LeetCode in Kotlin
3550. Smallest Index With Digit Sum Equal to Index
Easy
You are given an integer array nums.
Return the smallest index i such that the sum of the digits of nums[i] is equal to i.
If no such index exists, return -1.
Example 1:
Input: nums = [1,3,2]
Output: 2
Explanation:
- For
nums[2] = 2, the sum of digits is 2, which is equal to indexi = 2. Thus, the output is 2.
Example 2:
Input: nums = [1,10,11]
Output: 1
Explanation:
- For
nums[1] = 10, the sum of digits is1 + 0 = 1, which is equal to indexi = 1. - For
nums[2] = 11, the sum of digits is1 + 1 = 2, which is equal to indexi = 2. - Since index 1 is the smallest, the output is 1.
Example 3:
Input: nums = [1,2,3]
Output: -1
Explanation:
- Since no index satisfies the condition, the output is -1.
Constraints:
1 <= nums.length <= 1000 <= nums[i] <= 1000
Solution
class Solution {
private fun sum(num: Int): Int {
var num = num
var s = 0
while (num > 0) {
s += num % 10
num /= 10
}
return s
}
fun smallestIndex(nums: IntArray): Int {
for (i in nums.indices) {
if (i == sum(nums[i])) {
return i
}
}
return -1
}
}