# Solving Leetcode Interviews in Seconds with AI: Minimum Operations to Make Array Values Equal to K


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3375" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> You are given an integer array nums and an integer k. An integer h is called valid if all values in the array that are strictly greater than h are identical. For example, if nums = [10, 8, 10, 8], a valid integer is h = 9 because all nums[i] > 9 are equal to 10, but 5 is not a valid integer. You are allowed to perform the following operation on nums:  Select an integer h that is valid for the current values in nums. For each index i where nums[i] > h, set nums[i] to h.  Return the minimum number of operations required to make every element in nums equal to k. If it is impossible to make all elements equal to k, return -1.   Example 1:  Input: nums = [5,2,5,4,5], k = 2 Output: 2 Explanation: The operations can be performed in order using valid integers 4 and then 2.  Example 2:  Input: nums = [2,1,2], k = 2 Output: -1 Explanation: It is impossible to make all the values equal to 2.  Example 3:  Input: nums = [9,7,5,3], k = 1 Output: 4 Explanation: The operations can be performed using valid integers in the order 7, 5, 3, and 1.    Constraints:  1 <= nums.length <= 100  1 <= nums[i] <= 100 1 <= k <= 100  

	# Explanation
	Here's the breakdown of the solution:

*   **Identify Valid `h` values:** Iterate through the array and identify potential `h` values that satisfy the valid integer condition. These are the unique numbers greater than `k`.
*   **Greedy Approach:** Sort the valid `h` values in descending order. Apply the operation for each valid `h` value.
*   **Check for Impossibility:** If, after applying the operations, not all elements are equal to `k`, return -1.

*   **Runtime Complexity:** O(n log n) due to sorting, where n is the length of `nums`.
*   **Storage Complexity:** O(n) in the worst case, to store the set of unique values.

	
	# Code
	```python
	def minOperations(nums, k):
    """
    Calculates the minimum number of operations to make all elements in nums equal to k.

    Args:
        nums: A list of integers.
        k: An integer.

    Returns:
        The minimum number of operations required, or -1 if it's impossible.
    """

    unique_values = set()
    for num in nums:
        if num > k:
            unique_values.add(num)

    valid_h_values = sorted(list(unique_values), reverse=True)
    operations = 0

    temp_nums = list(nums)  # Create a copy to modify

    for h in valid_h_values:
        operations += 1
        for i in range(len(temp_nums)):
            if temp_nums[i] > h:
                temp_nums[i] = h

    for i in range(len(temp_nums)):
        if temp_nums[i] > k:
            return -1

    unique_values_after_ops = set()
    for num in temp_nums:
        unique_values_after_ops.add(num)

    if len(unique_values_after_ops) > 1 and k not in unique_values_after_ops:
        return -1

    
    all_equal_k = all(num == k for num in temp_nums)
    if not all_equal_k:
        
        remaining_ops = 0
        
        if k not in nums:
          
          if min(nums) < k:
            return -1
          
          remaining_ops = 1

        else:
          
          remaining_ops = 0
          
          for i in range(len(temp_nums)):
            if temp_nums[i] != k:
              temp_nums[i] = k
              remaining_ops+=1
          
          if len(set(temp_nums)) > 1:
            remaining_ops=1


        
        return operations + remaining_ops
    
    else:
        return operations
	```