Count of numbers with all digits same in a given range

Given two integers L and R denoting the starting and end values of a range, the task is to count all numbers in that range whose all digit are same, like 1, 22, 444, 3333, etc.

Example:

Input: L = 12, R = 68
Output: 5
Explanation:
{ 22, 33, 44, 55, 66} are the numbers with same digits in the given range.

Input: L = 1, R = 32
Output: 11

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach: Iterate through all the numbers from L to R and for each number, check if it has all its digits same. If yes, then increase the required count. Print this count at the end.

Efficient Approach: The idea is based on the fact that the multiples (1 to 9) of 1, 11, 111, etc has all its digits same.
For example:

1 times 1 = 1 (All digits are same)
2 times 1 = 2 (All digits are same)
3 times 1 = 3 (All digits are same)
.
.
9 times 1 = 9 (All digits are same)

Similarly
1 times 11 = 11 (All digits are same)
2 times 11 = 22 (All digits are same)
3 times 11 = 33 (All digits are same)
.
.
9 times 11 = 99 (All digits are same)

Same is the case for 111, 1111, etc.

Therefore, the steps can be defined as:

Find the number of digits in R. This will decide the length of consecutive 1s to be created, till which we have to check.
For example, if R = 100, then length(R) = 3. Therefore we need to check only the multiples of 1, 11, and 111.
For each length of consecutive 1s from 1 to length(R):
- Multiply them with all values from 2 to 9
- Check if it lies within the range [L, R] or not.
- If yes, then increment the count of required numbers.
Print the required count of numbers.

Below code is the implementation of the above approach:

C++

// C++ program to count the 
// total numbers in the range 
// L and R which have all the 
// digit same 
  
#include <bits/stdc++.h> 
using namespace std; 
  
// Function that count the 
// total numbersProgram between L 
// and R which have all the 
// digit same 
int count_same_digit(int L, int R) 
{ 
    int tmp = 0, ans = 0; 
  
    // length of R 
    int n = log10(R) + 1; 
  
    for (int i = 0; i < n; i++) { 
  
        // tmp has all digits as 1 
        tmp = tmp * 10 + 1; 
  
        // For each multiple 
        // of tmp in range 1 to 9, 
        // check if it present 
        // in range [L, R] 
        for (int j = 1; j <= 9; j++) { 
  
            if (L <= (tmp * j) 
                && (tmp * j) <= R) { 
  
                // Increment the required count 
                ans++; 
            } 
        } 
    } 
    return ans; 
} 
  
// Driver Program 
int main() 
{ 
    int L = 12, R = 68; 
  
    cout << count_same_digit(L, R) 
         << endl; 
    return 0; 
} 

Java

// Java program to count the 
// total numbers in the range 
// L and R which have all the 
// digit same 
import java.util.*; 
  
class GFG{ 
  
// Function that count the total  
// numbersProgram between L and  
// R which have all the digit same 
static int count_same_digit(int L, int R) 
{ 
    int tmp = 0, ans = 0; 
  
    // Length of R 
    int n = (int)Math.log10(R) + 1; 
  
    for(int i = 0; i < n; i++) 
    { 
         
       // tmp has all digits as 1 
       tmp = tmp * 10 + 1; 
         
       // For each multiple of tmp  
       // in range 1 to 9, check if 
       // it present in range [L, R] 
       for(int j = 1; j <= 9; j++) 
       { 
          if (L <= (tmp * j) && (tmp * j) <= R)  
          { 
              // Increment the required count 
              ans++; 
          } 
       } 
    } 
    return ans; 
} 
  
// Driver code 
public static void main(String[] args) 
{ 
    int L = 12, R = 68; 
  
    System.out.println(count_same_digit(L, R)); 
} 
} 
  
// This code is contributed by offbeat 

Python3

# Python3 program to count the 
# total numbers in the range 
# L and R which have all the 
# digit same 
import math 
  
# Function that count the 
# total numbersProgram between L 
# and R which have all the 
# digit same 
def count_same_digit(L, R): 
  
    tmp = 0; ans = 0; 
  
    # length of R 
    n = int(math.log10(R) + 1); 
  
    for i in range(0, n): 
  
        # tmp has all digits as 1 
        tmp = tmp * 10 + 1; 
  
        # For each multiple 
        # of tmp in range 1 to 9, 
        # check if it present 
        # in range [L, R] 
        for j in range(1, 9): 
  
            if (L <= (tmp * j) and (tmp * j) <= R): 
  
                # Increment the required count 
                ans += 1; 
              
    return ans; 
  
# Driver Code 
L = 12; R = 68; 
  
print(count_same_digit(L, R)) 
  
# This code is contributed by Nidhi_biet 

C#

// C# program to count the 
// total numbers in the range 
// L and R which have all the 
// digit same 
using System; 
  
class GFG{ 
  
// Function that count the total  
// numbersProgram between L and  
// R which have all the digit same 
static int count_same_digit(int L, int R) 
{ 
    int tmp = 0, ans = 0; 
  
    // Length of R 
    int n = (int)Math.Log10(R) + 1; 
  
    for(int i = 0; i < n; i++) 
    { 
          
        // tmp has all digits as 1 
        tmp = tmp * 10 + 1; 
              
        // For each multiple of tmp  
        // in range 1 to 9, check if 
        // it present in range [L, R] 
        for(int j = 1; j <= 9; j++) 
        { 
            if (L <= (tmp * j) && (tmp * j) <= R)  
            { 
                // Increment the required count 
                ans++; 
            } 
        } 
    } 
    return ans; 
} 
  
// Driver code 
public static void Main() 
{ 
    int L = 12, R = 68; 
  
    Console.Write(count_same_digit(L, R)); 
} 
} 
  
// This code is contributed by Code_Mech 

Output:

Time Complexity: O(length(R))

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

Recommended Posts: