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



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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


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#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP implementation of the approach 
  
// Function that returns true if n 
// contains digit m exactly k times 
function digitWell($n, $m, $k
    $cnt = 0; 
    while ($n > 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
?>

chevron_right


Output:

122


My Personal Notes arrow_drop_up

Discovering ways to develop a plane for soaring career goals

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : Ryuga, Code_Mech, Akanksha_Rai