LeetCode in Kotlin
2283. Check if Number Has Equal Digit Count and Digit Value
Easy
You are given a 0-indexed string num of length n consisting of digits.
Return true if for every index i in the range 0 <= i < n, the digit i occurs num[i] times in num, otherwise return false.
Example 1:
Input: num = “1210”
Output: true
Explanation:
num[0] = ‘1’. The digit 0 occurs once in num.
num[1] = ‘2’. The digit 1 occurs twice in num.
num[2] = ‘1’. The digit 2 occurs once in num.
num[3] = ‘0’. The digit 3 occurs zero times in num.
The condition holds true for every index in “1210”, so return true.
Example 2:
Input: num = “030”
Output: false
Explanation:
num[0] = ‘0’. The digit 0 should occur zero times, but actually occurs twice in num.
num[1] = ‘3’. The digit 1 should occur three times, but actually occurs zero times in num.
num[2] = ‘0’. The digit 2 occurs zero times in num.
The indices 0 and 1 both violate the condition, so return false.
Constraints:
n == num.length1 <= n <= 10numconsists of digits.
Solution
class Solution {
fun digitCount(num: String): Boolean {
val cnt = IntArray(11)
val arr = num.toCharArray()
for (d in arr) {
++cnt[d.code - '0'.code]
}
for (i in arr.indices) {
if (cnt[i] != arr[i].code - '0'.code) {
return false
}
}
return true
}
}