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 the hash table $m$ to store the array value and the corresponding subscript.

Traverse the array nums, when you find target - nums[i] in the hash table, it means that the target value is found, and the index of target - nums[i] and $i$ are returned.

The time complexity is $O(n)$ and the space complexity is $O(n)$. Where $n$ is the length of the array nums.

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

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

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

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

TypeScript Code
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function twoSum(nums: number[], target: number): number[] {
    const m: Map<number, number> = new Map();

    for (let i = 0; ; ++i) {
        const x = nums[i];
        const y = target - x;

        if (m.has(y)) {
            return [m.get(y)!, i];
        }

        m.set(x, i);
    }
}

Rust Code
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use std::collections::HashMap;

impl Solution {
    pub fn two_sum(nums: Vec<i32>, target: i32) -> Vec<i32> {
        let mut map = HashMap::new();
        for (i, item) in nums.iter().enumerate() {
            if map.contains_key(item) {
                return vec![i as i32, map[item]];
            } else {
                let x = target - nums[i];
                map.insert(x, i as i32);
            }
        }
        unreachable!()
    }
}

JavaScript Code
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/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function (nums, target) {
    const m = new Map();
    for (let i = 0; ; ++i) {
        const x = nums[i];
        const y = target - x;
        if (m.has(y)) {
            return [m.get(y), i];
        }
        m.set(x, i);
    }
};

C# Code
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public class Solution {
    public int[] TwoSum(int[] nums, int target) {
        var m = new Dictionary<int, int>();
        for (int i = 0, j; ; ++i) {
            int x = nums[i];
            int y = target - x;
            if (m.TryGetValue(y, out j)) {
                return new [] {j, i};
            }
            if (!m.ContainsKey(x)) {
                m.Add(x, i);
            }
        }
    }
}

PHP Code
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class Solution {
    /**
     * @param Integer[] $nums
     * @param Integer $target
     * @return Integer[]
     */
    function twoSum($nums, $target) {
        foreach ($nums as $key => $x) {
            $y = $target - $x;
            if (isset($hashtable[$y])) {
                return [$hashtable[$y], $key];
            }
            $hashtable[$x] = $key;
        }
    }
}

Scala Code
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import scala.collection.mutable

object Solution {
  def twoSum(nums: Array[Int], target: Int): Array[Int] = {
    var map = new mutable.HashMap[Int, Int]()
    for (i <- 0 to nums.length) {
      if (map.contains(target - nums(i))) {
        return Array(map(target - nums(i)), i)
      } else {
        map += (nums(i) -> i)
      }
    }
    Array(0, 0)
  }
}

Swift Code
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class Solution {
    func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
        var m = [Int: Int]()
        var i = 0
        while true {
            let x = nums[i]
            let y = target - nums[i]
            if let j = m[target - nums[i]] {
                return [j, i]
            }
            m[nums[i]] = i
            i += 1
        }
    }
}

Ruby Code
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# @param {Integer[]} nums
# @param {Integer} target
# @return {Integer[]}
def two_sum(nums, target)
  nums.each_with_index do |x, idx|
    if nums.include? target - x
      return [idx, nums.index(target - x)] if nums.index(target - x) != idx
    end
    next
  end
end

Nim Code
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import std/enumerate

proc twoSum(nums: seq[int], target: int): seq[int] =
    var
        bal: int
        tdx: int
    for idx, val in enumerate(nums):
        bal = target - val
        if bal in nums:
            tdx = nums.find(bal)
            if idx != tdx:
                return @[idx, tdx]