# 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:

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.
1. 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 ` `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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Improved By : offbeat, Code_Mech, nidhi_biet

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