Sum of Three Numbers Less than Target - Google Top Interview Questions

Problem Statement :

Given a list of integers nums and an integer target, return the number of triples i < j < k that exist such that nums[i] + nums[j] + nums[k] < target.


n ≤ 1,000 where n is the length of nums

Example 1


nums = [-3, 5, 3, 2, 7]

target = 9




Here are the different triples' values:

-3 + 5 + 3 = 5
-3 + 5 + 2 = 4
-3 + 3 + 2 = 2
-3 + 3 + 7 = 7
-3 + 2 + 7 = 6

Solution :


                        Solution in C++ :

int solve(vector<int>& nums, int target) {
    sort(nums.begin(), nums.end());
    int ans = 0;
    for (int i = 0; i < nums.size(); i++) {
        for (int j = i + 1, k = nums.size() - 1; j < k;) {
            if (nums[i] + nums[j] + nums[k] < target) {
                ans += k - j;
            } else {
    return ans;

                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int[] nums, int target) {
        int res = 0;
        for (int i = 0; i < nums.length; i++) res += twoSum(nums, target, i);
        return res;

    // Two pointer function to find all triplets less than target
    public int twoSum(int[] nums, int target, int i) {
        int j = i + 1, k = nums.length - 1, res = 0;

        // Do not do i <= j because the same index cannot be used twice
        while (j < k) {
            int total = nums[i] + nums[j] + nums[k];
            if (total >= target)

            // The array is sorted so all triplets using a smaller k can be added
            else {
                res += k - j;
        return res;

                        Solution in Python : 
class Solution:
    def two_sum(self, nums, start, val, target):
        a, b = start, len(nums) - 1

        ans = 0

        while a < b:
            if nums[a] + nums[b] + val < target:
                a += 1
                ans += b - a + 1
                b -= 1

        return ans

    def solve(self, nums, target):

        return sum(self.two_sum(nums, i + 1, nums[i], target) for i in range(len(nums)))

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