What is the primary strategy to solve the 'Minimum Number of Increments on Subarrays to Form a Target Array' problem?

The primary strategy is to use state transition dynamic programming, focusing on minimizing the number of operations by calculating the differences between adjacent target values.

How does the greedy approach help in solving this problem?

The greedy approach helps by incrementing the entire subarray by the minimum value in that range, which ensures that the total number of operations is minimized.

What role does the monotonic stack play in this problem?

The monotonic stack helps manage the subarrays efficiently, ensuring that increments are applied in the most optimal order.

What are the time and space complexities of the 'Minimum Number of Increments on Subarrays to Form a Target Array' problem?

The time complexity is O(n), where n is the length of the target array, and the space complexity is O(n), assuming dynamic programming or greedy approaches are used.

Can this problem be solved using a brute force approach?

While a brute force approach is possible, it would be inefficient, as it would involve checking all possible subarray increments, leading to high time complexity.

#1526

Hard

auto_awesome状态·转移·动态规划

LeetCode 题解工作台

形成目标数组的子数组最少增加次数

给你一个整数数组 target 和一个数组 initial ， initial 数组与 target 数组有同样的大小，且一开始全部为 0 。一次操作中，你可以从 initial 数组中选择任何子数组，并将每个值加 1 。返回从 initial 数组构造 target 数组的最少操作次数。 …

数组动态规划栈贪心单调栈

题目描述

给你一个整数数组 target 和一个数组 initial ，initial 数组与 target 数组有同样的大小，且一开始全部为 0 。

一次操作中，你可以从 initial 数组中选择任何子数组，并将每个值加 1。

返回从 initial 数组构造 target 数组的最少操作次数。

答案保证在 32 位整数以内。

示例 1：

输入：target = [1,2,3,2,1]
输出：3
解释：我们需要至少 3 次操作从 intial 数组得到 target 数组。
[0,0,0,0,0] 将下标为 0 到 4 的元素（包含二者）加 1 。
[1,1,1,1,1] 将下标为 1 到 3 的元素（包含二者）加 1 。
[1,2,2,2,1] 将下标为 2 的元素增加 1 。
[1,2,3,2,1] 得到了目标数组。

示例 2：

输入：target = [3,1,1,2]
输出：4
解释：(initial)[0,0,0,0] -> [1,1,1,1] -> [1,1,1,2] -> [2,1,1,2] -> [3,1,1,2] (target) 。

示例 3：

输入：target = [3,1,5,4,2]
输出：7
解释：(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)。

示例 4：

输入：target = [1,1,1,1]
输出：1

提示：

1 <= target.length <= 10⁵
1 <= target[i] <= 10⁵
输入保证答案在 32 位整数范围内。

lightbulb

解题思路

方法一：动态规划

我们定义 $f[i]$ 表示得到 $target[0,..i]$ 的最少操作次数，初始时 $f[0] = target[0]$ 。

对于 $target[i]$ ，如果 $target[i] \leq target[i-1]$ ，则 $f[i] = f[i-1]$ ；否则 $f[i] = f[i-1] + target[i] - target[i-1]$ 。

最终答案即为 $f[n-1]$ 。

我们注意到 $f[i]$ 只与 $f[i-1]$ 有关，因此可以只用一个变量来维护操作次数。

时间复杂度 $O(n)$ ，其中 $n$ 为数组 $target$ 的长度。空间复杂度 $O(1)$ 。

相似题目：

3229. 使数组等于目标数组所需的最少操作次数

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class Solution:
    def minNumberOperations(self, target: List[int]) -> int:
        return target[0] + sum(max(0, b - a) for a, b in pairwise(target))

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Does the candidate effectively use state transition dynamic programming for the problem?
question_mark
Is the candidate comfortable applying greedy strategies for range-based optimization?
question_mark
Does the candidate demonstrate proficiency in managing subarray operations with a monotonic stack?

warning

常见陷阱

外企场景

error
Incorrectly handling the subarray increments could lead to more operations than necessary.
error
Failing to optimize range increments might result in a solution that exceeds the required number of operations.
error
Overcomplicating the solution with unnecessary nested loops or complex data structures can increase time complexity.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Modifying the problem to work with a non-zero initial array, requiring adjustments to the dynamic programming approach.
arrow_right_alt
Allowing multiple increments on the same subarray, testing the candidate's ability to handle more complex operations.
arrow_right_alt
Changing the target array values to be random or non-sequential, requiring more adaptive range-based strategies.

help

常见问题

外企场景

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