1. Two Sum

Description

You are 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

Thinking

The first idea is a nested loop: pick \(x\), then scan for \(target-x\). Correct, but \(O(n^2)\). \(n \le 10^4\) might pass, yet the follow-up asks for faster.

The bottleneck is repeatedly searching for the complement among numbers already seen. A hash table makes that lookup expected \(O(1)\). Walk once, storing value \(\to\) index: when we see \(x\), ask whether \(target-x\) already appeared; if yes, return the two indices; if not, store \(x\).

Look up before inserting. Insert-then-lookup would reuse the same index on \([3,3]\), \(target=6\). Lookup-first guarantees the match is an earlier index.

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 \(\textit{nums}\).

Python3JavaC++GoTypeScriptRustJavaScriptC#PHPScalaSwiftRubyKotlinNimCangjieC

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

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);
        }
    }
}

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;
        }
    }
};

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
    }
}

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);
    }
}

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![]
    }
}

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function (nums, target) {
    const d = new Map();
    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);
    }
};

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

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

import scala.collection.mutable

object Solution {
    def twoSum(nums: Array[Int], target: Int): Array[Int] = {
        val d = mutable.Map[Int, Int]()
        var ans: Array[Int] = Array()
        for (i <- nums.indices if ans.isEmpty) {
            val x = nums(i)
            val y = target - x
            if (d.contains(y)) {
                ans = Array(d(y), i)
            } else {
                d(x) = i
            }
        }
        ans
    }
}

class Solution {
    func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
        var d = [Int: Int]()
        for (i, x) in nums.enumerated() {
            let y = target - x
            if let j = d[y] {
                return [j, i]
            }
            d[x] = i
        }
        return []
    }
}

# @param {Integer[]} nums
# @param {Integer} target
# @return {Integer[]}
def two_sum(nums, target)
    d = {}
    nums.each_with_index do |x, i|
      y = target - x
      if d.key?(y)
        return [d[y], i]
      end
      d[x] = i
    end
end

class Solution {
    fun twoSum(nums: IntArray, target: Int): IntArray {
        val m = mutableMapOf<Int, Int>()
        nums.forEachIndexed { i, x ->
            val y = target - x
            val j = m.get(y)
            if (j != null) {
                return intArrayOf(j, i)
            }
            m[x] = i
        }
        return intArrayOf()
    }
}

import std/enumerate
import std/tables

proc twoSum(nums: seq[int], target: int): seq[int] =
  var d = initTable[int, int]()
  for i, x in nums.pairs():
    let y = target - x
    if d.hasKey(y):
      return @[d[y], i]
    d[x] = i
  return @[]

class Solution {
    func twoSum(nums: Array<Int64>, target: Int64): Array<Int64> {
        let d = HashMap<Int64, Int64>()
        for (i in 0..nums.size) {
            if (d.contains(target - nums[i])) {
                return [d[target - nums[i]], i]
            }
            d[nums[i]] = i
        }
        []
    }
}

#include <stdlib.h>

int* twoSum(int* nums, int numsSize, int target, int* returnSize) {
    int capacity = 1;
    while (capacity < numsSize * 2) capacity <<= 1;
    int* keys = malloc(capacity * sizeof(int));
    int* vals = malloc(capacity * sizeof(int));
    char* used = calloc(capacity, sizeof(char));
    if (!keys || !vals || !used) {
        free(keys);
        free(vals);
        free(used);
        *returnSize = 0;
        return NULL;
    }
    for (int i = 0; i < numsSize; ++i) {
        int x = nums[i];
        int y = target - x;
        unsigned int h = (unsigned int) y & (capacity - 1);
        while (used[h]) {
            if (keys[h] == y) {
                int* res = malloc(2 * sizeof(int));
                res[0] = vals[h];
                res[1] = i;
                *returnSize = 2;
                free(keys);
                free(vals);
                free(used);
                return res;
            }
            h = (h + 1) & (capacity - 1);
        }
        unsigned int h2 = (unsigned int) x & (capacity - 1);
        while (used[h2]) h2 = (h2 + 1) & (capacity - 1);
        used[h2] = 1;
        keys[h2] = x;
        vals[h2] = i;
    }
    *returnSize = 0;
    free(keys);
    free(vals);
    free(used);
    return NULL;
}

1. Two Sum

Description

Solutions

Solution 1: Hash Table

Comments