- Difficulty: Easy
- Tags: LeetCode, Easy, Hash Table, String, Counting, leetcode-2283, O(n), O(1), Freq Table
Problem
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.length
1 <= n <= 10
num
consists of digits.