Minimum Number of Increments on Subarrays to Form a Target Array

Given an array of positive integers target and an array initial of same size with all zeros.

Return the minimum number of operations to form a target array from initial if you are allowed to do the following operation:

• Choose any subarray from initial and increment each value by one.

The answer is guaranteed to fit within the range of a 32-bit signed integer.

Example 1:

Input: target = [1,2,3,2,1]
Output: 3
Explanation: We need at least 3 operations to form the target array from the initial array.
[0,0,0,0,0] increment 1 from index 0 to 4 (inclusive).
[1,1,1,1,1] increment 1 from index 1 to 3 (inclusive).
[1,2,2,2,1] increment 1 at index 2.
[1,2,3,2,1] target array is formed.

Example 2:

Input: target = [3,1,1,2]
Output: 4
Explanation: (initial)[0,0,0,0] -> [1,1,1,1] -> [1,1,1,2] -> [2,1,1,2] -> [3,1,1,2] (target).

Example 3:

Input: target = [3,1,5,4,2]
Output: 7
Explanation: (initial)[0,0,0,0,0] -> [1,1,1,1,1] -> [2,1,1,1,1] -> [3,1,1,1,1] 
                                  -> [3,1,2,2,2] -> [3,1,3,3,2] -> [3,1,4,4,2] -> [3,1,5,4,2] (target).

Example 4:

Input: target = [1,1,1,1]
Output: 1

Constraints:

• 1 <= target.length <= 10^5

• 1 <= target[i] <= 10^5

class Solution {
    public int minNumberOperations(int[] target) {
        // Each operation increments by 1
        int operationsUsed = target[0];
        int maxHeight = target[0];
        // Now we need to see upto where our current operations extend upto
        for (int i = 1; i < target.length; i++) {
            if (target[i] < maxHeight)
                maxHeight = target[i];
            // We need new operations for new local maximas
            else {
                operationsUsed += target[i] - maxHeight;
                maxHeight = target[i];
            }
        }
        return operationsUsed;
    }
}