1, Two Sum

I 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: Becausenums[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?

Related Topics

• Array
• Hash Table

II Solution

Approach 1: Brute Force

pub fn two_sum(nums: Vec<i32>, target: i32) -> Vec<i32> {
    let len = nums.len();
    let mut res = Vec::with_capacity(2);

    'outer: for i in 0..len {
        for j in i + 1..len {
            if nums[i] + nums[j] == target {
                res.push(i as i32);
                res.push(j as i32);
                break 'outer;
            }
        }
    }

    res
}

public int[] twoSum(int[] nums, int target) {
    int[] res = new int[2];

    outer: for (int i = 0; i < nums.length; i++) {
        for (int j = i + 1; j < nums.length; j++) {
            if (nums[i] + nums[j] == target) {
                res[0] = i;
                res[1] = j;
                break outer;
            }
        }
    }

    return res;
}

Approach 2: Use Hash

pub fn two_sum(nums: Vec<i32>, target: i32) -> Vec<i32> {
    let len = nums.len();
    let mut map = HashMap::with_capacity(len);
    let mut res = Vec::with_capacity(2);

    for i in 0..len {
        let diff = target - nums[i];
        match map.get(&diff) {
            None => {
                map.insert(nums[i], i);
            }
            Some(idx) => {
                res.push(*idx as i32);
                res.push(i as i32);
                break;
            }
        }
    }

    res
}

public int[] twoSum(int[] nums, int target) {
    int[] res = new int[2];
    Map<Integer, Integer> map = new HashMap<>();

    for (int i = 0; i < nums.length; i++) {
        int diff = target - nums[i];
        Integer idx = map.getOrDefault(diff, Integer.MIN_VALUE);
        if (idx == Integer.MIN_VALUE) {
            map.put(nums[i], i);
        } else {
            res[0] = idx;
            res[1] = i;
            break;
        }
    }

    return res;
}