Q's asked by Top MNC's

 String Challenges  Challenge 1  UpperCase-LowerCase interconversion  Given a string s with both uppercase and lowercase latin characters (‘a’ - ‘z’).  Your task is convert whole string into   1. Lower Case  2. Upper Case    Base idea: ‘a’ - ‘A’ = 32  1. Lowercase to UpperCase Approach  1. Iterate over the string s and if s[i] is a lower case character, then update s[i] -= 32    2. UpperCase to LowerCase Approach  1. Iterate over the string s and if s[i] is a upper case character, then update s[i] +=32    Code:    Challenge 2 Form the biggest number  of integers, our task is to form the biggest number out of those numbers in the string.    Approach:  Sort the string in descending order using inbuilt sort function.    Code     Challenge 3 Max Frequency  s of latin characters, your task is to output the character which has  maximum f...

Recursion - II   Permutation  To print all the permutations of a string.  Idea: for each character s[i] in the given string, we add a character in the ans string and then solve solve s.substr(0,i) + s.substr(i+1)    Sample Input:   ABC  Sample Output:  ABC  ACB  BAC  BCA  CAB  CBA    Time Complexity: O(N*2n )   Space Complexity: O(2n )      permutation(s, “”) will give the required answer   CountPaths  Find the number of ways to reach e from s.  Idea:  We have 6 ways to go forward (1,2,3,4,5,6).  At the starting point s,   Current answer = countPath(s+1,e) + countPath(s+2,e) + countPath(s+3,e) + countPath(s+4,e) + countPath(s+5,e) + countPath(s+6,e)  Time Complexity: O(2n)  Space Complexity: O(2n)    CountPathMaze  Given a 2D grid, find the number of ways to reach (n-1, n-1).  You can go to (i,j) from (i-1,j) and (...

                        Recursion - I   Print String in Reverse Objective: To print the string in reverse.   Base Case: If the string is empty:                       return;   Reverse print the remaining string using recursion, and print then print the current character.  Time Complexity: O(N2) [ IMP]  Space Complexity: O(N2) [ IMP]  Time complexity will be O(N2) because s.substr(i) takes O(N) times and it is called O(N) times.  Space complexity will be O(N2) because s.substr(i) gives a string of O(N) size and it is called O(N) times.  FollowUp: Try to do this in O(N) time and space. Hint: Pass by reference and indices.  Move all ‘x’ to the end of the string If the string is empty:  return “ “;   If the current character ch is ‘x’, we add the resultant string  + ch,   Else ...

 Array Challenges  Maximum Sum Subarray Array   1. Brute Force:   Idea: For each subarray arr[i..j], calculate its sum.  Time Complexity: O(N3 )  Space Complexity: O(1)    2. Prefix Sum Technique:  Idea: For each subarray arr[i..j], calculate its sum. Using prefix sum can  reduce time to calculate the sum of arr[i..j] to O(1)  Time Complexity: O(N2 )  Space Complexity: O(N)     3. Kadane’s Algorithm:  Idea: Start taking the sum of the array, as soon as it gets negative, discard the current subarray, and start a new sum.  Time Complexity: O(N)  Space Complexity: O(1)      Maximum Sum Circular Subarray:    Idea: There will 2 cases,     To get the Min subarray we multiply the array by -1 and get the maximum sum subarray.  Time Complexity: O(N)     Pair Target Sum Problem  Find whether there exist 2 numbers that sum t...

            Arrays Challenge        Subarray with given sum    (Google, Amazon, Facebook) Problem Given an unsorted array A​o​ f size N​  ​of non-negative integers, find a continuous subarray which adds to a given number S​. Constraints 1 <= N <= 105​ 0 <= Ai <= 101​ 0 Example Input:​ N = 5, S = 12 A[] = {1,2,3,7,5} Output:​2​ 4 Explanation:​T​ he sum of elements from 2nd position to 4th position is 12. Solution Brute Force Solution ● Find sum of all possible subarrays. If any of the sum equates to S​ , o​ utput the starting and ending index of the subarray. Time Complexity : O​ (n2​ ​) Optimized Approach Steps: 1.    Keep the pointers st and en, and a variable currSum that stores the sum from st to en. 2.    alize st = 0, en = 0initi 3.    Increment en till currSum + a[en + 1] > S  4.    When 3rd condition occurs, start increasing st until currSum <= S. 5....