Solution: Find Two Numbers that Add up to "s"
This review provides a detailed analysis of the different ways to solve the previous challenge
Solution #1: Brute Force #
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#include <iostream>#include <vector>using namespace std;vector<int> findSum(int arr[], int arrSize, int s) {vector<int> elements;for (int i = 0; i < arrSize; i++) {for (int j = i + 1; j < arrSize; j++) {if (arr[i] + arr[j] == s) {elements.push_back(arr[i]);elements.push_back(arr[j]);}}}return elements;}int main() {int arr[] = {8,4,1,6,5,9};int sum = 14;vector<int> result = findSum(arr, 6, sum);if(!result.empty())cout << "Sum of " << sum << " found: " << result[0] << " " << result[1];elsecout << "Results not found" << endl;}
This is the most time-intensive but intuitive solution. It is a lot like a modified version of a linear search.
The whole array is traversed for each element and checked for any two elements that add up to the given number s
. If they are found, we print them out. To do all of the above, we use a nested for-loop and iterate over the entire array for each element
Time Complexity
Since we iterate over the entire array (size ) times, the time complexity is ...
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