# 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

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` `# Sanjit_Prasad`

## 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` `?>`

## Javascript

 ``

Output:

`4747`

Time Complexity: O(2n)
Auxiliary Space: O(1)

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!

Previous
Next