LeetCode: Two Sum (Brute Force)

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

Input: [2,7,11,15]

target=9

Output: [0,1]

Notes

There will always be a solution and we cannot use the same element twice (Ex: 11 can’t show up in the array twice)

Approach

1. Create two for loops
2. First loop will go over every element one at a time
3. Second loop that’s nested inside, will iterate over every element after the element from the first loop
4. Use an if-else statement to check if the number at first loop + number at second loop = target
5. If the two element’s sum matches the target value, we will return an array of the indices

Code

for(let p1 = 0; p1<nums.length; p1++){
   for(let p2 = p1+1; p2<nums.length; p2++){
       if(nums[p1]+nums[p2] === target){
           return [p1,p2];
       }
   }
}

Time Complexity

O(n²): Due to nested for loop iterating over the same array, it is O(n²)

Space Complexity

O(1): We don’t store any values into an array, set, map, etc.