18. 4Sum(Solution || Leetcode medium || Java)

Palakkgoyal
2 min readNov 18, 2022

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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

SOLUTION:

My solution also works on the testcase:-

Input

[1000000000,1000000000,1000000000,1000000000]
-294967296

Output

[[1000000000,1000000000,1000000000,1000000000]]

Expected

[]

class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
//First sort array as without sorting it will get too complex
Arrays.sort(nums);

//Create a list of list integer to store quadruplets
List<List<Integer>> output = new ArrayList<List<Integer>>();

//Special case for the above given testcase in the article
if(target < 0 && nums[0] > 0){
return output;
}

//Start looping through each element
for(int i = 0; i < nums.length - 3; i++){

//This condition is for skipping duplicates
if(i == 0 || (i > 0 && nums[i] != nums[i-1])){

//From here we will solve this question as we solved 3 Sum
//question using pointers
for(int j = i + 1; j < nums.length - 2; j++){

//This condition is same as above to skip duplicates
if(j == i+1 || (j > i+1 && nums[j] != nums[j-1])){

//First pointer
int low = j + 1;
//Second pointer
int high = nums.length - 1;

while(low < high){

//Sum of the quadruplets on which we are currently at
int currSum = nums[i] + nums[j] + nums[low] + nums[high];

//if any quad sum result in target add that quad in our list of list
if(currSum == target){
output.add(Arrays.asList(nums[i],nums[j],nums[low],nums[high]));

//Condition for skipping duplicates
while(low < high && nums[low] == nums[low+1]) low++;
while(low < high && nums[high] == nums[high -1])high--;

low++;
high--;
}
else if(currSum > target){
high--;
}
else{
low++;
}
}
}
}
}
}


return output;
}
}

Thank you for reading. If you have any queries then, please let me know in the comment section. I will surely be responsive toward it.

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Palakkgoyal
Palakkgoyal

Written by Palakkgoyal

Solutions to all your coding related problems at one point. DSA question on daily basis and much more.

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