Leetcode 1: Two Sum Solution

Problem statement could be found @https://leetcode.com/problems/two-sum/


Code Snippet -

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> retVec;
        int vSize = nums.size();
        for (int i=0;i<vSize;i++){
            int cur = nums[i];
            for (int j=(i+1);j<vSize;j++){
             
                if ((cur + nums[j]) == target){
                     retVec.push_back(i);
                    retVec.push_back(j);
                     return retVec;
                }
            }
        }
          return retVec;
    }
};

Status - Accepted
Runtime: 386 ms

Leetcode 437: Path Sum III Solution

Problem Statement could be found @https://leetcode.com/problems/path-sum-iii/

Solution -
Idea is to save the sum of all intermediate nodes  from Root Node to a given node. Try adding the value of current nodes to all the stored values and compare it with the required value. To store all required value we can use stack. Here, I'm using vector to implement Stack

Following is the code which is accepted by Leetcode. It took 138 msec to execute 126 test cases

class Solution {
    int count;
    int k;
public:
    void trav(TreeNode *root, vector<int> s){
        if (root == NULL){
            return;
        }
        if (root->val == k){
            count++;
        }
        vector<int> temp;
        for (int i=0;i<s.size();i++){
            if ((s[i] + root->val) == k){
                count ++;
               
            }
            temp.push_back(s[i] + root->val);
        }
        temp.push_back(root->val);
        trav(root->left, temp);
        trav(root->right, temp);
        temp.pop_back();
    }
    int pathSum(TreeNode* root, int sum) {
        count = 0;
        k = sum;
        vector<int> s;
        trav(root, s);
        return count;
    }
};