LeetCode in Kotlin
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].
Constraints:
1 <= nums.length <= 1000 <= nums[i] <= 9
Solution
class Solution {
fun smallestEqual(nums: IntArray): Int {
for (i in nums.indices) {
if (i % 10 == nums[i]) {
return i
}
}
return -1
}
}