Minimum Operations To Make The Array K Increasing - Leetcode Solution

Problem:

Given an array nums of integers, a move consists of choosing any element and increasing it by 1.

Return the minimum number of moves to make every value in nums greater than or equal to k.

Example 1:

Input: nums = [1,2,0,3], k = 3 Output: 3 Explanation: To make every value greater than or equal to 3, we can do the following three operations:

1. Increase the first element by 2: [3,2,0,3].
2. Increase the second element by 1: [3,3,0,3].
3. Increase the third element by 3: [3,3,3,3]. Example 2:

Input: nums = [2,3,4,5,6], k = 1 Output: 0 Explanation: Every value in nums already greater than or equal to 1. Constraints:

1 <= nums.length <= 105 0 <= nums[i] <= 109 1 <= k <= 109

Solution:

To solve this problem, we can first calculate the minimum number of operations required to make each element of the array greater than or equal to k. Then, we can return the sum of all these operations.

Let's define a function "minOp(x, k)" which calculates the minimum number of operations required to make x greater than or equal to k. We can calculate this by taking the maximum of 0 and (k - x) and then dividing the result by k - x, rounded up to the nearest integer.

The reason why we take the maximum of 0 and (k - x) is that we only need to increase the value if it is less than k. If x is already greater than or equal to k, we don't need to make any changes. The division part is to make sure that we round up to the nearest integer, since we can only increase the value by whole numbers.

Now, we can apply this function to each element of the array and sum up the results to get the minimum number of operations required to make every value greater than or equal to k.

Here is the Python code for the solution:

def minOp(x, k): return -(-(k - x) // (k - x))

class Solution: def minOperations(self, nums: List[int], k: int) -> int: return sum(minOp(nums[i], k) for i in range(len(nums)))

Time Complexity: O(n) where n is the length of the input array. We traverse the entire array once.

Space Complexity: O(1). We use constant amount of extra space for variables.

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