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.

 

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 <= 104
  • -109 <= nums[i] <= 109
  • -109 <= target <= 109
  • Only one valid answer exists.

 

Follow-up: Can you come up with an algorithm that is less than O(n2) time complexity?

submission

impl Solution {
    pub fn two_sum(nums: Vec<i32>, target: i32) -> Vec<i32> {
        // caching visited numbers and corresponding indices
        let mut cache = std::collections::HashMap::new();
        for (idx, num) in nums.iter().enumerate() {
            match cache.get(num) {
                // found cached index, return
                Some(&pos) => return vec![idx as i32, pos],
                // not found, cache it
                None => cache.insert(target - num, idx as i32),
            };
        }
        // !unreachable
        vec![-1, -1]
    }
}