Placement

Leet Code : Three (3) Sum Java | CPP | Python Solution

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


Solution :
Java :

Solution :

Java :

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
       
        Arrays.sort(nums);
        Set<List<Integer>> threeSum = new HashSet<>();
        
        for(int i = 0; i < nums.length - 2; i++){
            int j = i + 1;
            int k = nums.length - 1;
            while(j < k){
                int sum = nums[i] + nums[j] + nums[k];
                
                if(sum == 0){
                    List<Integer> temp = new ArrayList<>();
                    temp.add(nums[i]);
                    temp.add(nums[j]);
                    temp.add(nums[k]);
                    threeSum.add(temp);
                    j++;
                    k--;
                }else if(sum > 0){
                    k--;
                }else{
                    j++;
                }
            }
        }
        
        return new ArrayList<>(threeSum);
    }
}

cpp

class Solution {
public:
vector<vector> threeSum(vector& nums) {
vector<vector> res;
sort(nums.begin(),nums.end());
for(int i = 0; i < nums.size();i++) {
if((i > 0) && nums[i]==nums[i-1]){
continue;
}
int left = i + 1;
int right = nums.size() - 1;
while (left < right) {
int sum = nums[i] + nums[left] + nums[right];
if(sum < 0){
left++;
}
else if(sum > 0){
right--;
}
else if(sum == 0) {
res.push_back(vector{nums[i],nums[left],nums[right]});
while (left < right && nums[left] == nums[left+1]){
left++;
}
while (left < right && nums[right] == nums[right-1]) {
right--;
}
left++;
right--;
}
}
}
return res;
}
};

Python:

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        res=[]
        nums.sort()
        for i,a in enumerate(nums) :
            if i > 0 and a==nums[i-1] :
                continue
            l,r=i+1,len(nums) -1 
            while l < r:
                tsome = a + nums[l] +nums[r]
                if tsome > 0 :
                    r -=1
                elif tsome < 0 :
                    l +=1 
                else :
                    res.append([a,nums[l],nums[r]])
                    l +=1
                    while nums[l]==nums[l-1] and l<r :
                        l +=1
        return res
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