1.TwoSum.cs

//1. Two Sum
//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.

//O(n*n)
public class Solution
{

    public int[] TwoSum(int[] nums, int target)
    {
        int[] indices = new int[2];
        int flag = 0;
        for (int i = 0; i < nums.Length - 1; i++)
        {
            for (int j = i + 1; j < nums.Length; j++)
            {
                if (nums[i] + nums[j] == target)
                {
                    indices[0] = i;
                    indices[1] = j;
                    flag = 1;
                    break;
                }
            }
            if (flag == 1)
                break;
        }
        return indices;
    }
}
//HashMap O(n)
public class Solution {
    public int[] TwoSum(int[] nums, int target) {
        Dictionary<int,int> hm = new Dictionary<int,int>();
        for(int i = 0; i < nums.Length; i++){
            //If the difference is present in the hashmap, return the indices
            if(hm.ContainsKey(target - nums[i])){
                return new int[]{hm[target - nums[i]],i};
            }
            else
                hm[nums[i]] = i;
        }
        return new int[2];
    }
}