Check if Number Has Equal Digit Count and Digit Value

Check if Number Has Equal Digit Count and Digit Value - Problem

Imagine you have a special number string where each digit has a secret rule to follow! 🔢✨

You are given a 0-indexed string num of length n consisting only of digits. Your task is to check if this string follows a very specific pattern:

For every position i (from 0 to n-1), the digit at that position should appear in the entire string exactly as many times as its value indicates.

In other words, if num[i] = '3', then the digit '3' should appear exactly 3 times in the entire string.

Return true if the string follows this magical property for all positions, otherwise return false.

Example: In string "1210":
• Position 0 has digit '1', and '1' appears 1 time ✅
• Position 1 has digit '2', and '2' appears 2 times ❌ (actually appears 1 time)
• This would return false

Input & Output

example_1.py — Python

$ Input: num = "1210"

› Output: false

💡 Note: At position 0, digit is '1', but '1' appears 2 times in the string (positions 0 and 2). Since 2 ≠ 1, the condition fails.

example_2.py — Python

$ Input: num = "030"

› Output: true

💡 Note: Position 0: digit '0' appears 1 time ✅. Position 1: digit '3' appears 1 time, but should appear 3 times ❌. Wait, let me recheck: '0' appears 1 time, '3' appears 1 time. Actually this should be false! Let me correct: Position 0 needs '0' to appear 0 times, but it appears 1 time.

example_3.py — Python

$ Input: num = "1200"

› Output: true

💡 Note: Position 0: '1' appears 1 time ✅. Position 1: '2' appears 1 time but should appear 2 times ❌. Actually: '1' appears 1 time, '2' appears 1 time, '0' appears 2 times. Position 0: 1=1 ✅, Position 1: 1≠2 ❌. This should be false.

Visualization

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Understanding the Visualization

Scan the Badge

Count how many times each digit appears: '0':1, '1':2, '2':1

Check Position Rules

Position 0 has '1' - does '1' appear 1 time? No, it appears 2 times!

Rule Violation

Badge is invalid because the rule at position 0 is violated

Key Takeaway

🎯 Key Insight: Instead of counting the same digits multiple times (brute force), count each digit once and store in a hash table, then validate each position's rule against the pre-computed counts.

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass to count frequencies plus single pass to verify, both O(n)

✓ Linear Growth

Space Complexity

O(1)

Hash map stores at most 10 digits (0-9), so constant space

✓ Linear Space

Constraints

1 ≤ num.length ≤ 10
num consists of only digits.
Each digit in num is between 0 and 9

Asked in

G Google 15 a Amazon 12 ⊞ Microsoft 8 f Meta 6

The optimal solution uses a hash table to count digit frequencies in O(n) time. First, count how many times each digit appears in the string. Then, check if each position's digit appears exactly as many times as its numeric value indicates. This avoids the O(n²) complexity of counting digits repeatedly for each position.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Nested Loops)	O(n²)	O(1)	For each position, count occurrences of that digit in the entire string
Hash Table Frequency Count (Optimal)	O(n)	O(1)	Count all digit frequencies first, then verify each position

Brute Force (Nested Loops) — Algorithm Steps

Iterate through each index i from 0 to n-1
For each position i, get the digit at that position
Count how many times this digit appears in the entire string
Check if the count equals the digit value
If any position fails, return false; otherwise return true

Visualization

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Step-by-Step Walkthrough

Start at position 0

Check digit '1' and count its occurrences

Scan entire string

Count how many times '1' appears (found 1 time)

Compare with digit value

1 occurrence matches digit value 1 ✅

Move to next position

Repeat for position 1 with digit '2'

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <stdbool.h>

bool digitCount(char* num) {
    int n = strlen(num);
    
    // Check each position
    for (int i = 0; i < n; i++) {
        char digit = num[i];
        int count = 0;
        
        // Count occurrences of this digit in entire string
        for (int j = 0; j < n; j++) {
            if (num[j] == digit) {
                count++;
            }
        }
        
        // Check if count matches the digit value
        if (count != (digit - '0')) {
            return false;
        }
    }
    
    return true;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of the n positions, we scan the entire string of length n to count occurrences

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a constant amount of extra space for counting variables

✓ Linear Space

Constraints

1 ≤ num.length ≤ 10
num consists of only digits.
Each digit in num is between 0 and 9

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Nested Loops) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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