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.
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 -1Given an array of integers and a number K. Find the count of distinct elements in every window of size K in the array.
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; }}Given an array arr[] of length n. Find all possible unique permutations of the array.
from itertools import permutationsclass Solution: def uniquePerms(self, arr, n): # code here result = [] per = set(permutations(arr)) for i in per: result.append(i) return sorted(result)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.
109 like 231 and 232class Solution{ //Function to swap odd and even bits. public static int swapBits(int n) {\t // Your code\t String s = Integer.toBinaryString(n);\t StringBuilder sb = new StringBuilder(s);\t if(s.length()%2==1 && sb.length()<31){\t sb.insert(0,\"0\");\t }\t \t for(int i=0;i<s.length();i+=2){\t char temp = sb.charAt(i);\t sb.setCharAt(i,sb.charAt(i+1));\t sb.setCharAt(i+1,temp);\t }\t String e = sb.toString();\t \t int res = Integer.parseInt(e, 2);\t return res;\t} }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.
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 ansGiven 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.
00000class 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; }}Given an undirected graph with V vertices and E edges, check whether it contains any cycle or not.
class Solution: #Function to detect cycle in an undirected graph. #def isCycle(self, V, adj):\t\tdef isCycle(self, V, adj):\t #Code here\t\tvis = [0]*V\t\t\t\tdef dfs(cur, parent):\t\t vis[cur] = 1\t\t \t\t for i in adj[cur]:\t\t if not vis[i]:\t\t if dfs(i, cur): return True\t\t elif i != parent:\t\t return True\t\t \t\t return False\t\t\t\tfor i in range(V):\t\t if not vis[i]:\t\t if dfs(i, -1): return True\t\t \t\treturn FalseGiven 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.
```java
class Solution{ class Element implements Comparable
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:
3 2 5 1 4Understanding Edge Case
4 / \\ 2 5 \\ 6public 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; }}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; }}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.
//User function Template for Javaclass 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; }}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.
M and N are coprimes, so there is no chance of double counting a specific number.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.
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; }}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; }}Given a stack, the task is to sort it such that the top of the stack has the greatest element.
class GfG{\tpublic Stack<Integer> sort(Stack<Integer> s){\t\t//add code here.\t\tif (s.isEmpty() != true){\t\t Integer temp = s.pop();\t\t s = sort(s);\t\t sortInsert(s,temp);\t\t}\t\treturn s;\t}\t\tpublic void sortInsert(Stack<Integer> s, Integer val){\t if(s.isEmpty() == true || s.peek() <= val){\t s.push(val);\t }else{\t Integer temp = s.pop();\t sortInsert(s,val);\t s.push(temp);\t }\t}}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?
# cook your dish hereT = int(input())for _ in range(T): M,N,K = list(map(int,input().split(\" \"))) if M>(N*K): print(\"YES\") else: print(\"NO\")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 rupeesAnar, where each costs b rupeesChakri, where each costs c rupeesThe crackers have the following interesting properties:-
A Fuljhari can be lit on its ownTo light an Anar, you will need x FuljharisTo light a Chakri, you will need y FuljharisWhat’s the maximum total of Anars and Chakris that Chef can light up with his given budget.
# cook your dish hereT = 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)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.
# cook your dish hereT = 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)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 is1-> 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.
/* 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; }}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.
# cook your dish here2T = 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)Given a word pat and a text txt. Return the count of the occurences of anagrams of the word in the text.
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; }}Given an array having both positive and negative integers. The task is to compute the length of the largest subarray with sum 0.
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; }}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.
This Soln is By Ayush Modihttps://auth.geeksforgeeks.org/user/ritumodi5/profile
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; } }You are given a number N. Find the total count of set bits for all numbers from 1 to N(both inclusive).
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); } }}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.
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;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
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 maximumGiven an integer N, find its factorial.
1250class 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; }}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;Given two sorted arrays array1 and array2 of size m and n respectively. Find the median of the two sorted arrays.
5.0 -> 55.1 -> 5.1 Odd and Even number of elements. Check explicitly for even number of total elements41 2 3 465 6 7 8 9 101412class 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; } }}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];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.
10-5, this should give a hint of an option of no operation is also an operation Input: num = \"105\", target = 5Output: [\"1*0+5\",\"10-5\"]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; }}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.
arr1 method took 14ms, where as arr2 method only took 6msint[] 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);}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; }}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.
O(n)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; }}Given an m x n matrix, return all elements of the matrix in spiral order.
[[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]]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; }}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.
5 7 672 86 100 112 113119 256 265 349 445 770 892Code:
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]; } }}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?
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; }}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.
This is called as Dutch National Flag Algorithm
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]; }}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.
41 2 3 41Code:
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; }}