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2919. Minimum Increment Operations to Make Array Beautiful

SourceWeekly Contest 369 Q3DifficultyMediumRating2030

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[1] by 1 -> [2,4,0,0,2].
Choose index i = 4 and increase nums[4] by 1 -> [2,4,0,0,3].
Choose index i = 4 and increase nums[4] 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.
Hence, the answer is 3.

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[2] by 1 -> [0,1,4,3].
Choose index i = 2 and increase nums[2] 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.
Hence, the answer is 2.

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.
Hence, the answer is 0.

 

Constraints:

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

Solutions

Solution 1: Dynamic Programming

Thinking

Every window of length \(3\) must have a maximum of at least \(k\). \(n \le 10^5\) forbids charging each window separately. Incrementing index \(i\) by \(\max(k-nums[i],0)\) covers every window that contains \(i\), and each window needs only one such cover.

The state therefore tracks the cheapest way to cover using one of the last three positions. Rolling \(f,g,h\) store those three optima; reading \(x\) advances them by \(\min(f,g,h)+\max(k-x,0)\). The answer is the minimum of the three.

We define \(f\), \(g\), and \(h\) as the minimum number of increment operations needed to get the maximum value from the last three items in the first \(i\) items, initially \(f = 0\), \(g = 0\), \(h = 0\).

Next, we traverse the array \(nums\). For each \(x\), we need to update the values of \(f\), \(g\), and \(h\) to meet the requirements of the problem, that is:

\[ \begin{aligned} f' &= g \\ g' &= h \\ h' &= \min(f, g, h) + \max(k - x, 0) \end{aligned} \]

Finally, we only need to return the minimum value among \(f\), \(g\), and \(h\).

The time complexity is \(O(n)\), where \(n\) is the length of the array. The space complexity is \(O(1)\).

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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)

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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);
    }
}

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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});
    }
};

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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))
}

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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);
}

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public class Solution {
    public long MinIncrementOperations(int[] nums, int k) {
        long f = 0, g = 0, h = 0;
        foreach (int x in 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);
    }
}

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