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Coderust: Hacking the Coding Interview

Sum of Three Values

Explore algorithms to check if three numbers in an array sum to a specific target. Understand naive and optimized approaches with their time and space complexities, including sorting with two pointers and hashing methods.

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Statement

Given an array of integers and an integer value, determine if there are any three integers in the array whose sum equals the given value. Return true if three such integers from the array are found. Otherwise, return false.

Example

Consider this array and the target sums:

g matrix Target sum 20 5 + 7 + 8 = 20 Target sum 21 No 3 values sum upto 21 array 3 7 1 2 8 4 5

Sample input

([3, 7, 1, 2, 8, 4, 5], 20)

Expected output

true

Try it yourself

#include <iostream>
#include <vector>
using namespace std;
bool FindSumOfThree(vector<int> nums, int required_sum) {
	// TODO: Write - Your - Code
	return false;
}

Solution 1

The simple and naive solution is to have three nested loops iterate over all unordered triples to see if their sum is equal to the given integer or not. Although this works, it’s not efficient.

There exist other solutions with much better time complexities. We’ll look at them later in this lesson.

#include <iostream>
#include <vector>
using namespace std;
bool FindSumOfThree(vector<int> nums, int required_sum) {
	// A variable to store the sum of the current elements
	int sum = 0;
	for (int i = 0; i < nums.size() - 2; ++i) {
		for (int j = i + 1; j < nums.size() - 1; ++j) {
			for (int k = j + 1; k < nums.size(); ++k) {
				// i, j and k are indices to cater the scenario in
				// which same array element is used to calculate the sum.
				if ((i != j) && (j != k) && (k != i)) {
					sum = nums[i] + nums[j] + nums[k];
					if (sum == required_sum) {
						return true;
					}
				}
			}
		}
	}
	return false;
}
int main() {
	vector<int> nums_1 = {3, 7, 1, 2, 8, 4, 5};
	vector<int> test_list_1 = {10, 20, 21};
	cout << "1. Original array: ";
	// A custom PrintList function to print the vector
	PrintList(nums_1);
	for (int test : test_list_1) {
		if (FindSumOfThree(nums_1, test)) {
			cout << "   Sum for " << test << " exists " << endl;
		} else {
			cout << "   Sum for " << test << " does not exist " << endl;
		}
	}
	cout << "--------------------------------------------------------------------"
					"----------------------------------\n"
			 << endl;
	vector<int> nums_2 = {-1, 2, 1, -4, 5, -3};
	vector<int> test_list_2 = {-8, 0, 7};
	cout << "2. Original array: ";
	// A custom PrintList function to print the vector
	PrintList(nums_2);
	for (int test : test_list_2) {
		if (FindSumOfThree(nums_2, test)) {
			cout << "   Sum for " << test << " exists " << endl;
		} else {
			cout << "   Sum for " << test << " does not exist " << endl;
		}
	}
	cout << "------------------------------------------------------------------------------------------------------\n"<< endl;
}

Time complexity

The time complexity of this solution is cubic, O(n3)O(n^3) ...