Introduction

In this blog post, we will explore how to solve the LeetCode problem "2283" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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.

Explanation

• Frequency Counting: Iterate through the input string num and count the occurrences of each digit from 0 to 9.
  • Verification: Compare the count of each digit i with the digit at index i in the input string. If any mismatch is found, return False. Otherwise, return True.

• Time Complexity: O(n), where n is the length of the input string. Space Complexity: O(1), as we use a fixed-size array to store the frequencies (at most 10 elements).

Code

    def check_if_frequency_is_equal(num: str) -> bool:
    """
    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.

    For example:
    check_if_frequency_is_equal("1210") == True
    check_if_frequency_is_equal("030") == False
    """
    n = len(num)
    counts = [0] * 10
    for digit in num:
        counts[int(digit)] += 1

    for i in range(n):
        if counts[i] != int(num[i]):
            return False
    return True