Check if Number Has Equal Digit Count and Digit Value

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

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: If num = "1210", then:

Digit 0 should appear num[0] = 1 time (it appears 1 time ✓)
Digit 1 should appear num[1] = 2 times (it appears 2 times ✓)
Digit 2 should appear num[2] = 1 time (it appears 1 time ✓)
Digit 3 should appear num[3] = 0 times (it appears 0 times ✓)

All conditions are satisfied, so return true.

Input & Output

Example 1 — Valid Self-Describing Number

$ Input: num = "1210"

› Output: true

💡 Note: Digit 0 appears 1 time (num[0]=1), digit 1 appears 2 times (num[1]=2), digit 2 appears 1 time (num[2]=1), digit 3 appears 0 times (num[3]=0). All conditions satisfied.

Example 2 — Invalid Count

$ Input: num = "030"

› Output: false

💡 Note: Digit 0 should appear 0 times (num[0]=0), but it appears 2 times in the string. This violates the condition.

Example 3 — Single Digit

$ Input: num = "0"

› Output: false

💡 Note: Digit 0 should appear 0 times (num[0]=0), but it actually appears 1 time in the string. This violates the condition.

Constraints

1 ≤ num.length ≤ 10
num consists of digits only

Visualization

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Asked in

G Google 15 a Amazon 12 M Meta 8

The key insight is to verify the self-describing property by counting digit frequencies and checking if each position i has the correct count for digit i. Best approach uses frequency counting with two passes through the string. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force - Check Each Digit Individually

⏱️ Time: O(n²) Space: O(1)

For each index i from 0 to n-1, scan the entire string to count how many times digit i appears, then check if this count equals the digit at position i.

Frequency Count - One Pass Optimization

⏱️ Time: O(n) Space: O(1)

First pass through the string to count frequency of each digit (0-9), then second pass to check if each position i has the correct count for digit i.

Brute Force - Check Each Digit Individually — Algorithm Steps

Step 1: For each index i, convert i to character and count its occurrences in the string
Step 2: Compare the count with the digit value at position i
Step 3: Return false if any mismatch found, true otherwise

Visualization

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Step-by-Step Walkthrough

Check Position 0

Count digit '0' in entire string

Check Position 1

Count digit '1' in entire string

Verify All

Each count matches the digit at that position

Code -

solution.c — C

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

bool solution(char* num) {
    int n = strlen(num);
    for (int i = 0; i < n; i++) {
        char digitToFind = '0' + i;
        int count = 0;
        for (int j = 0; j < n; j++) {
            if (num[j] == digitToFind) {
                count++;
            }
        }
        int expected = num[i] - '0';
        if (count != expected) {
            return false;
        }
    }
    return true;
}

int main() {
    char num[1001];
    fgets(num, sizeof(num), stdin);
    num[strcspn(num, "\n")] = 0;
    bool result = solution(num);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

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

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables, no extra data structures

✓ Linear Space

23.0K Views

Medium Frequency

~8 min Avg. Time

845 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Check Each Digit Individually — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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