In this guide we will provide 16. 3Sum Closest LeetCode Solution with best time and space complexity. The solution to Sum Closest problem is provided in various programming languages like C++, Java and python. This will be helpful for you if you are preparing for placements, hackathon, interviews or practice purposes. The solutions provided here are very easy to follow and with detailed explanations.
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).
class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {
int ans = nums[0] + nums[1] + nums[2];
ranges::sort(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 number in the triplet, then search the
// remaining numbers 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;
}
}; /* code provided by PROGIEZ */
16. 3Sum Closest LeetCode Solution in Java
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 number in the triplet, then search the
// remaining numbers 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;
}
} // code provided by PROGIEZ
16. 3Sum Closest LeetCode Solution in Python
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
# Choose nums[i] as the first number in the triplet, then search the
# remaining numbers in [i + 1, n - 1].
l = i + 1
r = len(nums) - 1
while l < r:
summ = nums[i] + nums[l] + nums[r]
if summ == target:
return summ
if abs(summ - target) < abs(ans - target):
ans = summ
if summ < target:
l += 1
else:
r -= 1
return ans #code by PROGIEZ
