# Simple explanation O(n^2)

• Well the idea here is simple,we do a reduction for this problem
we iterate through the array picking each number at a time,and reducing it from the target
now the question is reducted to a simpler question,finding closest 2 sum.
and that can be done with 2 pointers.

here's a code for it

``````public static int threeSumClosest(int[] nums, int target) {
int minS = nums[0]+nums[1]+nums[2];
Arrays.sort(nums);
for (int k = 0; k < nums.length - 2; k++) {
int temp = twoSum(nums, target - nums[k], k + 1) + nums[k];
minS = Math.abs(temp - target) < Math.abs(minS - target) ? temp : minS;
}
return minS;
}

public static int twoSum(int[] numbers, int target, int start) {

int end = numbers.length - 1;
int minS = numbers[start]+numbers[start+1];
int minDiff = target;
while (start < end) {
if (Math.abs((numbers[end] + numbers[start]) - target) < Math.abs(minDiff)) {
minDiff = Math.abs((numbers[end] + numbers[start]) - target);
minS = numbers[end] + numbers[start];
}
if (numbers[end] + numbers[start] > target)
end = end - 1;
else if (numbers[end] + numbers[start] < target)
start = start + 1;
else
return numbers[end] + numbers[start];
}
return minS;
}
``````

Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.