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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 = 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
Constraints:
3 <= nums.length <= 1000
-1000 <= nums[i] <= 1000
-104 <= target <= 104
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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