# Partition a number into two divisible parts

Given a number (as string) and two integers a and b, divide the string in two non-empty parts such that the first part is divisible by a and second part is divisible by b. If string can not be divided into two non-empty parts, output “NO”, else print “YES” with the two parts.

Examples:

```Input  : str = "123", a = 12, b = 3
Output : YES
12, 3
"12" is divisible by a and "3" is
divisible by b.

Input  : str = "1200", a = 4, b = 3
Output : YES
12, 00

Input  : str = "125", a = 12, b = 3
Output : NO
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

A simple solution is to one by one partition array around all points. For every partition, check if left and right of it are divisible by a and b respectively. If yes, print the left and right parts and return.

An efficient solution is to do some preprocessing and save the division modulo by ‘a’ by scanning the string from left to right and division modulo by ‘b’ from right to left.

If we know the remainder of prefix from 0 to i, when divided by a, then we compute remainder of prefix from 0 to i+1 using below formula.
lr[i+1] = (lr[i]*10 + str[i] -‘0’)%a.

Same way, modulo by b can be found by scanning from right to left. We create another rl[] to store remainders with b from right to left.

Once we have precomputed two remainders, we can easily find the point that partition string in two parts.

## C++

 `// C++ program to check if a string can be splitted ` `// into two strings such that one is divisible by 'a' ` `// and other is divisible by 'b'. ` `#include ` `using` `namespace` `std; ` ` `  `// Finds if it is possible to partition str ` `// into two parts such that first part is ` `// divisible by a and second part is divisible ` `// by b. ` `void` `findDivision(string &str, ``int` `a, ``int` `b) ` `{ ` `    ``int` `len = str.length(); ` ` `  `    ``// Create an array of size len+1 and initialize ` `    ``// it with 0. ` `    ``// Store remainders from left to right when ` `    ``// divided by 'a' ` `    ``vector<``int``> lr(len+1, 0); ` `    ``lr = (str - ``'0'``)%a; ` `    ``for` `(``int` `i=1; i rl(len+1, 0); ` `    ``rl[len-1] = (str[len-1] - ``'0'``)%b; ` `    ``int` `power10 = 10; ` `    ``for` `(``int` `i= len-2; i>=0; i--) ` `    ``{ ` `        ``rl[i] = (rl[i+1] + (str[i]-``'0'``)*power10)%b; ` `        ``power10 = (power10 * 10) % b; ` `    ``} ` ` `  `    ``// Find a point that can partition a number ` `    ``for` `(``int` `i=0; i

## Java

 `// Java program to check if a string can be splitted  ` `// into two strings such that one is divisible by 'a'  ` `// and other is divisible by 'b'.  ` `class` `GFG ` `{ ` `     `  `// Finds if it is possible to partition str  ` `// into two parts such that first part is  ` `// divisible by a and second part is divisible  ` `// by b.  ` `static` `void` `findDivision(String str, ``int` `a, ``int` `b)  ` `{  ` `    ``int` `len = str.length();  ` ` `  `    ``// Create an array of size len+1 and initialize  ` `    ``// it with 0.  ` `    ``// Store remainders from left to right when  ` `    ``// divided by 'a'  ` `    ``int``[] lr = ``new` `int``[len + ``1``];  ` `     `  `    ``lr[``0``] = ((``int``)str.charAt(``0``) - (``int``)``'0'``)%a;  ` `    ``for` `(``int` `i = ``1``; i < len; i++)  ` `        ``lr[i] = ((lr[i - ``1``] * ``10``) % a +  ` `                ``((``int``)str.charAt(i)-(``int``)``'0'``)) % a;  ` ` `  `    ``// Compute remainders from right to left when  ` `    ``// divided by 'b'  ` `    ``int``[] rl = ``new` `int``[len + ``1``];  ` `    ``rl[len - ``1``] = ((``int``)str.charAt(len - ``1``) - ` `                            ``(``int``)``'0'``) % b;  ` `    ``int` `power10 = ``10``;  ` `    ``for` `(``int` `i= len - ``2``; i >= ``0``; i--)  ` `    ``{  ` `        ``rl[i] = (rl[i + ``1``] + ((``int``)str.charAt(i) -  ` `                        ``(``int``)``'0'``) * power10) % b;  ` `        ``power10 = (power10 * ``10``) % b;  ` `    ``}  ` ` `  `    ``// Find a point that can partition a number  ` `    ``for` `(``int` `i = ``0``; i < len - ``1``; i++)  ` `    ``{  ` `        ``// If split is not possible at this point  ` `        ``if` `(lr[i] != ``0``)  ` `            ``continue``;  ` ` `  `        ``// We can split at i if one of the following  ` `        ``// two is true.  ` `        ``// a) All charactes after str.charAt(i] are 0  ` `        ``// b) String after str.charAt(i] is divisible by b, i.e.,  ` `        ``// str.charAt(i+1..n-1] is divisible by b.  ` `        ``if` `(rl[i + ``1``] == ``0``)  ` `        ``{  ` `            ``System.out.println(``"YES"``);  ` `            ``for` `(``int` `k = ``0``; k <= i; k++)  ` `                ``System.out.print(str.charAt(k));  ` ` `  `            ``System.out.print(``", "``);  ` ` `  `            ``for` `(``int` `k = i + ``1``; k < len; k++)  ` `                ``System.out.print(str.charAt(k));  ` `            ``return``;  ` `        ``}  ` `    ``}  ` `    ``System.out.println(``"NO"``);  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `main (String[] args)  ` `{ ` `    ``String str = ``"123"``;  ` `    ``int` `a = ``12``, b = ``3``;  ` `    ``findDivision(str, a, b);  ` `}  ` `} ` ` `  `// This code is contributed by mits `

## Python3

 `# Python3 program to check if a can be splitted ` `# into two strings such that one is divisible by 'a' ` `# and other is divisible by 'b'. ` ` `  `# Finds if it is possible to partition str ` `# into two parts such that first part is ` `# divisible by a and second part is divisible ` `# by b. ` `def` `findDivision(``str``, a, b): ` `    ``lenn ``=` `len``(``str``) ` `     `  `    ``# Create an array of size lenn+1 and  ` `    ``# initialize it with 0. ` `    ``# Store remainders from left to right  ` `    ``# when divided by 'a' ` `    ``lr ``=` `[``0``] ``*` `(lenn ``+` `1``) ` `    ``lr[``0``] ``=` `(``int``(``str``[``0``]))``%``a ` `    ``for` `i ``in` `range``(``1``, lenn): ` `        ``lr[i] ``=` `((lr[i ``-` `1``] ``*` `10``) ``%` `a ``+` `\ ` `                     ``int``(``str``[i])) ``%` `a ` `                      `  `    ``# Compute remainders from right to left  ` `    ``# when divided by 'b' ` `    ``rl ``=` `[``0``] ``*` `(lenn ``+` `1``) ` `    ``rl[lenn ``-` `1``] ``=` `int``(``str``[lenn ``-` `1``]) ``%` `b ` `    ``power10 ``=` `10` `    ``for` `i ``in` `range``(lenn ``-` `2``, ``-``1``, ``-``1``): ` `        ``rl[i] ``=` `(rl[i ``+` `1``] ``+` `int``(``str``[i]) ``*` `power10) ``%` `b ` `        ``power10 ``=` `(power10 ``*` `10``) ``%` `b ` `         `  `    ``# Find a pothat can partition a number ` `    ``for` `i ``in` `range``(``0``, lenn ``-` `1``): ` `         `  `        ``# If split is not possible at this point ` `        ``if` `(lr[i] !``=` `0``): ` `            ``continue` `             `  `        ``# We can split at i if one of the following ` `        ``# two is true. ` `        ``# a) All charactes after str[i] are 0 ` `        ``# b) after str[i] is divisible by b, i.e., ` `        ``# str[i+1..n-1] is divisible by b. ` `        ``if` `(rl[i ``+` `1``] ``=``=` `0``): ` `            ``print``(``"YES"``) ` `            ``for` `k ``in` `range``(``0``, i ``+` `1``): ` `                ``print``(``str``[k], end ``=` `"") ` `             `  `            ``print``(``","``, end ``=` `" "``) ` `             `  `            ``for` `i ``in` `range``(i ``+` `1``, lenn): ` `                ``print``(``str``[k], end ``=` `"") ` `                ``return` `     `  `    ``print``(``"NO"``) ` ` `  `# Driver code ` `str` `=` `"123"` `a, b ``=` `12``, ``3` `findDivision(``str``, a, b) ` ` `  `# This code is contributed by SHUBHAMSINGH10 `

## C#

 `// C# program to check if a string can be splitted  ` `// into two strings such that one is divisible by 'a'  ` `// and other is divisible by 'b'.  ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `// Finds if it is possible to partition str  ` `// into two parts such that first part is  ` `// divisible by a and second part is divisible  ` `// by b.  ` `static` `void` `findDivision(``string` `str, ``int` `a, ``int` `b)  ` `{  ` `    ``int` `len = str.Length;  ` ` `  `    ``// Create an array of size len+1 and initialize  ` `    ``// it with 0.  ` `    ``// Store remainders from left to right when  ` `    ``// divided by 'a'  ` `    ``int``[] lr = ``new` `int``[len + 1];  ` `    ``lr = ((``int``)str - (``int``)``'0'``)%a; ` `     `  `    ``for` `(``int` `i = 1; i < len; i++)  ` `        ``lr[i] = ((lr[i - 1] * 10) % a +  ` `                ``((``int``)str[i] - (``int``)``'0'``)) % a;  ` ` `  `    ``// Compute remainders from right to left when  ` `    ``// divided by 'b'  ` `    ``int``[] rl = ``new` `int``[len + 1];  ` `    ``rl[len - 1] = ((``int``)str[len - 1] - (``int``)``'0'``) % b;  ` `     `  `    ``int` `power10 = 10;  ` `    ``for` `(``int` `i= len - 2; i >= 0; i--)  ` `    ``{  ` `        ``rl[i] = (rl[i + 1] + ((``int``)str[i] -  ` `                ``(``int``)``'0'``) * power10) % b;  ` `        ``power10 = (power10 * 10) % b;  ` `    ``}  ` ` `  `    ``// Find a point that can partition a number  ` `    ``for` `(``int` `i = 0; i < len - 1; i++)  ` `    ``{  ` `        ``// If split is not possible at this point  ` `        ``if` `(lr[i] != 0)  ` `            ``continue``;  ` ` `  `        ``// We can split at i if one of the following  ` `        ``// two is true.  ` `        ``// a) All charactes after str[i] are 0  ` `        ``// b) String after str[i] is divisible by b, i.e.,  ` `        ``// str[i+1..n-1] is divisible by b.  ` `        ``if` `(rl[i + 1] == 0)  ` `        ``{  ` `            ``Console.WriteLine(``"YES"``);  ` `            ``for` `(``int` `k = 0; k <= i; k++)  ` `                ``Console.Write(str[k]);  ` ` `  `            ``Console.Write(``", "``);  ` ` `  `            ``for` `(``int` `k = i + 1; k < len; k++)  ` `                ``Console.Write(str[k]);  ` `            ``return``;  ` `        ``}  ` `    ``}  ` `    ``Console.WriteLine(``"NO"``);  ` `}  ` ` `  `// Driver code  ` `static` `void` `Main()  ` `{  ` `    ``string` `str = ``"123"``;  ` `    ``int` `a = 12, b = 3;  ` `    ``findDivision(str, a, b);  ` `}  ` `} ` ` `  `// This code is contributed by mits `

Output :

```YES
12, 3```

Time Complexity : O(len) where len is length of input number string.

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