What is the brute force solution for Check if Number Has Equal Digit Count and Digit Value?

The brute force approach for Check if Number Has Equal Digit Count and Digit Value is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each digit's expected count by iterating through the string multiple times. It directly compares the expected frequency of each digit with its actual frequency, which is simple but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Check if Number Has Equal Digit Count and Digit Value?

Check if Number Has Equal Digit Count and Digit Value uses the following concepts: Hash Table, String, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Check if Number Has Equal Digit Count and Digit Value?

Always clarify the problem requirements and constraints before diving into coding. Think about the efficiency of your solution and whether it can be improved from a brute-force approach. Practice dry-running your code with sample inputs to catch logical errors before submitting.

What are common mistakes when solving Check if Number Has Equal Digit Count and Digit Value?

Forgetting to convert characters to integers when comparing counts. Not considering edge cases such as leading zeros or digits that exceed the string length.

#2283

Check if Number Has Equal Digit Count and Digit Value

Easy

Hash TableStringCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal approach uses a single pass to count the occurrences of each digit in the string and then checks if the counts match the expected values. This is efficient because it reduces the number of iterations needed.

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Algorithm

4 steps

1Step 1: Create an array of size 10 (for digits 0-9) to count occurrences of each digit.
2Step 2: Iterate through the string and populate the count array based on the digits found.
3Step 3: Iterate through the string again, checking if the count of each digit matches the expected value from the string.
4Step 4: If all checks pass, return true; otherwise, return false.

solution.py8 lines

1def digitCount(num):
2    count = [0] * 10
3    for digit in num:
4        count[int(digit)] += 1
5    for i in range(len(num)):
6        if count[i] != int(num[i]):
7            return False
8    return True

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Complexity note: The time complexity is O(n) because we make a single pass to count the digits and another pass to verify the counts, resulting in linear time. The space complexity is O(1) since the count array is fixed in size (10).

1The digit at each index directly represents the expected count of that index's digit in the string.
2Using a counting array allows for efficient frequency tracking without repeated string scans.

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