What is the primary pattern for solving Smallest Index With Equal Value?

The primary pattern is an array-driven solution strategy where we check each index for the condition `i mod 10 == nums[i]`.

What if no index satisfies the condition in Smallest Index With Equal Value?

If no index satisfies the condition, the correct answer is -1, indicating that no valid index was found.

How does GhostInterview help with the Smallest Index With Equal Value problem?

GhostInterview guides you through iterating the array and ensures that you handle early exits and edge cases properly.

What is the time complexity of Smallest Index With Equal Value?

The time complexity is O(n) because each index is checked once, where n is the length of the array.

Can the Smallest Index With Equal Value problem be optimized further?

The solution is already optimized for the problem as it checks each index only once and stops early once a valid index is found.

#2057

Easy

auto_awesome数组·driven

LeetCode 题解工作台

值相等的最小索引

给你一个下标从 0 开始的整数数组 nums ，返回 nums 中满足 i mod 10 == nums[i] 的最小下标 i ；如果不存在这样的下标，返回 -1 。 x mod y 表示 x 除以 y 的余数。示例 1：输入： nums = [0,1,2] 输出： 0 解释： i=0: 0…

数组

题目描述

给你一个下标从 0 开始的整数数组 nums ，返回 nums 中满足 i mod 10 == nums[i] 的最小下标 i ；如果不存在这样的下标，返回 -1 。

x mod y 表示 x 除以 y 的余数。

示例 1：

输入：nums = [0,1,2]
输出：0
解释：
i=0: 0 mod 10 = 0 == nums[0].
i=1: 1 mod 10 = 1 == nums[1].
i=2: 2 mod 10 = 2 == nums[2].
所有下标都满足 i mod 10 == nums[i] ，所以返回最小下标 0

示例 2：

输入：nums = [4,3,2,1]
输出：2
解释：
i=0: 0 mod 10 = 0 != nums[0].
i=1: 1 mod 10 = 1 != nums[1].
i=2: 2 mod 10 = 2 == nums[2].
i=3: 3 mod 10 = 3 != nums[3].
2 唯一一个满足 i mod 10 == nums[i] 的下标

示例 3：

输入：nums = [1,2,3,4,5,6,7,8,9,0]
输出：-1
解释：不存在满足 i mod 10 == nums[i] 的下标

示例 4：

输入：nums = [2,1,3,5,2]
输出：1
解释：1 是唯一一个满足 i mod 10 == nums[i] 的下标

提示：

1 <= nums.length <= 100
0 <= nums[i] <= 9

lightbulb

解题思路

方法一：遍历

我们直接遍历数组，对于每个下标 $i$ ，我们判断是否满足 $i \bmod 10 = \textit{nums}[i]$ ，如果满足则返回当前下标 $i$ 。

如果遍历完数组都没有找到满足条件的下标，则返回 $-1$ 。

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

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class Solution:
    def smallestEqual(self, nums: List[int]) -> int:
        for i, x in enumerate(nums):
            if i % 10 == x:
                return i
        return -1

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
Check the candidate's approach to early exits when a solution is found.
question_mark
Assess how they handle edge cases like no valid index or a very small array.
question_mark
Evaluate their understanding of the modulo operation and how it can be applied efficiently.

warning

常见陷阱

外企场景

error
Overlooking the possibility of no valid index and returning incorrect results.
error
Not returning the smallest valid index when multiple indices satisfy the condition.
error
Confusing the modulo operation and how it works with non-zero divisors.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Change the condition to `i mod k == nums[i]` where k is a parameter.
arrow_right_alt
Allow negative numbers in the array and modify the solution accordingly.
arrow_right_alt
Limit the size of the array to a specific value and optimize the approach.

help

常见问题

外企场景

继续练习

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