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

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:

1. 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.
2. 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.
3. 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:

   5
   

   Time Complexity: O(length(R))

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