README.md

Problem 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.

Test Case 1
Input: nums = [2,7,11,15], target = 9
Output: [0,1]

Test Case 2
Input: nums = [3,2,4], target = 6
Output: [1,2]

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        // Mapping of elements with their indexes
        unordered_map<int, int> M;
        int N = nums.size();
        for(int i = 0; i < N; i++) {
            int req = target - nums[i];
            if(M.count(req)) {
                return {i, M[req]};
            }
            M[nums[i]] = i;
        }
        
        return {};
    }
};

Problem 2

Valid Parentheses Given a string s containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid. An input string is valid if: Open brackets must be closed by the same type of brackets. Open brackets must be closed in the correct order.

Input: s = "()"
Output: true

Input: s = "()[]{}"
Output: true

class Solution {
public:
    bool isValid(string s) {
        stack<char> S;
        int N = s.length();
        for(int i = 0; i < N; i++) {
            char c = s[i];
            if(c == '(' || c == '{' || c == '[') {
                S.push(c);
            } else {
                if(S.empty()) return false;
                if(c == ')' && S.top() != '(') return false;
                if(c == '}' && S.top() != '{') return false;
                if(c == ']' && S.top() != '[') return false;
                S.pop();
            }
        }
        
        return S.empty();
    }
};

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This file is created by Kiranpal Singh
For Data Structures and Algorithm snippets, click here

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