1. Two Sum

Problem

Tags: Array, Hash Table

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.

Example 1:

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

Example 2:

Input: nums = [3,2,4], target = 6
Output: [1,2]

Example 3:

Input: nums = [3,3], target = 6
Output: [0,1]

Constraints:

• 2 <= nums.length <= 10^4
• -10^9 <= nums[i] <= 10^9
• -10^9 <= target <= 10^9
• Only one valid answer exists.

Follow-up: Can you come up with an algorithm that is less than O(n^2) time complexity?

Code

C

// 1. Two Sum (5/15/53896)
// Runtime: 72 ms (78.49%) Memory: 6.56 MB (21.30%) 

int* twoSum(int* nums, int nums_size, int target, int* return_size) {
    int* result = (int*)malloc(sizeof(int) * 2);
    int i, j;
    for (i = 0; i < nums_size; i++) {
        for (j = i + 1; j < nums_size; j++) {
            if (nums[i] + nums[j] == target) {
                result[0] = i;
                result[1] = j;
                *return_size = 2;
                return result;
            }
        }
    }
    return result;
}

GO

// 1. Two Sum (5/22/53846)
// Runtime: 4 ms (88.21%) Memory: 4.34 MB (15.47%) 

func twoSum(nums []int, target int) []int {
    table := make(map[int]int)
    
    for i := 0; i < len(nums); i++ {
        if val, exists := table[nums[i]]; exists {
            return []int{val, i}
            break
        }
        table[target - nums[i]] = i
    }
    return []int{0, 1}
}

JS

// 1. Two Sum (3/14/53371)
// Runtime: 72 ms (82.97%) Memory: 39.42 MB (94.91%) 

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */
const twoSum = function(nums, target) {
    let table = {};
    
    for(var i = 0; i < nums.length; i++) {
        if(table[nums[i]] >= 0) {
            return [table[nums[i]], i];
        }
        
        table[target - nums[i]] = i;
    }
};