README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/2000-2099/2057.Smallest%20Index%20With%20Equal%20Value/README_EN.md	1167	Weekly Contest 265 Q1	Array

2057. Smallest Index With Equal Value

中文文档

Description

Given a 0-indexed integer array nums, return the smallest index i of nums such that i mod 10 == nums[i], or -1 if such index does not exist.

x mod y denotes the remainder when x is divided by y.

 

Example 1:

Input: nums = [0,1,2]
Output: 0
Explanation: 
i=0: 0 mod 10 = 0 == nums[0].
i=1: 1 mod 10 = 1 == nums[1].
i=2: 2 mod 10 = 2 == nums[2].
All indices have i mod 10 == nums[i], so we return the smallest index 0.

Example 2:

Input: nums = [4,3,2,1]
Output: 2
Explanation: 
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 is the only index which has i mod 10 == nums[i].

Example 3:

Input: nums = [1,2,3,4,5,6,7,8,9,0]
Output: -1
Explanation: No index satisfies i mod 10 == nums[i].

 

Constraints:

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

Solutions

Solution 1

Python3

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

Java

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

C++

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

Go

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

TypeScript

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