LeetCode-in-Java

2919. Minimum Increment Operations to Make Array Beautiful

Medium

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

Solution

public class Solution {
    public long minIncrementOperations(int[] nums, int k) {
        long[] dp = new long[nums.length];
        dp[0] = Math.max(0, k - nums[0]);
        dp[1] = Math.max(0, k - nums[1]);
        dp[2] = Math.max(0, k - nums[2]);
        for (int i = 3; i < nums.length; i++) {
            dp[i] = Math.max(0, k - nums[i]) + Math.min(Math.min(dp[i - 3], dp[i - 2]), dp[i - 1]);
        }
        return Math.min(Math.min(dp[nums.length - 3], dp[nums.length - 2]), dp[nums.length - 1]);
    }
}

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