What does a balanced string mean in Check Balanced String?

A balanced string has the sum of digits at even indices equal to the sum at odd indices.

How do I efficiently implement this without extra memory?

Use two accumulators for even and odd sums and iterate once through the string, avoiding arrays or lists.

Can the string contain non-digit characters?

No, the input string consists only of digits according to the problem constraints.

Is a string of length 2 always balanced?

Not necessarily; it depends on whether the two digits at index 0 and 1 are equal.

What is the key pattern to recognize for this problem?

The string-driven solution strategy focuses on summing digits by index parity to detect balance efficiently.

#3340

Easy

auto_awesomeString-driven solution strategy

LeetCode 题解工作台

检查平衡字符串

给你一个仅由数字 0 - 9 组成的字符串 num 。如果偶数下标处的数字之和等于奇数下标处的数字之和，则认为该数字字符串是一个平衡字符串。如果 num 是一个平衡字符串，则返回 true ；否则，返回 false 。示例 1：输入： num = "1234" 输出： false 解释…

字符串

题目描述

给你一个仅由数字 0 - 9 组成的字符串 num。如果偶数下标处的数字之和等于奇数下标处的数字之和，则认为该数字字符串是一个 平衡字符串。

如果 num 是一个 平衡字符串，则返回 true；否则，返回 false。

示例 1：

输入：num = "1234"

输出：false

解释：

偶数下标处的数字之和为 1 + 3 = 4，奇数下标处的数字之和为 2 + 4 = 6。
由于 4 不等于 6，num 不是平衡字符串。

示例 2：

输入：num = "24123"

输出：true

解释：

偶数下标处的数字之和为 2 + 1 + 3 = 6，奇数下标处的数字之和为 4 + 2 = 6。
由于两者相等，num 是平衡字符串。

提示：

2 <= num.length <= 100
num 仅由数字 0 - 9 组成。

lightbulb

解题思路

方法一：模拟

我们可以用一个长度为 $2$ 的数组 $f$ 来记录偶数下标和奇数下标的数字之和，然后遍历字符串 $\textit{nums}$ ，根据下标的奇偶性将数字加到对应的位置上，最后判断 $f[0]$ 是否等于 $f[1]$ 即可。

时间复杂度 $O(n)$ ，其中 $n$ 为字符串 $\textit{nums}$ 的长度。空间复杂度 $O(1)$ 。

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class Solution:
    def isBalanced(self, num: str) -> bool:
        f = [0, 0]
        for i, x in enumerate(map(int, num)):
            f[i & 1] += x
        return f[0] == f[1]

speed

复杂度分析

指标	值
时间	complexity is O(n) as each character is visited once. Space complexity is O(1) because only two sum variables are needed, independent of string length.
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Checks understanding of string iteration and index handling.
question_mark
Evaluates ability to implement a numeric check without unnecessary data structures.
question_mark
Observes awareness of edge cases, such as strings with minimal length or equal digits.

warning

常见陷阱

外企场景

error
Confusing index parity, summing wrong positions for even vs odd indices.
error
Converting characters incorrectly, e.g., adding ASCII values instead of integers.
error
Using extra arrays or lists unnecessarily, increasing space complexity.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Balance check on strings with alphabetic characters representing numeric values.
arrow_right_alt
Check balanced condition for longer numeric strings with length up to 10^5.
arrow_right_alt
Modify the definition to consider weighted sums for even and odd positions.

help

常见问题

外企场景

继续练习

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