3Sum LeetCode

 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.

Example 3:


Input: nums = [0,0,0]

Output: [[0,0,0]]

Explanation: The only possible triplet sums up to 0.

 

code :1

import java.util.*;

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> ans = new ArrayList<>();
        int n = nums.length;
        Arrays.sort(nums); // Step 1: Sort the array

        for (int i = 0; i < n - 2; i++) {
            // Skip duplicate elements for 'i'
            if (i > 0 && nums[i] == nums[i - 1]) continue;

            int left = i + 1, right = n - 1;
            int target = 0;

            while (left < right) {
                int sum = nums[i] + nums[left] + nums[right];

                if (sum == target) {
                    // Store triplet
                    ans.add(Arrays.asList(nums[i], nums[left], nums[right]));

                    // Move left and right pointers while avoiding duplicates
                    while (left < right && nums[left] == nums[left + 1]) left++;
                    while (left < right && nums[right] == nums[right - 1]) right--;

                    left++;
                    right--;
                } else if (sum < target) {
                    left++; // Increase sum
                } else {
                    right--; // Decrease sum
                }
            }
        }
        return ans;
    }

    // public static void main(String[] args) {
    //     Solution sol = new Solution();
    //     int[] nums = {-1, 0, 1, 2, -1, -4};
    //     System.out.println(sol.threeSum(nums));
    // }
}



Code 2:

class Solution {

    // Should return true if there is a triplet with sum equal

    // to x in arr[], otherwise false

    public static boolean hasTripletSum(int arr[], int target) {

        // Your code Here

        Arrays.sort(arr);

        int n=arr.length;

        for(int i=0;i<n-2;i++)

        {

           int start=i+1;

            int end=arr.length-1;

            while(start < end)

            {

           int sum=arr[i]+arr[start]+arr[end];

            

            if(sum == target)

            {

                return true;

            }

            else if(sum < target)

            {

                start++;

            }

            else

            {

                end--;

            }

            }

        }

        return false;

    }

}




Comments

Post a Comment

Popular posts from this blog

Two Sum II - Input Array Is Sorted

 Given a 1-indexed array of integers numbers that is already sorted in non-decreasing order , find two numbers such that they add up to a specific target number. Let these two numbers be numbers[index1] and numbers[index2] where 1 <= index1 < index2 <= numbers.length. Return the indices of the two numbers, index1 and index2, added by one as an integer array [index1, index2] of length 2. The tests are generated such that there is exactly one solution. You may not use the same element twice. Your solution must use only constant extra space .   Example 1: Input: numbers = [2,7,11,15], target = 9 Output: [1,2] Explanation: The sum of 2 and 7 is 9. Therefore, index1 = 1, index2 = 2. We return [1, 2]. Example 2: Input: numbers = [2,3,4], target = 6 Output: [1,3] Explanation: The sum of 2 and 4 is 6. Therefore index1 = 1, index2 = 3. We return [1, 3]. Example 3: Input: numbers = [-1,0], target = -1 Output: [1,2] Explanation: The sum of -1 and 0 is -1. Therefore index1 = 1,...
Read more

Comparable Vs. Comparator in Java

  In Java , Comparable and Comparator both are interfaces used to sort objects. Comparable It defines  natural ordering  within a class. It is implemented  inside  the class being sorted. It uses the compareTo() method. It allows sorting  by one one . Comparable provides a  single sorting sequence . In other words, we can sort the collection on the basis of a single element such as id, name, and price. Comparable  affects the original class , i.e., the actual class is modified. Comparable is present in  java.lang  package. We can sort the list elements of Comparable type by  Collections.sort(List)  method. Only one compareTo() method can be implemented, restricting flexibility. Comparator It defines  custom sorting logic It is implemented  in a separate class  or using lambda expressions . It uses the compare() method. It allows sorting  by multiple criteria . The Comparator provides  multiple sorting...
Read more

Increasing Triplet Subsequence

 Given an integer array nums, return true if there exists a triple of indices (i, j, k) such that i < j < k and nums[i] < nums[j] < nums[k]. If no such indices exists, return false.   Example 1: Input: nums = [1,2,3,4,5] Output: true Explanation: Any triplet where i < j < k is valid. Example 2: Input: nums = [5,4,3,2,1] Output: false Explanation: No triplet exists. Example 3: Input: nums = [2,1,5,0,4,6] Output: true Explanation: The triplet (3, 4, 5) is valid because nums[3] == 0 < nums[4] == 4 < nums[5] == 6. Brute Force : Failed for some testcase class Solution {     public boolean increasingTriplet ( int [] nums ) {         int n = nums . length ;         int i = 0 ;         int j =i+ 1 ;         int k =j+ 1 ;         while (i <n && j<n && k<n)         {         if (nums...
Read more