DEV Community

Alkesh Ghorpade
Alkesh Ghorpade

Posted on

LeetCode Two Sum

Problem statement

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.

Problem statement taken from:

Example 1:

Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Output: Because nums[0] + nums[1] == 9, we return [0, 1].
Enter fullscreen mode Exit fullscreen mode

Example 2:

Input: nums = [3,2,4], target = 6
Output: [1,2]
Enter fullscreen mode Exit fullscreen mode

Example 3:

Input: nums = [8,2,6,9,10], target = 7
Output: []
Enter fullscreen mode Exit fullscreen mode


Brute force

A brute force solution is to use two nested loops.

We use an outer loop with iterator i to visit each element of the array and
inner loop with iterator j to check if there is any other element that adds up to the target.

A small code block in C++ will look like this.

for (int i = 0; i < nums.length; i++){
    for (int j = i + 1; j < nums.length; j++){
        if (nums[i] + nums[j] == target){
            return new int[] { i, j };
Enter fullscreen mode Exit fullscreen mode

The time complexity for the above code is O(NĀ²) and hence the solution is
not efficient.

Hash Map

The problem can be solved in O(N) time, using extra space.

We can use a Hash table / HashMap. The HashMap
will store the array element as a key and the value will be the index at which the element is stored.

- Create a new HashMap
- Iterate over the array
  - Subtract current array element from target
  - Check if the above difference exists in HashMap
    - If YES, then return the current index and the value of the difference in HashMap
    - If NO, then store the current element as HashMap key and current index as its value
- Return empty array as a result if nums array iteration is completed in the above step
Enter fullscreen mode Exit fullscreen mode
C++ solution
class Solution {
    vector<int> twoSum(vector<int>& nums, int target) {
        map<int, int> umap;
        int difference;

        for(int i = 0; i < nums.size(); i++ ){
            difference = target -;
            if(umap.find(difference) != umap.end()) {
                vector<int> v{umap[difference], i};
                return v;
            } else {
                umap[] = i;

        return vector<int> {};
Enter fullscreen mode Exit fullscreen mode
Golang solution
func twoSum(nums []int, target int) []int {
    record := make(map[int]int)

    for index, num := range nums {
        difference := target - num
        if res, ok := record[difference]; ok {
            return []int{index, res}
        record[num] = index

    return []int{}
Enter fullscreen mode Exit fullscreen mode

Originally published at

Top comments (2)

pinkflyod1 profile image
pinkflyod1 • Edited

Great explanation. This article along with another nice one on code recipe helped me to solve this problem. Leaving link in case someone finds it useful:

Keep up the good work guys.

vilskyviktor profile image
Viktor Vilsky

Here is a nice video about all the possible solutions of Two Sum algorithm -