3Sum Closest LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the 3Sum Closest in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the Problem3Sum 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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