Leetcode 15 3Sum Solution in java | 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.

Example 2:

Input: nums = [0,1,1]

Output: []

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

class Solution {

public List<List<Integer>> threeSum(int[] nums) {

Arrays.sort(nums);

List<List<Integer>> res = new LinkedList<>();

for(int i = 0 ; i < nums.length - 2; i++)

{

if( i==0 || (i > 0 && nums[i] != nums[i - 1]))

{

int lo = i + 1, hi = nums.length - 1, sum = 0 - nums[i];

while(lo < hi)

{

if(nums[lo] + nums[hi] == sum)

{

res.add(Arrays.asList(nums[i], nums[lo], nums[hi]));

while(lo < hi && nums[lo] == nums[lo + 1]) lo++;

while(lo < hi && nums[hi] == nums[hi - 1]) hi--;

lo++;

hi--;

}

else if(nums[lo] + nums[hi] < sum) lo++;

else hi--;

}

}

}

return res;

}

}