# Next greater Number than N with the same quantity of digits A and B

Given a number and two digits and . The task is to find the least number not less than N which contains the equal number of digits A and B.

Note: N <= 107

Examples:

Input : N = 4500, A = 4, B = 7
Output : 4747
The number greater than 4500 which has the same quantity of number ‘4’ and number ‘7’ is 4747.

Input : N = 99999999, A = 6, B = 7
Output : 6666677777

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

Below is the step by step algorithm to solve this problem:

1. If the length of ‘N’ is odd then the resulting number will be of length ‘N+1’ as both ‘a’ and ‘b’ has to be in equal quantity.
2. If the length of ‘N’ is even then the resulting number will either be of length ‘N’ or ‘N+2’.
3. We will generate the number recursively by appending both A and B one by one and take the minimum of the two for the next recursive call.
4. At last return the smallest number greater than or equal to ‘N’.

Below is the implementation of the above idea:

## C++

 // C++ program to find next greater Number // than N with the same quantity of // digits A and B    #include using namespace std;    // Recursive function to find the required number long findNumUtil(long res, int a, int aCount, int b, int bCount, int n) {     if (res > 1e11)         return 1e11;        // If the resulting number is >= n and     // count of a = count of b, return the number     if (aCount == bCount && res >= n)         return res;        // select minimum of two and call the function again     return min(findNumUtil(res * 10 + a, a, aCount + 1, b, bCount, n),                findNumUtil(res * 10 + b, a, aCount, b, bCount + 1, n)); }    // Function to find the number next greater Number // than N with the same quantity of // digits A and B int findNum(int n, int a, int b) {     int result = 0;     int aCount = 0;     int bCount = 0;        return findNumUtil(result, a, aCount, b, bCount, n); }    // Driver code int main() {     int N = 4500;     int A = 4;     int B = 7;        cout << findNum(N, A, B);        return 0; }

## Java

 // Java program to find next greater Number // than N with the same quantity of // digits A and B    public class GFG {            // Recursive function to find the required number     static long findNumUtil(long res, int a, int aCount, int b, int bCount, int n)     {         if (res > 1e11)             return (long) 1e11;            // If the resulting number is >= n and         // count of a = count of b, return the number         if (aCount == bCount && res >= n)             return res;            // select minimum of two and call the function again         return Math.min(findNumUtil(res * 10 + a, a, aCount + 1, b, bCount, n),                    findNumUtil(res * 10 + b, a, aCount, b, bCount + 1, n));     }        // Function to find the number next greater Number     // than N with the same quantity of     // digits A and B     static int findNum(int n, int a, int b)     {         int result = 0;         int aCount = 0;         int bCount = 0;            return (int) findNumUtil(result, a, aCount, b, bCount, n);     }               // Driver code     public static void main(String args[])     {            int N = 4500;             int A = 4;             int B = 7;                System.out.println(findNum(N, A, B));           }     // This Code is contributed by ANKITRAI1 }

## Python3

# Python 3 program to find next greater
# Number than N with the same quantity of
# digits A and B

# Recursive function to find the
# required number
def findNumUtil(res, a, aCount, b, bCount, n):
if (res > 1e11):
return 1e11

# If the resulting number is >= n
# and count of a = count of b,
# return the number
if (aCount == bCount and res >= n):
return res

# select minimum of two and call
# the function again
return min(findNumUtil(res * 10 + a,
a, aCount + 1, b, bCount, n),
findNumUtil(res * 10 + b, a,
aCount, b, bCount + 1, n))

# Function to find the number next
# greater Number than N with the
# same quantity of digits A and B
def findNum(n, a, b):
result = 0
aCount = 0
bCount = 0

return findNumUtil(result, a, aCount,
b, bCount, n)

# Driver code
if __name__ == ‘__main__’:
N = 4500
A = 4
B = 7

print(findNum(N, A, B))

# This code is contributed by

## C#

 // C# program to find next greater Number // than N with the same quantity of // digits A and B using System;    class GFG {    // Recursive function to find the required number static long findNumUtil(long res, int a, int aCount,                          int b, int bCount, int n) {     if (res > 1e11)         return (long) 1e11;        // If the resulting number is >= n and     // count of a = count of b, return the number     if (aCount == bCount && res >= n)         return res;        // select minimum of two and call      // the function again     return Math.Min(findNumUtil(res * 10 + a, a,                                  aCount + 1, b, bCount, n),             findNumUtil(res * 10 + b, a, aCount,                               b, bCount + 1, n)); }    // Function to find the number next  // greater Number than N with the  // same quantity of digits A and B static int findNum(int n, int a, int b) {     int result = 0;     int aCount = 0;     int bCount = 0;        return (int) findNumUtil(result, a, aCount,                                       b, bCount, n); }    // Driver code public static void Main() {     int N = 4500;     int A = 4;     int B = 7;        Console.WriteLine(findNum(N, A, B)); } }    // This code is contributed by Shashank

## PHP

 100000000000)         return 10000000000;        // If the resulting number is >= n and     // count of a = count of b, return the number     if (\$aCount == \$bCount && \$res >= \$n)         return \$res;        // select minimum of two and call the function again     return min(findNumUtil(\$res * 10 + \$a, \$a, \$aCount + 1, \$b, \$bCount, \$n),             findNumUtil(\$res * 10 + \$b, \$a, \$aCount, \$b, \$bCount + 1, \$n)); }    // Function to find the number next greater Number // than N with the same quantity of // digits A and B function findNum(\$n, \$a, \$b) {     \$result = 0;     \$aCount = 0;     \$bCount = 0;        return findNumUtil(\$result, \$a, \$aCount, \$b, \$bCount, \$n); }    // Driver code        \$N = 4500;     \$A = 4;     \$B = 7;        echo findNum(\$N, \$A, \$B);    // This Code is contributed by mits ?>

Output:

4747

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