Number of Beautiful Integers in the Range - Problem

You are given positive integers low, high, and k.

A number is beautiful if it meets both of the following conditions:

The count of even digits in the number is equal to the count of odd digits.
The number is divisible by k.

Return the number of beautiful integers in the range [low, high].

Input & Output

Example 1 — Basic Range

$ Input: low = 10, high = 20, k = 6

› Output: 2

💡 Note: Numbers 12 (even=1, odd=1, 12%6=0) and 18 (even=1, odd=1, 18%6=0) are beautiful

Example 2 — Larger k

$ Input: low = 1, high = 10, k = 1

› Output: 1

💡 Note: Only number with equal even/odd digits in range [1,10] is none for single digits, but all are divisible by 1

Example 3 — No Valid Numbers

$ Input: low = 5, high = 5, k = 1

› Output: 0

💡 Note: Single digit 5 has 0 even and 1 odd digit, so counts are not equal

Constraints

1 ≤ low ≤ high ≤ 10⁹
1 ≤ k ≤ 20

Visualization

Tap to expand

Asked in

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

The key insight is using digit DP to efficiently count valid numbers without generating them. We track digit balance (even-odd count) and remainder modulo k simultaneously while building numbers digit by digit. Best approach is digit DP with memoization. Time: O(d × b × k), Space: O(d × b × k)

Common Approaches

✓ Brute Force - Check Every Number

⏱️ Time: O(n × d) Space: O(1)

For each number from low to high, count even and odd digits, then check if counts are equal and number is divisible by k.

Digit DP with Memoization

⏱️ Time: O(d × b × k) Space: O(d × b × k)

Build numbers digit by digit using dynamic programming, tracking digit balance and remainder modulo k simultaneously. Cache results for overlapping subproblems.

Brute Force - Check Every Number — Algorithm Steps

Iterate from low to high
For each number, count even/odd digits
Check if counts equal and divisible by k
Count valid numbers

Visualization

Tap to expand

Step-by-Step Walkthrough

Iterate Range

Go through each number from low to high

Count Digits

Count even and odd digits in each number

Check Conditions

Verify equal counts and divisibility by k

Code -

solution.c — C

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

int isBeautiful(int num, int k) {
    char s[20];
    sprintf(s, "%d", num);
    int evenCount = 0;
    int len = strlen(s);
    
    for (int i = 0; i < len; i++) {
        int digit = s[i] - '0';
        if (digit % 2 == 0) {
            evenCount++;
        }
    }
    
    int oddCount = len - evenCount;
    return evenCount == oddCount && num % k == 0;
}

int solution(int low, int high, int k) {
    int count = 0;
    for (int num = low; num <= high; num++) {
        if (isBeautiful(num, k)) {
            count++;
        }
    }
    return count;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n × d)

n numbers in range, d digits per number to count

✓ Linear Growth

Space Complexity

O(1)

Only variables for counting

✓ Linear Space

23.4K Views

Medium Frequency

~45 min Avg. Time

892 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Number of Beautiful Integers in the Range - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check Every Number — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler