# 2919. Minimum Increment Operations to Make Array Beautiful

## Description

You are given a 0-indexed integer array nums having length n, and an integer k.

You can perform the following increment operation any number of times (including zero):

• Choose an index i in the range [0, n - 1], and increase nums[i] by 1.

An array is considered beautiful if, for any subarray with a size of 3 or more, its maximum element is greater than or equal to k.

Return an integer denoting the minimum number of increment operations needed to make nums beautiful.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [2,3,0,0,2], k = 4
Output: 3
Explanation: We can perform the following increment operations to make nums beautiful:
Choose index i = 1 and increase nums by 1 -> [2,4,0,0,2].
Choose index i = 4 and increase nums by 1 -> [2,4,0,0,3].
Choose index i = 4 and increase nums by 1 -> [2,4,0,0,4].
The subarrays with a size of 3 or more are: [2,4,0], [4,0,0], [0,0,4], [2,4,0,0], [4,0,0,4], [2,4,0,0,4].
In all the subarrays, the maximum element is equal to k = 4, so nums is now beautiful.
It can be shown that nums cannot be made beautiful with fewer than 3 increment operations.


Example 2:

Input: nums = [0,1,3,3], k = 5
Output: 2
Explanation: We can perform the following increment operations to make nums beautiful:
Choose index i = 2 and increase nums by 1 -> [0,1,4,3].
Choose index i = 2 and increase nums by 1 -> [0,1,5,3].
The subarrays with a size of 3 or more are: [0,1,5], [1,5,3], [0,1,5,3].
In all the subarrays, the maximum element is equal to k = 5, so nums is now beautiful.
It can be shown that nums cannot be made beautiful with fewer than 2 increment operations.


Example 3:

Input: nums = [1,1,2], k = 1
Output: 0
Explanation: The only subarray with a size of 3 or more in this example is [1,1,2].
The maximum element, 2, is already greater than k = 1, so we don't need any increment operation.


Constraints:

• 3 <= n == nums.length <= 105
• 0 <= nums[i] <= 109
• 0 <= k <= 109

## Solutions

• class Solution {
public long minIncrementOperations(int[] nums, int k) {
long f = 0, g = 0, h = 0;
for (int x : nums) {
long hh = Math.min(Math.min(f, g), h) + Math.max(k - x, 0);
f = g;
g = h;
h = hh;
}
return Math.min(Math.min(f, g), h);
}
}

• class Solution {
public:
long long minIncrementOperations(vector<int>& nums, int k) {
long long f = 0, g = 0, h = 0;
for (int x : nums) {
long long hh = min({f, g, h}) + max(k - x, 0);
f = g;
g = h;
h = hh;
}
return min({f, g, h});
}
};

• class Solution:
def minIncrementOperations(self, nums: List[int], k: int) -> int:
f = g = h = 0
for x in nums:
f, g, h = g, h, min(f, g, h) + max(k - x, 0)
return min(f, g, h)


• func minIncrementOperations(nums []int, k int) int64 {
var f, g, h int
for _, x := range nums {
f, g, h = g, h, min(f, g, h)+max(k-x, 0)
}
return int64(min(f, g, h))
}

• function minIncrementOperations(nums: number[], k: number): number {
let [f, g, h] = [0, 0, 0];
for (const x of nums) {
[f, g, h] = [g, h, Math.min(f, g, h) + Math.max(k - x, 0)];
}
return Math.min(f, g, h);
}