4Sum — LeetCode #18

Given an array nums of n integers, return an array of all the unique quadruplets [nums[a], nums[b], nums[c], nums[d]] such that:

• 0 <= a, b, c, d < n
• a, b, c, and d are distinct.
• nums[a] + nums[b] + nums[c] + nums[d] == target

You may return the answer in any order.

Example 1:

Input: nums = [1,0,-1,0,-2,2], target = 0
Output: [[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]

Example 2:

Input: nums = [2,2,2,2,2], target = 8
Output: [[2,2,2,2]]

Constraints:

• 1 <= nums.length <= 200
• -109 <= nums[i] <= 109
• -109 <= target <= 109

Solutions:

Python

class Solution:
    def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
        # Sort the list
        nums.sort()

        # Initialize result list
        result = []

        # Fix the first two numbers using two pointers
        for i in range(len(nums)-3):
            # Skip duplicates
            if i > 0 and nums[i] == nums[i-1]:
                continue

            # Fix the second number
            for j in range(i+1, len(nums)-2):
                # Skip duplicates
                if j > i+1 and nums[j] == nums[j-1]:
                    continue

                # Use two pointers to find the last two numbers
                left, right = j+1, len(nums)-1
                while left < right:
                    curr_sum = nums[i] + nums[j] + nums[left] + nums[right]
                    if curr_sum == target:
                        result.append([nums[i], nums[j], nums[left], nums[right]])
                        left += 1
                        right -= 1
                        # Skip duplicates
                        while left < right and nums[left] == nums[left-1]:
                            left += 1
                        while left < right and nums[right] == nums[right+1]:
                            right -= 1
                    elif curr_sum < target:
                        left += 1
                    else:
                        right -= 1

        return result

Javascript

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[][]}
 */
var fourSum = function(nums, target) {
    // Resultant list
    const quadruplets = [];
    // Base condition
    if (nums == undefined || nums.length < 4) {
        return quadruplets;
    }
    // Sort the array
    nums.sort((a, b) => a - b);
    // Length of the array
    const n = nums.length;
    // Loop for each element of the array
    for (let i = 0; i < n - 3; i++) {
        // Check for skipping duplicates
        if (i > 0 && nums[i] == nums[i - 1]) {
            continue;
        }
        // Reducing to three sum problem
        for (let j = i + 1; j < n - 2; j++) {
            // Check for skipping duplicates
            if (j != i + 1 && nums[j] == nums[j - 1]) {
                continue;
            }
            // Left and right pointers
            let k = j + 1;
            let l = n - 1;
            // Reducing to two sum problem
            while (k < l) {
                const currentSum = nums[i] + nums[j] + nums[k] + nums[l];
                if (currentSum < target) {
                    k++;
                } else if (currentSum > target) {
                    l--;
                } else {
                    quadruplets.push([nums[i], nums[j], nums[k], nums[l]]);
                    k++;
                    l--;
                    // Check for skipping duplicates
                    while (k < l && nums[k] == nums[k - 1]) {
                        k++;
                    }
                    while (k < l && nums[l] == nums[l + 1]) {
                        l--;
                    }
                }
            }
        }
    }
    return quadruplets;
};

I hope this helps! Let me know if you have any questions. Don’t forget to follow and give some claps and comments