = currentOperand * 10 + Character.getNumericValue(nums.charAt(index));
    String current_val_rep = Long.toString(currentOperand);
    int length = nums.length();

    // To avoid cases where we have 1 + 05 or 1 * 05 since 05 won't be a
    // valid operand. Hence this check
    if (currentOperand > 0) {

      // NO OP recursion
      recurse(index + 1, previousOperand, currentOperand, value, ops);
    }

    // ADDITION
    ops.add("+");
    ops.add(current_val_rep);
    recurse(index + 1, currentOperand, 0, value + currentOperand, ops);
    ops.remove(ops.size() - 1);
    ops.remove(ops.size() - 1);

    if (ops.size() > 0) {

      // SUBTRACTION
      ops.add("-");
      ops.add(current_val_rep);
      recurse(index + 1, -currentOperand, 0, value - currentOperand, ops);
      ops.remove(ops.size() - 1);
      ops.remove(ops.size() - 1);

      // MULTIPLICATION
      ops.add("*");
      ops.add(current_val_rep);
      recurse(
          index + 1,
          currentOperand * previousOperand,
          0,
          value - previousOperand + (currentOperand * previousOperand),
          ops);
      ops.remove(ops.size() - 1);
      ops.remove(ops.size() - 1);
    }
  }

  public List<String> addOperators(String num, int target) {

    if (num.length() == 0) {
      return new ArrayList<String>();
    }

    this.target = target;
    this.digits = num;
    this.answer = new ArrayList<String>();
    this.recurse(0, 0, 0, 0, new ArrayList<String>());
    return this.answer;
  }
}

Intersection of Two Arrays II

2021-09-18T00:00:00-05:00

Problem Statement

Given a linked list of N nodes such that it may contain a loop. A loop here means that the last node of the link list is connected to the node at position X. If the link list does not have any loop, X=0. Remove the loop from the linked list, if it is present.

Appris

Streams in Java take much time then simple array value insertion. arr1 method took 14ms, where as arr2 method only took 6ms

int[] arr1 = lst.stream().mapToInt(Integer::intValue).toArray();

int arr2[] = new int[lst.size()];
for(int i=0;i<lst.size();i++) {
    arr2[i] = lst.get(i);
}

Solution

class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {

        // Map to count frequeny of numbers
        Map<Integer,Integer> map = new HashMap<>();
        
        for(int i:nums1){
            map.put(i,map.getOrDefault(i,0)+1);
        }
        
        // Use Resizeable array, since there may n or m numbers 
        // of numbers in results 
        ArrayList<Integer> lst = new ArrayList<>();
        
        for(int i:nums2){
            // for each number in array 2, update the character 
            // frequency of initial map 
            if(map.containsKey(i)==true && map.get(i)>0){
                lst.add(i);
                map.put(i,map.get(i)-1);
            }
        }
        
        // ArrayList -> Array 
        int arr[] = new int[lst.size()];
        for(int i=0;i<lst.size();i++) {
            arr[i] = lst.get(i);
        }

        return arr;
    }
}

Sum of Query II

2021-09-18T00:00:00-05:00

Problem Statement

You are given an array arr[] of n integers and q queries in an array queries[] of length 2*q containing l, r pair for all q queries. You need to compute the following sum over q queries.

Example

Need to find the sum of numbers between two points
- why not find complete sum and remove
  - the staring and
  - the ending sum
- starting and ending sum can be found out in O(n)

Solution

class Solution{
    List<Integer> querySum(int n, int arr[], int q, int queries[])
    {
        // code here
        // find prefix_sum and suffix_sum
        int[] preSum = new int[n+1];
        int[] sufSum = new int[n+1];
        
        preSum[0] = 0;
        sufSum[n] = 0;
        int t = 0;
        for(int i=0;i<n;i++){
            preSum[i+1]=arr[i]+preSum[i];
            t+=arr[i];
        }
        
        for(int i=n-1;i>=0;i--){
            sufSum[i]=arr[i]+sufSum[i+1];
        }
        
        List<Integer> res = new ArrayList<>();
        
        for(int i=0;i<2*q;i+=2){
            res.add(t-preSum[queries[i]-1]-sufSum[queries[i+1]]);
        }
        
        return res;
    }
}

Spiral order in matrix.

2021-09-17T00:00:00-05:00

Problem Statement

Given an m x n matrix, return all elements of the matrix in spiral order.

Edge Cases

Even and Odd Number of Rows and Columns

[[1,2,3],[4,5,6],[7,8,9],[7,8,9]]
[[1,2,3,4],[5,6,7,8],[9,10,11,12],[9,10,11,12]]

Solution

Code:

class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        List<Integer> res = new ArrayList<>();
        
        int n = matrix.length;
        int m = matrix[0].length;
        
        int x0 =0,x1=m-1;
        int y0=0,y1=n-1;
        
        
        while(x0<=x1 && y0<=y1){
            
            for(int i=x0;i<=x1;i++){
                res.add(matrix[y0][i]);
            }
            y0++;
            
            for(int j=y0;j<=y1;j++){
                res.add(matrix[j][x1]);
            }
            x1--;
            
            if(y0<=y1){
                for(int i=x1;i>=x0;i--){
                    res.add(matrix[y1][i]);
                }
            }
            y1--;

            if(x0<=x1){
                for(int j=y1;j>=y0;j--){
                    res.add(matrix[j][x0]);
                }
            }
            x0++;
        }
        return res;
    }
}

K-th element of two sorted Arrays

2021-09-17T00:00:00-05:00

Problem Statement

Given two sorted arrays arr1 and arr2 of size N and M respectively and an element K. The task is to find the element that would be at the k’th position of the final sorted array.

Edge Cases

What if arrays are sorted and given.

7 6
86 100 112 113
256 265 349 445 770 892

Solution

Code:

class Solution {
    public long kthElement( int arr1[], int arr2[], int n, int m, int k) {
        int a1 = 0;
        int a2 = 0;
        boolean flag = true;
        
        for(int i=0;i<k;i++){
            if(a1>=n){
                flag = false;
                a2++;
            }else if (a2>=m){
                flag = true;
                a1++;
            }else{
                if(arr1[a1]<=arr2[a2]){
                    flag=true;
                    a1++;
                }else{
                    flag=false;
                    a2++;
                }
            }
        }
        if(flag){
            return  arr1[a1-1];
        }else{
            return arr2[a2-1];
        }
    }
}

N meetings in one room

2021-09-16T00:00:00-05:00

Problem Statement

There is one meeting room in a firm. There are N meetings in the form of (start[i], end[i]) where start[i] is start time of meeting i and end[i] is finish time of meeting i. What is the maximum number of meetings that can be accommodated in the meeting room when only one meeting can be held in the meeting room at a particular time?

Solution

class Solution 
{
    //Function to find the maximum number of meetings that can
    //be performed in a meeting room.
    public static int maxMeetings(int start[], int end[], int n)
    {
        // add your code here
        ArrayList<Meet> schd = new ArrayList<>();
        for(int i=0;i<n;i++){
            schd.add(new Meet(start[i],end[i]));
        }
        Collections.sort(schd,(Meet m1, Meet m2) -> {
            if(m1.e==m2.e){
                return m2.s-m1.s;
            }
            return m1.e-m2.e;
        });
        
        int cnt = 0;
        // System.out.println(schd);
        int l = Integer.MIN_VALUE;
        for(int i=0;i<n;i++){
            if(schd.get(i).s>l){
                l = schd.get(i).e;
                cnt++;
            }
        }
        return cnt;
    }
}

Sort an array of 0s, 1s and 2s

2021-09-15T00:00:00-05:00

Problem Statement

Given an array A[] consisting 0s, 1s and 2s. The task is to write a function that sorts the given array. The functions should put all 0s first, then all 1s and all 2s in last.

Trivia

This is called as Dutch National Flag Algorithm

Solution

Code:

class Solution
{
    public static void sort012(int a[], int n)
    {
        // code here 
        int l =0;
        int m =0;
        int f = n-1;
        while(m<=f){
            if(a[m]==0){
                swap(a,m,l);
                m++;
                l++;
            }else if(a[m]==1){
                m++;
            }else if(a[m]==2){
                swap(a,m,f);
                f--;
            }
        }
    }
    private static void swap(int[] arr, int i, int j){
        //3      1        2
        arr[i]=arr[i]+arr[j];
        //1     3        2
        arr[j]=arr[i]-arr[j];
        //2     3        1
        arr[i]=arr[i]-arr[j];
    }
}

Remove loop in Linked List

2021-09-14T00:00:00-05:00

Problem Statement

Edge Cases

Even Length Circular Linkedlist.

4
1 2 3 4
1

Solution

Code:

class Solution
{
    //Function to remove a loop in the linked list.
    public static void removeLoop(Node head){
        // code here
        // remove the loop without losing any nodes
        if(head==null || head.next==null){
            return;
        }

        Node fast = head;
        Node slow = head;
        Node prev = head;
        do{
            fast = fast.next.next;
            prev = slow;
            slow = slow.next;
        }while(slow != fast && fast!=null && fast.next!=null);
        
        // check if no loop
        if(fast==null || fast.next == null){
            return;
        }
        
        if(prev.next==head){
            prev.next=null;
            return;
        }

        // even length Circular Linked List Edge Case
        fast = head;
        Node temp=null;
        while(fast!=slow){
            fast=fast.next;
            temp = slow;
            slow=slow.next;
        }
        temp.next = null;
        return;
    }
}