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    LeetCode link: 1. Two Sum, difficulty: Easy.

    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 <= 10^4
    • -10^9 <= nums[i] <= 10^9
    • -10^9 <= target <= 10^9
    • Only one valid answer exists.
    Hint 1

    A really brute force way would be to search for all possible pairs of numbers but that would be too slow. Again, it's best to try out brute force solutions for just for completeness. It is from these brute force solutions that you can come up with optimizations.

    Hint 2

    So, if we fix one of the numbers, say x, we have to scan the entire array to find the next number y which is value - x where value is the input parameter. Can we change our array somehow so that this search becomes faster?

    Hint 3

    The second train of thought is, without changing the array, can we use additional space somehow? Like maybe a hash map to speed up the search?

    Intuition
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    1. The time complexity of the brute force solution is O(n^2). To improve efficiency, you can sort the array, and then use two pointers, one pointing to the head of the array and the other pointing to the tail of the array, and decide left += 1 or right -= 1 according to the comparison of sum and target.

    2. After sorting an array of numbers, if you want to know the original index corresponding to a certain value, there are two solutions:

    Click to view the answer


    - Solution 1: Bring the index when sorting, that is, the object to be sorted is an array of tuples of (num, index). This technique must be mastered, as it will be used in many questions.
    - Solution 2: Use index() method to find it. I have discussed this in another solution.

    Complexity

    Time complexity

    O(N * log N)

    Space complexity

    O(N)

    Python # 

    class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        num_index_list = [(num, i) for i, num in enumerate(nums)]
        num_index_list.sort()

        left = 0
        right = len(nums) - 1

        while left < right:
            sum_ = num_index_list[left][0] + num_index_list[right][0]

            if sum_ == target:
                return [num_index_list[left][1], num_index_list[right][1]]

            if sum_ < target:
                left += 1
                continue

            right -= 1

    Other languages

    Welcome to contribute code to our GitHub repository, thanks! The location of this solution is 1. Two Sum.

    Intuition of solution 2: Using Map

    1. In Map, key is num, and value is array index.
    2. Traverse the array, if target - num is in Map, return it. Otherwise, add num to Map.

    Steps

    1. In Map, key is num, and value is array index.

      let numToIndex = new Map()

for (let i = 0; i < nums.length; i++) {
  numToIndex.set(nums[i], i)
}

    2. Traverse the array, if target - num is in Map, return it. Otherwise, add num to Map.

      let numToIndex = new Map()

for (let i = 0; i < nums.length; i++) {
  if (numToIndex.has(target - nums[i])) { // 1
    return [numToIndex.get(target - nums[i]), i] // 2
  }

  numToIndex.set(nums[i], i)
}

    Complexity

    Time complexity

    O(N)

    Space complexity

    O(N)

    Solution in languages: Java Python C++ JavaScript C# Go Ruby

    Java # 

    class Solution {
    public int[] twoSum(int[] nums, int target) {
        var numToIndex = new HashMap<Integer, Integer>();

        for (var i = 0; i < nums.length; i++) {
            if (numToIndex.containsKey(target - nums[i])) {
                return new int[]{numToIndex.get(target - nums[i]), i};
            }

            numToIndex.put(nums[i], i);
        }

        return null;
    }
}

    Python # 

    class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        num_to_index = {}

        for i, num in enumerate(nums):
            if target - num in num_to_index:
                return [num_to_index[target - num], i]

            num_to_index[num] = i

    C++ # 

    class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int, int> num_to_index;

        for (auto i = 0; i < nums.size(); i++) {
            if (num_to_index.contains(target - nums[i])) {
                return {num_to_index[target - nums[i]], i};
            }

            num_to_index[nums[i]] = i;
        }

        return {};
    }
};

    JavaScript # 

    var twoSum = function (nums, target) {
  let numToIndex = new Map()

  for (let i = 0; i < nums.length; i++) {
    if (numToIndex.has(target - nums[i])) {
      return [numToIndex.get(target - nums[i]), i]
    }

    numToIndex.set(nums[i], i)
  }
};

    C# # 

    public class Solution {
    public int[] TwoSum(int[] nums, int target) {
        var numToIndex = new Dictionary<int, int>();

        for (int i = 0; i < nums.Length; i++) {
            if (numToIndex.ContainsKey(target - nums[i])) {
                return [numToIndex[target - nums[i]], i];
            }

            numToIndex[nums[i]] = i;
        }

        return null;
    }
}

    Go # 

    func twoSum(nums []int, target int) []int {
    numToIndex := map[int]int{}

    for i, num := range nums {
        if index, ok := numToIndex[target - num]; ok {
            return []int{index, i}
        }

        numToIndex[num] = i
    }

    return nil
}

    Ruby # 

    def two_sum(nums, target)
  num_to_index = {}

  nums.each_with_index do |num, i|
    if num_to_index.key?(target - num)
      return [num_to_index[target - num], i]
    end

    num_to_index[num] = i
  end
end
    Solution in languages: Java Python C++ JavaScript C# Go Ruby

    Other languages

    Welcome to contribute code to our GitHub repository, thanks! The location of this solution is 1. Two Sum.

    Intuition of solution 3: Two Pointers 2

    1. The time complexity of the brute force solution is O(n^2). To improve efficiency, you can sort the array, and then use two pointers, one pointing to the head of the array and the other pointing to the tail of the array, and decide left += 1 or right -= 1 according to the comparison of sum and target.
    2. After finding the two values which sum is target, you can use the index() method to find the index corresponding to the value.

    Complexity

    Time complexity

    O(N * log N)

    Space complexity

    O(N)

    Python # 

    class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        original_nums = nums.copy()
        nums.sort()

        left = 0
        right = len(nums) - 1

        while left < right:
            sum_ = nums[left] + nums[right]

            if sum_ == target:
                break

            if sum_ < target:
                left += 1
                continue

            right -= 1

        return [
            original_nums.index(nums[left]),
            len(nums) - 1 - original_nums[::-1].index(nums[right])
        ]

    Other languages

    Welcome to contribute code to our GitHub repository, thanks! The location of this solution is 1. Two Sum.
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