The problem boils down to:

Given a target array ( a ) and an initial array of all zeros, where you can pick any segment ([l,r]) and add any positive integer ( x ) to all elements in ([l,r]), find the minimum number of operations needed to reach ( a ).

Initially, it feels like we should just compute:

because it counts the "upward movements" — but this formula only works when ( x = 1 ) per operation.

Since here we can pick any (x > 0), the task is different:
We want to cover the "histogram" defined by ( a ) with as few horizontal rectangles as possible.

Solution Idea (Monotonic Stack Approach):

• Traverse (a) from left to right.
• Maintain a monotonically increasing stack of previous heights.
• At each position (i):
• If the current height (a[i]) is less than the previous, pop until the stack is valid.
• If the current height (a[i]) is greater than the previous, push it onto the stack and increment the answer.

Algorithm (in pseudocode):

ans = 0
stack = empty

for i = 1 to n:
    while stack not empty and stack.top > a[i]:
        stack.pop()
    if stack.empty or stack.top < a[i]:
        if a[i] > 0:
            ans += 1
            stack.push(a[i])

Why does it work? - Popping ensures that we properly close any rectangles that can’t extend further. - If (a[i]) is strictly larger than what’s on top, we need a new rectangle starting here. - Only non-zero heights are considered.

Example:

Given ( a = [1, 2, 1, 2] )

Final answer = 3 operations.

Time Complexity:

• Each element is pushed and popped at most once → (O(n)) overall.

Summary:
If (x) is free to be any positive integer, use the monotonic stack method to count new necessary rectangle "starts."
If (x=1) is fixed, then summing positive differences $$$\max(a_i - a_{i-1}, 0)$$$ would work.