LeetCode 2919: Minimum Increment Operations to Make Array Beautiful

Explanation

To make the array beautiful, we need to ensure that for any subarray of size 3 or more, the maximum element is greater than or equal to k. To achieve this, we can iterate over the array and for each element, we calculate the required increment to make it greater than or equal to k while maintaining the minimum overall increment operations required. We can do this by keeping track of the current maximum element in the subarray so far.

Algorithm

1. Initialize ops to 0 to keep track of the total increment operations needed.
2. Initialize maxSeen to 0 to keep track of the current maximum element in the subarray so far.
3. Iterate over the array:
   • If the current element is less than k, calculate the increment required to make it greater than or equal to k.
   • If the current element is greater than or equal to k, update maxSeen.
   • Increment ops by the difference between maxSeen and the current element.
4. Return ops.

Time Complexity

The time complexity of this algorithm is O(n), where n is the length of the input array.

Space Complexity

The space complexity of this algorithm is O(1) as we are using only a constant amount of extra space.

class Solution {
    public int minOperations(int[] nums, int k) {
        int ops = 0;
        int maxSeen = 0;
        
        for (int num : nums) {
            if (num < k) {
                ops += k - num;
            } else {
                maxSeen = Math.max(maxSeen, num);
            }
            ops += Math.max(0, maxSeen - num);
        }
        
        return ops;
    }
}