1 . Two Sum Given an array of integers nums and an integer target , return indices of the two numbers such that they add up to target . You may assume that each input would have exactly one solution , and you may not use the same element twice. You can return the answer in any order. Example 1: Input: nums = [2,7,11,15], target = 9 Output: [0,1] Output: Because nums[0] + nums[1] == 9, we return [0, 1]. Example 2: Input: nums = [3,2,4], target = 6 Output: [1,2] Example 3: Input: nums = [3,3], target = 6 Output: [0,1] Constraints: 2 <= nums.length <= 10 3 -10 9 <= nums[i] <= 10 9 -10 9 <= target <= 10 9 Only one valid answer exists. solution - class Solution { public: vector<int> twoSum(vector<int>& nums, int target) { unordered_map<int,int> mp; v...
Love for Characters Send Feedback Ayush loves the characters ‘a’, ‘s’, and ‘p’. He got a string of lowercase letters and he wants to find out how many times characters ‘a’, ‘s’, and ‘p’ occurs in the string respectively. Help him find it out. Input: First line contains an integer denoting length of the string. Next line contains the string. Constraints: 1<=n<=10^5 ‘a’<= each character of string <= ‘z’ Output: Three space separated integers denoting the occurrence of letters ‘a’, ‘s’ and ‘p’ respectively. Sample Input: 6 aabsas Sample output: 3 2 0 solution --> #include<bits/stdc++.h> using namespace std; #include<unordered_map> int main() { int n; cin>>n; string s; cin>>s; int count =0,vount=0,bount=0; for(int i =0; i<=s.size(); i++) { if(s[i]=='a' ) { count++; } else if( s[i]=='s') ...