# 2057. Smallest Index With Equal Value

## 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

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

• 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;
}
};

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


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

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