Digit Frequency Histogram - Problem

Given a positive integer, count the frequency of each digit (0-9) and display a horizontal histogram using '*' characters.

For each digit that appears in the number, print the digit followed by a colon, then print '*' characters equal to the frequency of that digit.

Rules:

Only display digits that actually appear in the number
Display digits in ascending order (0, 1, 2, ..., 9)
Each histogram bar should be on a separate line
Format: digit: **** (digit, colon, space, then stars)

Input & Output

Example 1 — Basic Case

$ Input: number = 1223

› Output: 1: * 2: ** 3: *

💡 Note: Digit 1 appears 1 time, digit 2 appears 2 times, digit 3 appears 1 time. Display in ascending order with histogram bars.

Example 2 — Repeated Digits

$ Input: number = 112233

› Output: 1: ** 2: ** 3: **

💡 Note: Each digit 1, 2, and 3 appears exactly 2 times, so each gets a 2-star histogram bar.

Example 3 — Single Digit

$ Input: number = 7

› Output: 7: *

💡 Note: Only digit 7 appears once, so output shows just '7: *'

Constraints

1 ≤ number ≤ 10¹⁸
Number is always positive

Visualization

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Asked in

a Amazon 15 G Google 12

The key insight is to count digit frequencies in a single pass using an array, then generate the histogram. The optimal approach uses frequency counting with Time: O(n), Space: O(1) where n is the number of digits.

Common Approaches

✓ Brute Force - Check Each Digit Separately

⏱️ Time: O(n × 10) = O(n) Space: O(1)

Convert number to string, then for each possible digit (0-9), iterate through the entire string counting how many times that digit appears. Only display digits with non-zero counts.

Single Pass Frequency Counting

⏱️ Time: O(n) Space: O(1)

Convert number to string and make a single pass through it. For each character, increment the count in a frequency array. Then iterate through the array to build histogram output.

Brute Force - Check Each Digit Separately — Algorithm Steps

Convert number to string
For each digit 0-9, count occurrences in string
If count > 0, print digit with histogram bar
Repeat for all 10 digits

Visualization

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Step-by-Step Walkthrough

Convert to String

Convert 1223 to "1223"

Check Each Digit

For digit 0: scan "1223" → count=0, skip

Found Matches

For digit 1: count=1 → "1: *", digit 2: count=2 → "2: **"

Code -

solution.c — C

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

void solution(int number) {
    char numStr[20];
    sprintf(numStr, "%d", number);
    int len = strlen(numStr);
    int first = 1;
    
    for (int digit = 0; digit <= 9; digit++) {
        int count = 0;
        for (int i = 0; i < len; i++) {
            if ((numStr[i] - '0') == digit) {
                count++;
            }
        }
        
        if (count > 0) {
            if (!first) printf("\n");
            printf("%d: ", digit);
            for (int i = 0; i < count; i++) {
                printf("*");
            }
            first = 0;
        }
    }
}

int main() {
    int number;
    scanf("%d", &number);
    solution(number);
    printf("\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × 10) = O(n)

For each of 10 digits, scan through n-digit number

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables, no extra data structures

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Digit Frequency Histogram - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check Each Digit Separately — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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