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++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function that returns true if n contains // digit m exactly k timesbool 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 timesint findInt(int n, int m, int k){ int i = n + 1; while (true) { if (digitWell(i, m, k)) return i; i++; }}// Driver codeint main(){ int n = 111, m = 2, k = 2; cout << findInt(n, m, k); return 0;} |
Java
// Java implementation of the approachimport java.io.*;class GFG{ // Function that returns true if n contains // digit m exactly k timesstatic 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 timesstatic int findInt(int n, int m, int k){ int i = n + 1; while (true) { if (digitWell(i, m, k)) return i; i++; }}// Driver codepublic 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 timesdef 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 timesdef findInt(n, m, k): i = n + 1; while (True): if (digitWell(i, m, k)): return i; i = i + 1;# Driver coden = 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 timesstatic 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 timesstatic int findInt(int n, int m, int k){ int i = n + 1; while (true) { if (digitWell(i, m, k)) return i; i++; }}// Driver codepublic static void Main(){ int n = 111, m = 2, k = 2; Console.WriteLine(findInt(n, m, k));}} // This code is contributed // by Akanksha Rai |
PHP
<?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?> |
Javascript
<script>// Javascript implementation of the approach// Function that returns true if n contains // digit m exactly k timesfunction digitWell(n, m, k){ var cnt = 0; while (n > 0) { if (n % 10 == m) ++cnt; n = Math.floor(n/10); } if(cnt == k) return true; else return false;}// Function to return the smallest integer > n // with digit m occurring exactly k timesfunction findInt(n, m, k){ var i = n + 1; while (true) { if (digitWell(i, m, k)) return i; i++; }}// Driver code var n = 111, m = 2, k = 2; document.write(findInt(n, m, k));</script> |
Output:
122
Time Complexity: O(n * log10n)
Auxiliary Space: O(1), since no extra space has been taken.
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