求将全零源数组转换为目标数组的最少操作步数算法优化

Hey there! Let’s break down this problem and find that efficient solution your interviewer was looking for. First, let’s restate the problem clearly to make sure we’re on the same page:

Given a source array filled entirely with zeros and a target array of non-negative integers, find the minimum number of operations needed to turn the source into the target. Each operation lets you increment every element from index L to R (inclusive) by 1.

Your Current Approach #

You mentioned splitting the array into non-zero subarrays, subtracting the minimum value of each subarray from all its elements, and adding that minimum to your count. For your example [4,2,3,5], your steps looked like this:

1. First pass: The whole array is one non-zero subarray with minimum 2. Subtract 2 from all elements (array becomes [2,0,1,3]), count += 2.
2. Second pass: Two non-zero subarrays ([2] and [1,3]) with minima 2 and 1. Subtract those values, count += 2 + 1 = 3 (total now 5), array becomes [0,0,0,2].
3. Third pass: Subtract 2 from the last element, count += 2 (total 7), array is all zeros.

This approach works, but it’s not optimal—worst-case time complexity can hit O(n²) (like for a strictly increasing array, you’d have to process each element multiple times). Let’s look at a better way.

The Optimal O(n) Solution #

Here’s a linear-time approach that’s both simple and efficient:

• Start with prev = 0 and result = 0.
• Iterate through each element curr in the target array:
  • If curr > prev, add curr - prev to result.
  • Update prev to curr.
• The final result is your minimum number of operations.

Let’s test this with your example [4,2,3,5]:

1. prev = 0, curr = 4 → 4 > 0, so result += 4 (result = 4), prev = 4.
2. curr = 2 → 2 < 4, no addition, prev = 2.
3. curr = 3 → 3 > 2, result += 1 (result = 5), prev = 3.
4. curr = 5 → 5 > 3, result += 2 (result = 7), prev = 5.

That’s the same total as your approach, but we’re done in a single pass through the array.

Why This Works #

Think of the problem as tracking the "rising steps" between consecutive elements. Every time the current element is larger than the previous one, that difference represents new operations we need to start at (or before) this index—these operations will cover from this point onward until the value drops again.

When the current element is smaller than the previous, it just means some of the earlier operations stop before this index—we don’t need to add anything to the count for those cases, since we already accounted for the initial start of those operations.

Segment Tree/Binary Indexed Tree (Fenwick Tree) Context #

You mentioned considering segment trees or BITs, and while they’re not necessary for this static problem, they could be useful if the target array had dynamic updates (like elements changing after your initial calculation).

For a static array, using a segment tree would be overkill, but if you wanted to implement it, you’d build a tree to track the differences between consecutive elements, then sum all the positive differences (which is exactly what the linear approach does). But again, the linear method is far simpler and faster here.

Final Takeaway #

The linear-time approach is the optimal solution your interviewer was likely expecting—it’s O(n) time, O(1) extra space, and trivial to code. Your original approach works, but this is the more efficient, elegant solution.

内容的提问来源于stack exchange，提问作者danish sodhi