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

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++ 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 <bits/stdc++.h> using namespace std; // Finds if it is possible to paritiion 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[0] = (str[0] - '0')%a; for (int i=1; i<len; i++) lr[i] = ((lr[i-1]*10)%a + (str[i]-'0'))%a; // Compute remainders from right to left when // divided by 'b' vector<int> 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<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) { cout << "YES\n"; for (int k=0; k<=i; k++) cout << str[k]; cout << ", "; for (int k=i+1; k<len; k++) cout << str[k]; return; } } cout << "NO\n"; } // Driver code int main() { string str = "123"; int a = 12, b = 3; findDivision(str, a, b); return 0; }

Output :

YES 12, 3

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

This article is contributed by **Ekta Goel**. 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.

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