How does dynamic programming help in solving the Build Array Where You Can Find The Maximum Exactly K Comparisons problem?

Dynamic programming helps by breaking down the problem into subproblems. It tracks possible states for array length, maximum element, and number of comparisons, ensuring an optimal solution.

What is the role of prefix sums in solving this problem?

Prefix sums optimize the DP solution by allowing efficient calculation of valid configurations, reducing redundant computations and improving runtime.

What is the time complexity of the Build Array Where You Can Find The Maximum Exactly K Comparisons problem?

The time complexity is O(n * m * k) due to the need to compute the DP table with multiple states for array length, maximum element, and comparisons.

How does the modulo operation affect the solution to this problem?

The modulo operation ensures that the solution stays within the bounds of 10^9 + 7, preventing overflow and ensuring correctness in large inputs.

What is the most common mistake when solving the Build Array Where You Can Find The Maximum Exactly K Comparisons problem?

A common mistake is not managing the DP transitions correctly or failing to apply optimizations like prefix sums, leading to inefficient solutions.

#1420

Hard

auto_awesome状态·转移·动态规划

LeetCode 题解工作台

生成数组

给定三个整数 n 、 m 和 k 。考虑使用下图描述的算法找出正整数数组中最大的元素。请你构建一个具有以下属性的数组 arr ： arr 中包含确切的 n 个整数。 1 其中 (0 。将上面提到的算法应用于 arr 之后， search_cost 的值等于 k 。返回在满足上述条件的情况下构建…

动态规划前缀和

题目描述

给定三个整数 n、m 和 k 。考虑使用下图描述的算法找出正整数数组中最大的元素。

请你构建一个具有以下属性的数组 arr ：

arr 中包含确切的 n 个整数。
1 <= arr[i] <= m 其中 (0 <= i < n) 。
将上面提到的算法应用于 arr 之后，search_cost 的值等于 k 。

返回在满足上述条件的情况下构建数组 arr 的 方法数量 ，由于答案可能会很大，所以必须对 10^9 + 7 取余。

示例 1：

输入：n = 2, m = 3, k = 1
输出：6
解释：可能的数组分别为 [1, 1], [2, 1], [2, 2], [3, 1], [3, 2] [3, 3]

示例 2：

输入：n = 5, m = 2, k = 3
输出：0
解释：没有数组可以满足上述条件

示例 3：

输入：n = 9, m = 1, k = 1
输出：1
解释：唯一可能的数组是 [1, 1, 1, 1, 1, 1, 1, 1, 1]

提示：

1 <= n <= 50
1 <= m <= 100
0 <= k <= n

lightbulb

解题思路

方法一：动态规划

假设 $dp[i][c][j]$ 表示长度为 $i$ ，搜索代价为 $c$ ，且最大值为 $j$ 的方案数。考虑第 $i$ 个数：

若第 $i$ 个数没有改变搜索代价，说明它不严格大于前 $i-1$ 个数，也就是说， $dp[i][c][j]$ 是从 $dp[i-1][c][j]$ 转移而来，即数组的前 $i-1$ 个数的最大值已经是 $j$ ，并且第 $i$ 个数没有改变最大值，因此第 $i$ 个数的可选范围是 $[1,..j]$ ，共有 $j$ 种可选方案。即

dp[i][c][j]=dp[i-1][c][j] \times j

若第 $i$ 个数改变了搜索代价，说明数组前 $i-1$ 个数的最大值小于 $j$ ，并且第 $i$ 个数恰好为 $j$ 。此时 $dp[i][c][j]$ 是从所有 $dp[i-1][c-1][j']$ 转移而来，其中 $j'<j$ 。即

dp[i][c][j] = \sum_{j'=1}^{j-1} dp[i-1][c-1][j']

综上，可得

dp[i][c][j] = dp[i-1][c][j] \times j + \sum_{j'=1}^{j-1} dp[i-1][c-1][j']

答案为

\sum_{j=1}^{m}dp[n][k][j]

时间复杂度 $O(nkm^2)$ 。

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class Solution:
    def numOfArrays(self, n: int, m: int, k: int) -> int:
        if k == 0:
            return 0
        dp = [[[0] * (m + 1) for _ in range(k + 1)] for _ in range(n + 1)]
        mod = 10**9 + 7
        for i in range(1, m + 1):
            dp[1][1][i] = 1
        for i in range(2, n + 1):
            for c in range(1, min(k + 1, i + 1)):
                for j in range(1, m + 1):
                    dp[i][c][j] = dp[i - 1][c][j] * j
                    for j0 in range(1, j):
                        dp[i][c][j] += dp[i - 1][c - 1][j0]
                        dp[i][c][j] %= mod
        ans = 0
        for i in range(1, m + 1):
            ans += dp[n][k][i]
            ans %= mod
        return ans

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
Can the candidate break down the dynamic programming transition states?
question_mark
How well does the candidate optimize the solution using prefix sums?
question_mark
Is the candidate aware of the need to use modulo operations to prevent overflow?

warning

常见陷阱

外企场景

error
Overlooking the constraints and not managing the dynamic programming transitions correctly.
error
Failing to use prefix sum optimization, leading to excessive computation time.
error
Not correctly handling the modulo operation, which can lead to overflow or incorrect results.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Change the value of k to test the candidate's ability to adapt the DP table to different constraints.
arrow_right_alt
Vary the range of possible values for m to test how the solution scales.
arrow_right_alt
Increase the size of the array n and ask for optimizations to handle larger inputs efficiently.

help

常见问题

外企场景

继续练习

#1444 切披萨的方案数 #1524 和为奇数的子数组数目 #1687 从仓库到码头运输箱子