1 Two Sum

Sep 2020

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]
Output: 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 <= 10⁵
-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?

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def twoSum(nums: 'List[int]', target: 'int') -> 'List[int]':
    map = {}
    for i, n in enumerate(nums):
        if n in map:
            return [map[n], i]
        map[target-n] = i

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func twoSum(nums []int, target int) []int {
    d := make(map[int]int)
    for idx, num := range nums {
        if v, ok := d[num]; ok {
            return []int{v, idx}
        }
        d[target-num] = idx
    }
    return nil
}

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public int[] twoSum(int[] nums, int target) {
    Map<Integer, Integer> map = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
        int comp = target - nums[i];
        if (map.containsKey(comp)) {
            return new int[]{map.get(comp), i};
        }
        map.put(nums[i], i);
    }
    return new int[]{};
}

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class Solution {
    fun twoSum(nums: IntArray, target: Int): IntArray {
        val diffMap = mutableMapOf<Int, Int>()
        nums.forEachIndexed { i, num ->
            diffMap[num]?.let {
                return intArrayOf(it, i)
            }
            diffMap[target - num] = i
        }
        return intArrayOf()
    }
}

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vector<int> twoSum(vector<int>& nums, int target) {
    unordered_map<int, int> imap;

    for (int i = 0;; ++i) {
        auto it = imap.find(target - nums[i]);

        if (it != imap.end()) 
            return vector<int> {i, it->second};

        imap[nums[i]] = i;
    }
}

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class Solution {
    func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
        var dict = [Int: Int]()

        for (index, value) in nums.enumerated() {
          if let i = dict[value] {
            return [i, index]
          } else {
            dict[target - value] = index
          }
        }

        return []
    }
}

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pub fn two_sum(nums: Vec<i32>, target: i32) -> Vec<i32> {
    use std::collections::HashMap;
    
    let mut m: HashMap<i32, i32> = HashMap::new();
    for (i, v) in nums.iter().enumerate() {
        match m.get(&(target - *v)) {
            Some(&i2) => return vec![i as i32, i2],
            None => m.insert(*v, i as i32),
        };
    }
    vec![]
}

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

    for (let i = 0; i < nums.length; i++) {
      const comp = target - nums[i];
      if (map[comp] !== undefined) {
        return [map[comp], i]
      }
      map[nums[i]] = i
    }

    return []
};

let input [] psh $input 2 7 11 15 cal two_sum $input 22 prt $ret

def two_sum let _list $0 let _target $1 let _result [] let _map {} let _idx 0

for _num $_list get $_map $_num _v ife $_v $nil sub _key $_target $_num put $_map $_key $_idx els psh $_result $_v $_idx ret $_result fin add _idx $_idx 1 nxt ret $_result end