NeetCode 150: Two Sum

Table of contents

The Problem

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

You can return the answer in any order.

The Approach

The goal is to find the very first pair’s indices that add up to target. Using a nested for-loop, we can loop across nums & every other element for each index of nums.

By getting the desiredNumber = target - currentNumber, we can check if this number is present in the other indices of nums. If this number is found, the valid pair is found & returned.

My Solution

class Solution {
    int[] pair = new int[2];

    public int[] twoSum(int[] nums, int target) {
        // First loop gets the current number & the desired value to reach target.
        for (int i = 0; i < nums.length; i++) {
            int currentNumber = nums[i];
            int desiredNumber = target - currentNumber;

            // Checks remaining elements to locate desired value if present.
            for (int j = i + 1; j < nums.length; j++) {                
                if (nums[j] == desiredNumber) {
                    pair[0] = i;
                    pair[1] = j;

                    return pair;
                }
            }
        }
        return pair;
    }
}

The first for-loop loops over all the elements in nums. The desiredNumber is then calculated.

The nested loop then loops over the elements from i + 1 (as the same element cannot be used twice), checking if the current index’s element is equal to desiredNumber. When it is found, the pair’s indices are returned.