-leetcode-

Two Sum
Easy
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.

Input: nums = [2,7,11,15], target = 9

Output: [0,1]

Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].

Input: nums = [3,2,4], target = 6

Output: [1,2]

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(n^2) time complexity?
Brute force Solution 
                  
public static int[] TwoSum_V1(int[] nums, int target)       // Brute force Solution, time complexity of O(n^2)
{
  for (int i = 0; i < nums.Length; i++)                     // outter loop of nums array
  {
    for (int j = 0; j < nums.Length; j++)                   // inner loop of nums array
    { 
      if (nums[i] + nums[j] == target && i!=j)              // if num[i] + num[j] == target, we found the required values
      { 
        return new int[] { nums[j], nums[i] };
      }
    }
  }
  return null;                                              // If no pair of numbers adds up to the target, return an empty array
}
Better Solution 
                
public static int[] TwoSum_V2(int[] nums, int target)                             // Better Solution, time complexity of O(n)
{
Dictionary numIndices = new Dictionary();                                       // Create a dictionary to store the indices of each number              
for (int i = 0; i < nums.Length; i++)                                           // Iterate through the array
{   
  int complement = target - nums[i];                                            // Calculations
  if (numIndices.ContainsKey(complement))                                       // if the dictionary contains the complement, we found the required values
  {
    return new int[] { nums[numIndices[complement]], nums[i]};
  }
  numIndices[nums[i]] = i;                                                      // Otherwise, add the current number to the dictionary
}
return null;                                                                    // If no pair of numbers adds up to the target, return an empty array
}
Sign of the Product of an Array
Easy
Description
There is a function signFunc(x) that returns:
  • 1 if x is positive.
  • -1 if x is negative.
  • 0 if x is equal to 0.
You are given an integer array nums. Let product be the product of all values in the array nums.
Return signFunc(product).

Input: nums = [-1,-2,-3,-4,3,2,1]

Output: 1

Explanation: The product of all values in the array is 144, and signFunc(144) = 1

Input: nums = [1,5,0,2,-3]

Output: 0

Explanation: The product of all values in the array is 0, and signFunc(0) = 0

Input: nums = [-1,1,-1,1,-1]

Output: -1

Explanation: The product of all values in the array is -1, and signFunc(-1) = -1

Constraints
  • 1 <= nums.length <= 1000
  • -100 <= nums[i] <= 100
Solution 
                  
public static int ArraySign(int[] nums)
{
  int product = 1;                                // initialize product variable = 1
  foreach (int num in nums)                       // multiply values or return 0 if 0 is found
  {
    if (num == 0)
      return 0;                
    else
      product *= num;
  }       
  return SignFunc(product);                       // call SignFunc to determine the sign
}
                
public static int SignFunc(int x)
{
  return (x > 0 ? 1 : -1);                        // equivalent to if(x>0) {return 1;} else {return -1;}
}