You are given an integer array target. You have an integer array initial of the same size as target with all elements initially zeros.
In one operation you can choose any subarray from initial and increment each value by one.
Return the minimum number of operations to form a target array from initial.
The test cases are generated so that the answer fits in a 32-bit 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: [0,0,0,0] -> [1,1,1,1] -> [1,1,1,2] -> [2,1,1,2] -> [3,1,1,2]
Example 3:
Input: target = [3,1,5,4,2] Output: 7 Explanation: [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].
Constraints:
1 <= target.length <= 1051 <= target[i] <= 105The key observation in #1526 Minimum Number of Increments on Subarrays to Form a Target Array is that each operation increases all elements in a chosen subarray by 1. Instead of simulating operations, we analyze how the target array grows from left to right. If the current element is greater than the previous one, additional increments are required to build that height difference.
A common greedy insight is that the minimum number of operations equals the sum of all positive increases between consecutive elements. In other words, whenever target[i] > target[i-1], the difference must be created using new increment operations. If the height decreases or stays the same, previous operations already cover it.
Another perspective involves using a monotonic stack to track increasing segments and compute required increments while maintaining order. Both approaches avoid brute-force simulation and run efficiently in linear time. This makes the problem manageable even for large arrays.
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Greedy Difference Traversal | O(n) | O(1) |
| Monotonic Stack Approach | O(n) | O(n) |
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Use these hints if you're stuck. Try solving on your own first.
For a given range of values in target, an optimal strategy is to increment the entire range by the minimum value. The minimum in a range could be obtained with Range minimum query or Segment trees algorithm.
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A monotonic stack can be used to track increasing heights and manage segments where increments begin or end. While not strictly necessary for the optimal solution, it provides another structured way to reason about increasing subarray heights.
Each operation effectively builds vertical layers across subarrays. When the next element is higher than the previous one, additional layers must start there. Counting these positive differences ensures we add exactly the required layers without redundant operations.
Yes, this problem is considered interview-relevant because it tests greedy reasoning and array pattern recognition. Variants of this question often appear in technical interviews to evaluate optimization skills and understanding of incremental operations.
The optimal approach uses a greedy observation: only increases between consecutive elements require new operations. By summing all positive differences between target[i] and target[i-1], we can compute the minimum number of increments in O(n) time.