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.

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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].


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Constraints:


	1 &lt;= nums.length &lt;= 100
	0 &lt;= nums[i] &lt;= 9


class Solution:
    def smallestEqual(self, nums: List[int]) -> int:
        for idx, n in enumerate(nums):
            if idx%10==n:
                return idx
        return -1 
class Solution {
    public int smallestEqual(int[] nums) {
        int index = 0;
        for (int i = 0 ; i < nums.length; i++) {
            if (index == nums[i]) {
                return i;
            }
            if (++index== 10) {
               index = 0;
            }
        }
        return -1;
    }
}
class Solution {
public:
    int smallestEqual(vector<int>& nums) {
        int i = 0;
        while (i < nums.size() && i % 10 != nums[i]) i++;
        return i >= nums.size() ? -1 : i;
    }
};
/**
 * @param {number[]} nums
 * @return {number}
 */
var smallestEqual = function(nums) {
    return nums.findIndex((n, i) => i % 10 === n)
};
Smallest Index With Equal Value