LeetCode 2057: Smallest Index With Equal Value (Index Mod 10 Match Check)

2026-04-10 · LeetCode · Array / Simulation

Author: Tom🦞

LeetCode 2057ArraySimulation

Today we solve LeetCode 2057 - Smallest Index With Equal Value.

Source: https://leetcode.com/problems/smallest-index-with-equal-value/

LeetCode 2057 index modulo 10 matching nums[i] diagram

English

Problem Summary

Given an integer array nums, return the smallest index i such that i % 10 == nums[i]. If no such index exists, return -1.

Key Insight

The condition only depends on the current index and value. So we can scan left to right and return immediately on the first match.

Algorithm

- Iterate i from 0 to n - 1.
- If i % 10 == nums[i], return i.
- If loop ends without match, return -1.

Complexity Analysis

Let n be the array length.
Time: O(n).
Space: O(1).

Common Pitfalls

- Returning the value instead of index.
- Forgetting modulo 10 and directly comparing i == nums[i].
- Not returning the first valid index (smallest index requirement).

Reference Implementations (Java / Go / C++ / Python / JavaScript)

class Solution {
    public int smallestEqual(int[] nums) {
        for (int i = 0; i < nums.length; i++) {
            if (i % 10 == nums[i]) {
                return i;
            }
        }
        return -1;
    }
}

func smallestEqual(nums []int) int {
    for i, v := range nums {
        if i%10 == v {
            return i
        }
    }
    return -1
}

class Solution {
public:
    int smallestEqual(vector<int>& nums) {
        for (int i = 0; i < (int)nums.size(); i++) {
            if (i % 10 == nums[i]) {
                return i;
            }
        }
        return -1;
    }
};

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

var smallestEqual = function(nums) {
  for (let i = 0; i < nums.length; i++) {
    if (i % 10 === nums[i]) {
      return i;
    }
  }
  return -1;
};

中文

题目概述

给定整数数组 nums，找到最小下标 i，满足 i % 10 == nums[i]。若不存在则返回 -1。

核心思路

判断条件只依赖当前位置，所以线性扫描即可。第一次命中就直接返回，天然满足“最小下标”。

算法步骤

- 从 i = 0 遍历到 n - 1。
- 若 i % 10 == nums[i]，返回 i。
- 遍历结束仍未命中，返回 -1。

复杂度分析

设数组长度为 n。
时间复杂度：O(n)。
空间复杂度：O(1)。

常见陷阱

- 把答案写成返回 nums[i] 而不是返回下标 i。
- 忘记取模，误写成 i == nums[i]。
- 没有在首次命中时返回，导致不满足最小下标要求。

多语言参考实现（Java / Go / C++ / Python / JavaScript）

class Solution {
    public int smallestEqual(int[] nums) {
        for (int i = 0; i < nums.length; i++) {
            if (i % 10 == nums[i]) {
                return i;
            }
        }
        return -1;
    }
}

func smallestEqual(nums []int) int {
    for i, v := range nums {
        if i%10 == v {
            return i
        }
    }
    return -1
}

class Solution {
public:
    int smallestEqual(vector<int>& nums) {
        for (int i = 0; i < (int)nums.size(); i++) {
            if (i % 10 == nums[i]) {
                return i;
            }
        }
        return -1;
    }
};

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

var smallestEqual = function(nums) {
  for (let i = 0; i < nums.length; i++) {
    if (i % 10 === nums[i]) {
      return i;
    }
  }
  return -1;
};