Check if Number Has Equal Digit Count and Digit Value Python/Java solution

In Python or Java, iterate through the string and increment a frequency array indexed by digit value. After counting, loop through the string again and compare num[i] with freq[i]. If any mismatch occurs, return false; otherwise return true after the loop finishes.

How to solve Check if Number Has Equal Digit Count and Digit Value in O(n)?

Traverse the string once to build a frequency array for digits 0–9. Then iterate through each index i and check whether the integer value of num[i] equals the frequency of digit i. If every index matches, return true; otherwise return false. The algorithm performs two linear passes, giving O(n) time.

What is the best approach for Check if Number Has Equal Digit Count and Digit Value?

The most efficient approach uses a frequency array to count how often each digit appears. First iterate through the string and record digit frequencies, then compare each index i with the stored count of digit i. This runs in O(n) time with O(1) space because the digit range is fixed (0–9).

Is Check if Number Has Equal Digit Count and Digit Value asked at Google/Amazon/Meta?

Problems like this commonly appear in coding interviews focused on string processing and counting logic. While not always asked under the exact problem name, similar frequency validation questions appear in screening rounds at companies such as Amazon and Google.

What data structure is used in Check if Number Has Equal Digit Count and Digit Value?

The typical solution uses a fixed-size array (or occasionally a hash map) to store digit frequencies. Because digits range from 0 to 9, an integer array of size 10 is sufficient and provides constant-time updates and lookups.

What is the time complexity of Check if Number Has Equal Digit Count and Digit Value?

The optimal solution runs in O(n) time where n is the length of the numeric string. One pass counts digit frequencies and another pass verifies each index against the frequency array. Space complexity is O(1) because the counting structure has a constant size of 10.

Description Hints 1 Solution Video 4:53

2283. Check if Number Has Equal Digit Count and Digit Value

Q: Is Check if Number Has Equal Digit Count and Digit Value easy or hard?

The problem is classified as Easy. The main concept is digit frequency counting and index validation, which can be implemented with simple loops and a small array while maintaining O(n) time complexity.

Easy73.1% acceptance2 companies

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: