Leetcode 15 3Sum Solution in c++ | Hindi Coding Community

 

Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.

Notice that the solution set must not contain duplicate triplets.

Example 1:

Input: nums = [-1,0,1,2,-1,-4]

Output: [[-1,-1,2],[-1,0,1]]

Explanation: 

nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.

nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.

nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.

The distinct triplets are [-1,0,1] and [-1,-1,2].

Notice that the order of the output and the order of the triplets does not matter.

class Solution {

public:

vector<vector<int>> threeSum(vector<int>& nums) {

vector<vector<int>> ans;

sort(nums.begin(), nums.end());

for(int i=0; i<nums.size(); i++) {

if(i > 0 and nums[i] == nums[i-1]) continue;

int l = i + 1, r = nums.size() - 1;

while (l < r) {

int sum = nums[i] + nums[l] + nums[r];

if(sum == 0) {

ans.push_back({nums[i], nums[l], nums[r]});

while (l < r and nums[l] == nums[l+1]) l++;

while (l < r and nums[r] == nums[r-1]) r--;

l++;

r--;

}

else if(sum < 0) l++;

else if(sum > 0) r--;

}

}

return ans;

}

};

Example 2:

Input: nums = [0,1,1]

Output: []

Explanation: The only possible triplet does not sum up to 0.