LeetCode-in-Java

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.length
• 1 <= n <= 10
• num consists of digits.

Solution

public class Solution {
    public boolean digitCount(String num) {
        int[] cnt = new int[11];
        char[] arr = num.toCharArray();
        for (char d : arr) {
            ++cnt[d - '0'];
        }
        for (int i = 0; i < arr.length; i++) {
            if (cnt[i] != arr[i] - '0') {
                return false;
            }
        }
        return true;
    }
}

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