p0001_two_sum.rs

/**
 * [0001] Two Sum
 *
 * 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)<font face="monospace"> </font>time complexity?
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/two-sum/
// discuss: https://leetcode.com/problems/two-sum/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

use std::collections::HashMap;

impl Solution {
    pub fn two_sum(nums: Vec<i32>, target: i32) -> Vec<i32> {
        let ans: Vec<i32> = vec![];
        let mut exist = HashMap::with_capacity(nums.len());
        
        for (index, num) in nums.iter().enumerate() {
            match exist.get(&(target - num)) {
                None => {
                    exist.insert(num, index);
                }
                Some(idx) => return vec![*idx as i32, index as i32],
            }
        }
        ans
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_0001() {
    }
}