2057

2057. Smallest Index With Equal Value (Easy)

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].

Example 4:

Input: nums = [2,1,3,5,2]
Output: 1
Explanation: 1 is the only index with i mod 10 == nums[i].

 

Constraints:

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

Solution 1.

// OJ: https://leetcode.com/problems/smallest-index-with-equal-value/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int smallestEqual(vector<int>& A) {
        for (int i = 0; i < A.size(); ++i) {
            if (i % 10 == A[i]) return i;
        }
        return -1;
    }
};