Count Numbers With Unique Digits II - Practice Coding Problems

Count Numbers With Unique Digits II - Problem

Given two positive integers a and b, return the count of numbers having unique digits in the range [a, b] (inclusive).

A number has unique digits if no digit appears more than once in the number. For example, 123 has unique digits, but 121 does not because the digit 1 appears twice.

Input & Output

Example 1 — Small Range

$ Input: a = 20, b = 25

› Output: 5

💡 Note: Numbers 20, 21, 23, 24, 25 all have unique digits. Number 22 has repeated digit '2', so it's not counted.

Example 2 — Single Digits

$ Input: a = 1, b = 10

› Output: 10

💡 Note: All single digit numbers (1-9) and 10 have unique digits. Total count is 10.

Example 3 — Range with Repeats

$ Input: a = 80, b = 120

› Output: 27

💡 Note: Count numbers from 80-120 where all digits are unique. Excludes numbers like 88, 99, 100, 101, 110, 111, etc.

Constraints

1 ≤ a ≤ b ≤ 10⁸
Both a and b are positive integers

Visualization

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Understanding the Visualization

Input Range

Given range [20, 25] to check

Check Uniqueness

Test each number for repeated digits

Count Valid

Return count of numbers with unique digits

Key Takeaway

🎯 Key Insight: Use hash sets for O(1) digit duplicate detection instead of nested loops

Asked in

G Google 35 a Amazon 28 M Microsoft 22 f Facebook 18

The key insight is to efficiently check digit uniqueness using hash sets for O(1) lookup. For large ranges, mathematical approaches using combinatorics provide O(log n) time complexity. Best approach depends on range size: brute force for small ranges, mathematical for large ranges. Time: O(n×d) to O(log n), Space: O(1) to O(d).

Common Approaches

Approach	Time	Space	Notes
✓ Hash Map Optimization	O(n × d)	O(d)	Use hash set for faster digit uniqueness checking
Brute Force - Check Each Number	O(n × d)	O(d)	Iterate through range and check each number for unique digits
Mathematical Approach	O(log n)	O(1)	Calculate valid numbers using combinatorics and digit placement rules

Hash Map Optimization — Algorithm Steps

Iterate through range [a,b]
For each number, use hash set to track seen digits
Return early if duplicate found, otherwise increment count

Visualization

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Step-by-Step Walkthrough

Process Number

Take number 123 and check digits

Hash Set Check

Use set for O(1) duplicate detection

Result

All digits unique, count it

Code -

solution.c — C

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

int solution(int a, int b) {
    int count = 0;
    for (int num = a; num <= b; num++) {
        char numStr[20];
        sprintf(numStr, "%d", num);
        
        bool seenDigits[10] = {false};
        bool hasUnique = true;
        
        for (int i = 0; numStr[i] != '\0'; i++) {
            int digit = numStr[i] - '0';
            if (seenDigits[digit]) {
                hasUnique = false;
                break;
            }
            seenDigits[digit] = true;
        }
        
        if (hasUnique) {
            count++;
        }
    }
    return count;
}

int main() {
    int a, b;
    scanf("%d", &a);
    scanf("%d", &b);
    
    int result = solution(a, b);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × d)

n numbers in range, d average digits per number

✓ Linear Growth

Space Complexity

O(d)

Hash set stores up to d digits per number (max 10)

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Hash Map Optimization — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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