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.


n ≤ 1,000 where n is length of nums.

Example 1


nums = [2, 4, 25, 7]

k = 15




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]);
            } else {
                ans = max(ans, nums[i] + nums[l] + nums[r]);
    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;

        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)

        return min;

                        Solution in Python : 
class Solution:
    def solve(self, nums, k):
        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
                    q -= 1
        return m

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