**Sum of Three Numbers Sequel - Google Top Interview Questions**

### Problem Statement :

Given a list of integers nums and an integer k, find three distinct elements in nums, a, b, c, such that abs(a + b + c - k) is minimized and return the absolute difference. Constraints n ≤ 1,000 where n is length of nums. Example 1 Input nums = [2, 4, 25, 7] k = 15 Output 2 Explanation Taking [2, 4, 7] will get us closest to 15 and the absolute difference is abs(13 - 15) = 2.

### Solution :

` ````
Solution in C++ :
int solve(vector<int>& nums, int k) {
int i, j, sum = 0, ans = INT_MIN, ans1 = INT_MAX;
sort(nums.begin(), nums.end());
for (i = 0; i < nums.size() - 2; i++) {
int l = i + 1, r = nums.size() - 1;
while (l < r) {
if (nums[i] + nums[l] + nums[r] > k) {
ans1 = min(ans1, nums[i] + nums[l] + nums[r]);
r--;
} else {
ans = max(ans, nums[i] + nums[l] + nums[r]);
l++;
}
}
}
if (ans == INT_MIN) {
return abs(ans1 - k);
}
if (ans1 == INT_MAX) {
return abs(ans - k);
}
return min(abs(ans1 - k), abs(ans - k));
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[] nums, int k) {
int min = Integer.MAX_VALUE;
Arrays.sort(nums);
for (int i = 0, len = nums.length; i < len - 2; i++) {
int left = i + 1;
int right = len - 1;
while (left < right) {
int runningSum = nums[i] + nums[left] + nums[right];
int diff = Math.abs(k - runningSum);
if (diff < min)
min = diff;
if (runningSum < k)
left++;
else
right--;
}
}
return min;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, nums, k):
nums.sort()
n = len(nums)
m = math.inf
for i in range(n):
p = i + 1
q = n - 1
while p < q:
s = nums[p] + nums[q] + nums[i]
m = min(m, abs(k - s))
if s < k:
p += 1
else:
q -= 1
return m
```

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