Problem

Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice.

Examples

Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].

Approach 1: Brute Force

The naive approach to solving this problem is to loop through each element x in nums and then loop through each element again to find if there is a value that equals target - x

Algorithm

for i = 0 to length of nums 
    for j = i + 1 to length of nums 
        if nums[j] == target - nums[i] 
            return [i, j]

Implementation

Complexity

• Time complexity: O(n2) since for each element we loop through of the array
• Space complexity: O(1)

Approach 2: Two-pass Hash Table

The time complexity can be reduced, by converting the input array into a hash map that stores the value nums[i] as the key and the index i as the value. This would involve two iterations, one to build the hash map and another to check if the complement - target - nums[i] - exists in the hash map but is not the same index.

Algorithm

map = new HashMap 
for i = 0 to length of nums 
    map[nums[i]] = i 

for i = 0 to length of nums 
    complement = target - nums[i] 
    if map.contains(complement) and map[complement] != i 
        return [i, map[complement]]

Implementation

#leetcode #data-structures #rust #algorithms