What is the best approach for solving the "Construct Smallest Number From DI String" problem?

The best approach is using a stack for state management combined with a greedy technique to construct the lexicographically smallest number.

How can stack-based algorithms help in solving string pattern problems like "Construct Smallest Number From DI String"?

Stack-based algorithms allow for efficient state management, ensuring correct order by pushing and popping elements based on pattern conditions ('I' and 'D').

What is the time and space complexity for "Construct Smallest Number From DI String"?

The time complexity is O(n) because each character in the pattern is processed once. The space complexity is O(n) due to the stack used for state management.

Can I solve the "Construct Smallest Number From DI String" problem using recursion?

While recursion can be used, an iterative stack-based approach is more efficient and easier to manage, especially for handling the lexicographical minimization.

How does GhostInterview help with the "Construct Smallest Number From DI String" problem?

GhostInterview helps by guiding candidates through the stack-based approach and providing immediate feedback on common mistakes and algorithmic efficiency.

#2375

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LeetCode 题解工作台

根据模式串构造最小数字

给你下标从 0 开始、长度为 n 的字符串 pattern ，它包含两种字符， 'I' 表示上升， 'D' 表示下降。你需要构造一个下标从 0 开始长度为 n + 1 的字符串，且它要满足以下条件： num 包含数字 '1' 到 '9' ，其中每个数字至多使用一次。如果 patter…

字符串回溯栈贪心

题目描述

给你下标从 0 开始、长度为 n 的字符串 pattern ，它包含两种字符，'I' 表示上升，'D' 表示下降。

你需要构造一个下标从 0 开始长度为 n + 1 的字符串，且它要满足以下条件：

num 包含数字 '1' 到 '9' ，其中每个数字至多使用一次。
如果 pattern[i] == 'I' ，那么 num[i] < num[i + 1] 。
如果 pattern[i] == 'D' ，那么 num[i] > num[i + 1] 。

请你返回满足上述条件字典序最小的字符串 num。

示例 1：

输入：pattern = "IIIDIDDD"
输出："123549876"
解释：
下标 0 ，1 ，2 和 4 处，我们需要使 num[i] < num[i+1] 。
下标 3 ，5 ，6 和 7 处，我们需要使 num[i] > num[i+1] 。
一些可能的 num 的值为 "245639871" ，"135749862" 和 "123849765" 。
"123549876" 是满足条件最小的数字。
注意，"123414321" 不是可行解因为数字 '1' 使用次数超过 1 次。

示例 2：

输入：pattern = "DDD"
输出："4321"
解释：
一些可能的 num 的值为 "9876" ，"7321" 和 "8742" 。
"4321" 是满足条件最小的数字。

提示：

1 <= pattern.length <= 8
pattern 只包含字符 'I' 和 'D' 。

lightbulb

解题思路

方法一：DFS

定义 $dfs(u)$ ，其中 $u$ 表示当前答案字符串的长度。从 $u=0$ 开始搜索，直至找到第一个符合条件的字符串。

时间复杂度 $O(n!)$ ，空间复杂度 $O(n)$ 。其中 $n$ 表示字符串 $pattern$ 的长度。

class Solution:
    def smallestNumber(self, pattern: str) -> str:
        def dfs(u):
            nonlocal ans
            if ans:
                return
            if u == len(pattern) + 1:
                ans = ''.join(t)
                return
            for i in range(1, 10):
                if not vis[i]:
                    if u and pattern[u - 1] == 'I' and int(t[-1]) >= i:
                        continue
                    if u and pattern[u - 1] == 'D' and int(t[-1]) <= i:
                        continue
                    vis[i] = True
                    t.append(str(i))
                    dfs(u + 1)
                    vis[i] = False
                    t.pop()

        vis = [False] * 10
        t = []
        ans = None
        dfs(0)
        return ans

speed

复杂度分析

指标	值
时间	O(n)
空间	O(n)

psychology

面试官常问的追问

外企场景

question_mark
The candidate should demonstrate an understanding of stack-based algorithms and how they can be used for state management in this context.
question_mark
Look for a clear explanation of the greedy approach, ensuring that the candidate can justify how it minimizes the lexicographical order.
question_mark
The candidate should efficiently explain the time and space complexity of their solution, ensuring that they are aware of the optimal performance for this problem.

warning

常见陷阱

外企场景

error
Forgetting to pop the stack when encountering an 'I' pattern may result in an incorrect number sequence.
error
Incorrect handling of 'I' and 'D' patterns could lead to not achieving the smallest lexicographical number.
error
Not managing the stack properly may lead to inefficient space or time complexity, making the solution suboptimal.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Handling larger inputs with longer patterns can add complexity, though the algorithm remains O(n) in time and space.
arrow_right_alt
If the pattern is all 'I' or all 'D', the stack's behavior changes, but the approach still applies with minimal modification.
arrow_right_alt
The pattern length could affect the stack usage and result in different behavior, so optimizing space usage in the stack could be a potential focus.

help

常见问题

外企场景

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