What is the core strategy for Remove Digit From Number to Maximize Result?

The core strategy is a greedy approach: iterate through all occurrences of the given digit and remove the one that results in the largest number.

Can I just remove the first occurrence of the digit?

No, removing the first occurrence may not always yield the maximum value. You must check all positions where the digit appears.

Does this problem require numeric conversion of the string?

Not necessarily. You can compare candidate strings lexicographically since all strings are of equal or shorter length and contain only digits.

How does the greedy choice ensure the correct result?

By always selecting the removal that produces the largest immediate candidate, we maintain the invariant that no other single removal produces a higher number.

What is the time complexity for this problem?

Time complexity is O(n), where n is the length of the number string, since we examine each occurrence of the target digit once.

#2259

Easy

auto_awesome贪心·invariant

LeetCode 题解工作台

移除指定数字得到的最大结果

给你一个表示某个正整数的字符串 number 和一个字符 digit 。从 number 中恰好移除一个等于 digit 的字符后，找出并返回按十进制表示最大的结果字符串。生成的测试用例满足 digit 在 number 中出现至少一次。示例 1：输入： number = "1…

字符串贪心枚举

题目描述

给你一个表示某个正整数的字符串 number 和一个字符 digit 。

从 number 中恰好移除一个等于 digit 的字符后，找出并返回按 十进制 表示最大的结果字符串。生成的测试用例满足 digit 在 number 中出现至少一次。

示例 1：

输入：number = "123", digit = '3'
输出："12"
解释："123" 中只有一个 '3' ，在移除 '3' 之后，结果为 "12" 。

示例 2：

输入：number = "1231", digit = '1'
输出："231"
解释：可以移除第一个 '1' 得到 "231" 或者移除第二个 '1' 得到 "123" 。
由于 231 > 123 ，返回 "231" 。

示例 3：

输入：number = "551", digit = '5'
输出："51"
解释：可以从 "551" 中移除第一个或者第二个 '5' 。
两种方案的结果都是 "51" 。

提示：

2 <= number.length <= 100
number 由数字 '1' 到 '9' 组成
digit 是 '1' 到 '9' 中的一个数字
digit 在 number 中出现至少一次

lightbulb

解题思路

方法一：暴力枚举

我们可以枚举字符串 $\textit{number}$ 的所有位置 $\textit{i}$ ，如果 $\textit{number}[i] = \textit{digit}$ ，那么我们取 $\textit{number}$ 的前缀 $\textit{number}[0:i]$ 和后缀 $\textit{number}[i+1:]$ 拼接起来，即为移除 $\textit{number}[i]$ 后的结果。我们取所有可能的结果中最大的即可。

时间复杂度 $O(n^2)$ ，空间复杂度 $O(n)$ 。其中 $n$ 为字符串 $\textit{number}$ 的长度。

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class Solution:
    def removeDigit(self, number: str, digit: str) -> str:
        return max(
            number[:i] + number[i + 1 :] for i, d in enumerate(number) if d == digit
        )

speed

复杂度分析

指标	值
时间	complexity is O(n) because each digit position is checked and candidates are compared in linear time. Space complexity is O(n) for storing positions and temporary candidate strings, where n is the length of number.
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Checks if you recognize the greedy pattern for maximizing after single removal.
question_mark
Observes whether you validate all possible positions of the target digit.
question_mark
Wants to see that you efficiently handle comparisons without unnecessary conversions.

warning

常见陷阱

外企场景

error
Removing the first occurrence without comparing other positions can fail on numbers like "1231" with digit '1'.
error
Comparing string values lexicographically without considering numeric meaning may lead to incorrect selection.
error
Not handling multiple consecutive target digits properly can miss the optimal removal.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Remove two digits to maximize result, extending greedy comparisons to pairs of removals.
arrow_right_alt
Minimize the number instead of maximizing, reversing the greedy logic to remove the largest digit first.
arrow_right_alt
Allow multiple different digits to be removed, requiring combination enumeration and invariant validation.

help

常见问题

外企场景

继续练习

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