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1. Two Sum

Description

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?

Solutions

Solution 1: Hash Table

We can use a hash table $\textit{d}$ to store each element and its corresponding index.

Traverse the array $\textit{nums}$, for the current element $\textit{nums}[i]$, we first check if $\textit{target} - \textit{nums}[i]$ is in the hash table $\textit{d}$. If it is in $\textit{d}$, it means the $\textit{target}$ value has been found, and we return the indices of $\textit{target} - \textit{nums}[i]$ and $i$.

Time complexity is $O(n)$, and space complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$.

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class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        d = {}
        for i, x in enumerate(nums):
            y = target - x
            if y in d:
                return [d[y], i]
            d[x] = i

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class Solution {
    public int[] twoSum(int[] nums, int target) {
        Map<Integer, Integer> d = new HashMap<>();
        for (int i = 0;; ++i) {
            int x = nums[i];
            int y = target - x;
            if (d.containsKey(y)) {
                return new int[] {d.get(y), i};
            }
            d.put(x, i);
        }
    }
}

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class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int, int> d;
        for (int i = 0;; ++i) {
            int x = nums[i];
            int y = target - x;
            if (d.contains(y)) {
                return {d[y], i};
            }
            d[x] = i;
        }
    }
};

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func twoSum(nums []int, target int) []int {
    d := map[int]int{}
    for i := 0; ; i++ {
        x := nums[i]
        y := target - x
        if j, ok := d[y]; ok {
            return []int{j, i}
        }
        d[x] = i
    }
}

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function twoSum(nums: number[], target: number): number[] {
    const d = new Map<number, number>();
    for (let i = 0; ; ++i) {
        const x = nums[i];
        const y = target - x;
        if (d.has(y)) {
            return [d.get(y)!, i];
        }
        d.set(x, i);
    }
}

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use std::collections::HashMap;

impl Solution {
    pub fn two_sum(nums: Vec<i32>, target: i32) -> Vec<i32> {
        let mut d = HashMap::new();
        for (i, &x) in nums.iter().enumerate() {
            let y = target - x;
            if let Some(&j) = d.get(&y) {
                return vec![j as i32, i as i32];
            }
            d.insert(x, i);
        }
        vec![]
    }
}

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/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function (nums,