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

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Check if Number Has Equal Digit Count and Digit Value?

This problem uses Hash Table, String, Counting. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Check if Number Has Equal Digit Count and Digit Value?

Insight 1: The problem requires iterating through the string twice. The outer loop iterates through each digit's position, and the inner loop counts the occurrences of that position's index. Insight 2: A simple counting approach is sufficient. No sophisticated data structures are needed. Directly comparing the count of each digit with the digit's value at the corresponding index works effectively.

Which companies ask LeetCode 2283. Check if Number Has Equal Digit Count and Digit Value in interviews?

This problem is frequently asked at J.P. Morgan, PornHub.

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Check if Number Has Equal Digit Count and Digit Value - LeetCode 2283 Solution

Q: How to solve Check if Number Has Equal Digit Count and Digit Value?

The provided code uses a brute-force approach. It iterates through each digit of the input string `num`. For each digit at index `i`, it counts how many times the digit `i` appears in the entire string. Then, it compares this count with the numerical value of the digit at index `i`. If they are different at any point, it means the condition is violated, and the function immediately returns `false`. If the loop completes without finding any discrepancies, it means the condition holds for all digits, and the function returns `true`.

Check if Number Has Equal Digit Count and Digit Value - Complete Solution Guide

Check if Number Has Equal Digit Count and Digit Value is LeetCode problem 2283, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 r

Detailed Explanation

The problem asks you to check if a given string `num` representing a number satisfies a specific condition: for each digit at index `i`, the digit `i` itself must appear in `num` exactly as many times as the value of the digit at index `i`. For example, in "1210", the digit at index 0 is '1', and the digit '0' appears once in the string. The digit at index 1 is '2', and the digit '1' appears twice. This pattern must hold true for all indices. The input is a string of digits, and the output is a boolean indicating whether this condition is met.

Solution Approach

The provided code uses a brute-force approach. It iterates through each digit of the input string `num`. For each digit at index `i`, it counts how many times the digit `i` appears in the entire string. Then, it compares this count with the numerical value of the digit at index `i`. If they are different at any point, it means the condition is violated, and the function immediately returns `false`. If the loop completes without finding any discrepancies, it means the condition holds for all digits, and the function returns `true`.

Step-by-Step Algorithm

Step 1: Iterate through each digit of the input string `num` using a loop (outer loop).
Step 2: For each digit at index `i`, initialize a counter `count` to 0.
Step 3: Iterate through the string `num` again (inner loop) to count the occurrences of the digit `i`.
Step 4: Compare the `count` with the numerical value of the digit at index `i`. If they are not equal, return `false`.
Step 5: If the outer loop completes without returning `false`, return `true`.

Key Insights

Insight 1: The problem requires iterating through the string twice. The outer loop iterates through each digit's position, and the inner loop counts the occurrences of that position's index.
Insight 2: A simple counting approach is sufficient. No sophisticated data structures are needed. Directly comparing the count of each digit with the digit's value at the corresponding index works effectively.
Insight 3: The constraints (length of the string <= 10) significantly limit the input size. This means the O(n^2) solution is acceptable, as optimization might not be necessary for such small inputs.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(1)

Topics

This problem involves: Hash Table, String, Counting.

Companies

Asked at: J.P. Morgan, PornHub.

Check if Number Has Equal Digit Count and Digit Value | LeetCode 2283

Easy

Hash Table String Counting

Time: O(n^2)Space: O(1)

Solution Code

Available in:

class Solution {
    public boolean digitCount(String num) {
        int n = num.length();
        for (int i = 0; i < n; i++) {
            int count = 0;
            for (int j = 0; j < n; j++) {
                if (Character.getNumericValue(num.charAt(j)) == i) {
                    count++;
                }
            }
            if (count != Character.getNumericValue(num.charAt(i))) {
                return false;
            }
        }
        return true;
    }
}

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Problem Info

Description

Problem 2283: Check if Number Has Equal Digit Count and Digit Value - Detailed explanation and solution approaches.

Topics

Hash Table String Counting

Companies

J.P. MorganPornHub

Stats

Difficulty:Easy

Languages:4

Companies:2

Problem #:2283

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

5 steps

1
Step 1: Iterate through each digit of the input string `num` using a loop (outer loop).
2
Step 2: For each digit at index `i`, initialize a counter `count` to 0.
3
Step 3: Iterate through the string `num` again (inner loop) to count the occurrences of the digit `i`.
4
Step 4: Compare the `count` with the numerical value of the digit at index `i`. If they are not equal, return `false`.
5
Step 5: If the outer loop completes without returning `false`, return `true`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The outer loop iterates `n` times, and for each iteration, the inner loop iterates `n` times. This leads to a nested loop structure with O(n^2) time complexity. The space complexity is O(1) because only a few integer variables are used, regardless of the input size.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using a Frequency Array

Create an array to store the frequency of each digit (0-9). Then iterate through the input string, updating the frequencies. Finally, iterate through the string again and check if the frequency of digit `i` matches the digit at index `i`.

Time: O(n)Space: O(1)

Trade-offs: This approach reduces the time complexity to O(n) compared to the O(n^2) solution, making it more efficient for larger inputs. However, the space complexity remains constant because the frequency array size is always 10.

✨

Key Insights

1Insight 1: The problem requires iterating through the string twice. The outer loop iterates through each digit's position, and the inner loop counts the occurrences of that position's index.
2Insight 2: A simple counting approach is sufficient. No sophisticated data structures are needed. Directly comparing the count of each digit with the digit's value at the corresponding index works effectively.
3Insight 3: The constraints (length of the string <= 10) significantly limit the input size. This means the O(n^2) solution is acceptable, as optimization might not be necessary for such small inputs.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: The input string contains leading zeros. The solution correctly handles this because it converts each character to its numeric value.
Edge case 2: The input string contains only one digit. The solution will correctly check if that digit is equal to its index (which is 0).

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling the conversion of characters to their integer values. Failing to account for the ASCII difference between '0' and the digits 0-9 can lead to inaccurate counting.

✗Common mistake 2: Off-by-one errors in the loop indices. Incorrect loop boundaries can cause the program to miss checking certain digits or produce incorrect counts.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Data validation checks. This type of algorithm could be used to verify the consistency of data entered into a system, for example, in a form where the user provides the count of items and the system validates it against actual data.

🏦Real-world scenario 2: Code generation or pattern matching. Certain types of code or data might have a self-referential nature where the position of an element is related to its value. This algorithm could check the validity of such patterns.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 2, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 29

❓

Frequently Asked Questions

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

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

For this easy-level Hash Table problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?