# 16. 3Sum Closest

## Description

Given an integer array nums of length n and an integer target, find three integers in nums such that the sum is closest to target.

Return the sum of the three integers.

You may assume that each input would have exactly one solution.

Example 1:

Input: nums = [-1,2,1,-4], target = 1
Output: 2
Explanation: The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).


Example 2:

Input: nums = [0,0,0], target = 1
Output: 0
Explanation: The sum that is closest to the target is 0. (0 + 0 + 0 = 0).


Constraints:

• 3 <= nums.length <= 500
• -1000 <= nums[i] <= 1000
• -104 <= target <= 104

## Solutions

Solution 1: Sorting + Two Pointers

We sort the array first, then traverse the array. For each element $nums[i]$, we use pointers $j$ and $k$ to point to $i+1$ and $n-1$ respectively, calculate the sum of the three numbers. If the sum of the three numbers equals $target$, we directly return $target$. Otherwise, we update the answer based on the difference from $target$. If the sum of the three numbers is greater than $target$, we move $k$ one place to the left, otherwise, we move $j$ one place to the right.

The time complexity is $O(n^2)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array.

• class Solution {
public int threeSumClosest(int[] nums, int target) {
Arrays.sort(nums);
int ans = 1 << 30;
int n = nums.length;
for (int i = 0; i < n; ++i) {
int j = i + 1, k = n - 1;
while (j < k) {
int t = nums[i] + nums[j] + nums[k];
if (t == target) {
return t;
}
if (Math.abs(t - target) < Math.abs(ans - target)) {
ans = t;
}
if (t > target) {
--k;
} else {
++j;
}
}
}
return ans;
}
}

• class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {
sort(nums.begin(), nums.end());
int ans = 1 << 30;
int n = nums.size();
for (int i = 0; i < n; ++i) {
int j = i + 1, k = n - 1;
while (j < k) {
int t = nums[i] + nums[j] + nums[k];
if (t == target) return t;
if (abs(t - target) < abs(ans - target)) ans = t;
if (t > target)
--k;
else
++j;
}
}
return ans;
}
};

• class Solution:
def threeSumClosest(self, nums: List[int], target: int) -> int:
nums.sort()
n = len(nums)
ans = inf
for i, v in enumerate(nums):
j, k = i + 1, n - 1
while j < k:
t = v + nums[j] + nums[k]
if t == target:
return t
if abs(t - target) < abs(ans - target):
ans = t
if t > target:
k -= 1
else:
j += 1
return ans


• func threeSumClosest(nums []int, target int) int {
sort.Ints(nums)
ans := 1 << 30
n := len(nums)
for i, v := range nums {
j, k := i+1, n-1
for j < k {
t := v + nums[j] + nums[k]
if t == target {
return t
}
if abs(t-target) < abs(ans-target) {
ans = t
}
if t > target {
k--
} else {
j++
}
}
}
return ans
}

func abs(x int) int {
if x < 0 {
return -x
}
return x
}

• function threeSumClosest(nums: number[], target: number): number {
nums.sort((a, b) => a - b);
let ans: number = 1 << 30;
const n = nums.length;
for (let i = 0; i < n; ++i) {
let j = i + 1;
let k = n - 1;
while (j < k) {
const t: number = nums[i] + nums[j] + nums[k];
if (t === target) {
return t;
}
if (Math.abs(t - target) < Math.abs(ans - target)) {
ans = t;
}
if (t > target) {
--k;
} else {
++j;
}
}
}
return ans;
}


• /**
* @param {number[]} nums
* @param {number} target
* @return {number}
*/
var threeSumClosest = function (nums, target) {
nums.sort((a, b) => a - b);
let ans = 1 << 30;
const n = nums.length;
for (let i = 0; i < n; ++i) {
let j = i + 1;
let k = n - 1;
while (j < k) {
const t = nums[i] + nums[j] + nums[k];
if (t === target) {
return t;
}
if (Math.abs(t - target) < Math.abs(ans - target)) {
ans = t;
}
if (t > target) {
--k;
} else {
++j;
}
}
}
return ans;
};


• class Solution {
/**
* @param int[] $nums * @param int$target
* @return int
*/

function threeSumClosest($nums,$target) {
$n = count($nums);
$closestSum =$nums[0] + $nums[1] +$nums[2];
$minDiff = abs($closestSum - $target); sort($nums);

for ($i = 0;$i < $n - 2;$i++) {

$left =$i + 1;
$right =$n - 1;

while ($left <$right) {
$sum =$nums[$i] +$nums[$left] +$nums[$right];$diff = abs($sum -$target);

if ($diff <$minDiff) {
$minDiff =$diff;
$closestSum =$sum;
} elseif ($sum <$target) {
$left++; } elseif ($sum > $target) {$right--;
} else {
return $sum; } } } return$closestSum;
}
}