Spent probably too long trying to brute force a greedy approach before realizing you can think about it in terms of the 'deficit' at each position relative to the previous element.
First, clarify that the goal is to make the array non-decreasing with minimum operations, where each operation increments a contiguous subarray by 1. Then, observe that the problem reduces to summing the positive differences between consecutive elements, as each increase in value from left to right requires at least that many increments, and these can be achieved independently.
Pro tip: Mention that this is equivalent to the 'minimum number of increments to make array non-decreasing' problem, which is a known pattern in competitive programming. Also, note that the answer is simply the sum of max(0, a[i] - a[i-1]) for i from 1 to n-1, and explain why this is optimal.
Restate the problem: we can increment any contiguous subarray by 1, and we want the array to be non-decreasing (each element ≤ the next). The goal is to minimize the number of operations.
For each adjacent pair (a[i-1], a[i]), if a[i-1] > a[i], we need to increase a[i] (and possibly elements to its right) to at least a[i-1]. The minimum increase needed at position i is max(0, a[i-1] - a[i]).
Each required increase can be achieved by an operation on a subarray starting at i and extending to the right, without affecting earlier elements. These operations do not interfere with each other's requirements.
The total minimum operations is the sum over i=1 to n-1 of max(0, a[i-1] - a[i]). This is because each operation can only increase elements, and the required increases are additive.
Test with simple arrays (e.g., [3,2,1], [1,2,3], [5,5,5]) to confirm the formula. Consider edge cases like empty array or single element (answer 0).
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