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** Link for the Problem** – 3Sum Closest– LeetCode Problem

3Sum Closest– LeetCode Problem

**Problem:**

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 = 1Output:2Explanation:The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

**Example 2:**

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

**Constraints:**

`3 <= nums.length <= 1000`

`-1000 <= nums[i] <= 1000`

`-10`

^{4}<= target <= 10^{4}

3Sum Closest– LeetCode Solutions

class Solution { public: int threeSumClosest(vector<int>& nums, int target) { int ans = nums[0] + nums[1] + nums[2]; sort(begin(nums), end(nums)); for (int i = 0; i + 2 < nums.size(); ++i) { if (i > 0 && nums[i] == nums[i - 1]) continue; // choose nums[i] as the first num in the triplet, // and search the remaining nums in [i + 1, n - 1] int l = i + 1; int r = nums.size() - 1; while (l < r) { const int sum = nums[i] + nums[l] + nums[r]; if (sum == target) return sum; if (abs(sum - target) < abs(ans - target)) ans = sum; if (sum < target) ++l; else --r; } } return ans; } };

class Solution { public int threeSumClosest(int[] nums, int target) { int ans = nums[0] + nums[1] + nums[2]; Arrays.sort(nums); for (int i = 0; i + 2 < nums.length; ++i) { if (i > 0 && nums[i] == nums[i - 1]) continue; // choose nums[i] as the first num in the triplet, // and search the remaining nums in [i + 1, n - 1] int l = i + 1; int r = nums.length - 1; while (l < r) { final int sum = nums[i] + nums[l] + nums[r]; if (sum == target) return sum; if (Math.abs(sum - target) < Math.abs(ans - target)) ans = sum; if (sum < target) ++l; else --r; } } return ans; } }

class Solution: def threeSumClosest(self, nums: List[int], target: int) -> int: ans = nums[0] + nums[1] + nums[2] nums.sort() for i in range(len(nums) - 2): if i > 0 and nums[i] == nums[i - 1]: continue l = i + 1 r = len(nums) - 1 while l < r: sum = nums[i] + nums[l] + nums[r] if sum == target: return sum if abs(sum - target) < abs(ans - target): ans = sum if sum < target: l += 1 else: r -= 1 return ans

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