# Smallest integer greater than n such that it consists of digit m exactly k times

Given three integer n, m and k, the task is to find the smallest integer > n such that digit m appears exactly k times in it.

Examples:

Input: n = 111, m = 2, k = 2
Output: 122

Input: n = 111, m = 2, k = 3
Output: 222

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

Approach: Start iterating from n + 1 and for each integer i check whether it consists of digit m exactly k times. This way smallest integer > n with digit m occurring exactly k times can be found.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function that returns true if n contains  ` `// digit m exactly k times ` `bool` `digitWell(``int` `n, ``int` `m, ``int` `k) ` `{ ` `    ``int` `cnt = 0; ` `    ``while` `(n > 0) { ` `        ``if` `(n % 10 == m) ` `            ``++cnt; ` `        ``n /= 10; ` `    ``} ` `    ``return` `cnt == k; ` `} ` ` `  `// Function to return the smallest integer > n  ` `// with digit m occurring exactly k times ` `int` `findInt(``int` `n, ``int` `m, ``int` `k) ` `{ ` ` `  `    ``int` `i = n + 1; ` ` `  `    ``while` `(``true``) { ` `        ``if` `(digitWell(i, m, k)) ` `            ``return` `i; ` `        ``i++; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 111, m = 2, k = 2; ` `    ``cout << findInt(n, m, k); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` `class` `GFG ` `{ ` `     `  `// Function that returns true if n contains  ` `// digit m exactly k times ` `static` `boolean` `digitWell(``int` `n, ``int` `m, ``int` `k) ` `{ ` `    ``int` `cnt = ``0``; ` `    ``while` `(n > ``0``)  ` `    ``{ ` `        ``if` `(n % ``10` `== m) ` `            ``++cnt; ` `        ``n /= ``10``; ` `    ``} ` `    ``return` `cnt == k; ` `} ` ` `  `// Function to return the smallest integer > n  ` `// with digit m occurring exactly k times ` `static` `int` `findInt(``int` `n, ``int` `m, ``int` `k) ` `{ ` ` `  `    ``int` `i = n + ``1``; ` ` `  `    ``while` `(``true``)  ` `    ``{ ` `        ``if` `(digitWell(i, m, k)) ` `            ``return` `i; ` `        ``i++; ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `n = ``111``, m = ``2``, k = ``2``; ` `    ``System.out.println(findInt(n, m, k)); ` `} ` `}  ` ` `  `// This code is contributed by Code_Mech `

## Python3

# Python3 implementation of the approach

# Function that returns true if n
# contains digit m exactly k times
def digitWell(n, m, k):

cnt = 0
while (n > 0):

if (n % 10 == m):
cnt = cnt + 1;
n = (int)(n / 10);

return cnt == k;

# Function to return the smallest integer > n
# with digit m occurring exactly k times
def findInt(n, m, k):

i = n + 1;

while (True):
if (digitWell(i, m, k)):
return i;
i = i + 1;

# Driver code
n = 111; m = 2; k = 2;
print(findInt(n, m, k));

# This code is contributed
# by Akanksha Rai

## C#

 `// C# implementation of the approach  ` `using` `System; ` `class` `GFG ` `{ ` `     `  `// Function that returns true if n contains  ` `// digit m exactly k times ` `static` `bool` `digitWell(``int` `n, ``int` `m, ``int` `k) ` `{ ` `    ``int` `cnt = 0; ` `    ``while` `(n > 0)  ` `    ``{ ` `        ``if` `(n % 10 == m) ` `            ``++cnt; ` `        ``n /= 10; ` `    ``} ` `    ``return` `cnt == k; ` `} ` ` `  `// Function to return the smallest integer > n  ` `// with digit m occurring exactly k times ` `static` `int` `findInt(``int` `n, ``int` `m, ``int` `k) ` `{ ` ` `  `    ``int` `i = n + 1; ` ` `  `    ``while` `(``true``)  ` `    ``{ ` `        ``if` `(digitWell(i, m, k)) ` `            ``return` `i; ` `        ``i++; ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main() ` `{ ` `    ``int` `n = 111, m = 2, k = 2; ` `    ``Console.WriteLine(findInt(n, m, k)); ` `} ` `}  ` ` `  `// This code is contributed  ` `// by Akanksha Rai `

## PHP

 ` 0)  ` `    ``{  ` `        ``if` `(``\$n` `% 10 == ``\$m``)  ` `            ``++``\$cnt``;  ` `        ``\$n` `= ``floor``(``\$n` `/ 10);  ` `    ``}  ` `    ``return` `\$cnt` `== ``\$k``;  ` `}  ` ` `  `// Function to return the smallest integer > n  ` `// with digit m occurring exactly k times  ` `function` `findInt(``\$n``, ``\$m``, ``\$k``)  ` `{  ` `    ``\$i` `= ``\$n` `+ 1;  ` ` `  `    ``while` `(true)  ` `    ``{  ` `        ``if` `(digitWell(``\$i``, ``\$m``, ``\$k``))  ` `            ``return` `\$i``;  ` `        ``\$i``++;  ` `    ``}  ` `}  ` ` `  `// Driver code  ` `\$n` `= 111; ` `\$m` `= 2; ` `\$k` `= 2;  ` ` `  `echo` `findInt(``\$n``, ``\$m``, ``\$k``);  ` ` `  `// This code is contributed by Ryuga ` `?> `

Output:

```122
```

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Improved By : Ryuga, Code_Mech, Akanksha_Rai