Find all possible ways to Split the given string into Primes

Given a string str that represents a number. The task is to find all possible ways to split the given string such that each segment is a prime number in the range of 1 to 106.

Examples:

Input: str = “3175”
Output:
[317, 5]
[31, 7, 5]
[3, 17, 5]

Explanation:
There can be 8 possible ways to split:
[3175]
[317, 5] – All primes
[31, 75]
[31, 7, 5] – All primes
[3, 175]
[3, 17, 5] – All primes
[3, 1, 75]
[3, 1, 7, 5]

Input: str = “11373”
Output:
[113, 73]
[113, 7, 3]
[11, 373]
[11, 37, 3]
[11, 3, 73]
[11, 3, 7, 3]



Approach:

  1. The idea is to generate all possible splits of a string of size N by counting binary number from 0 to 2(N – 1) – 1. Where every 1 indicates that the string should split at that point.

    For example:

     S = "3175"
     0 0 0   3175
     0 0 1   317, 5
     0 1 0   31, 75
     0 1 1   31, 7, 5
     1 0 0   3, 175
     1 0 1   3, 17, 5
     1 1 0   3, 1, 75
     1 1 1   3, 1, 7, 5
    
  2. To check the prime number efficiently we will pre-process prime number in a boolean array using Sieve of Eratosthenes.

Below is the implementation of the above approach.

C++

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// C++ program to Find all the 
// ways to split the given string
// into Primes.
#include<bits/stdc++.h>
using namespace std;
  
bool primes[1000000];
const int maxn = 1000000;
  
// Sieve of Eratosthenes
void sieve()
{
    memset(primes,true,sizeof(primes));
    primes[0] = primes[1] = 0;
      
    for(int i = 2; i * i <= maxn; i++)
    {
        if(primes[i])
        {
            for(int j = i * i ;
                   j <= maxn ; j += i)
            primes[j] = false;
        }
    }
}
  
// Function Convert integer
// to binary string
string toBinary(int n)
{
    string r = "";
    while(n != 0) 
    {
        r = (n % 2 == 0 ?"0":"1") + r;
        n /= 2;
    }
    return (r == "")?"0":r;
}
  
// Function print all the all the 
// ways to split the given string
// into Primes.
void PrimeSplit(string str)
{
    string temp;
    int cnt=0;
      
    // To store all possible strings
    vector<string> ans;
    int bt = 1<<(str.size()-1);
    int n = str.size();
  
  
    // Exponetnital complexity n*(2^(n-1))
    // for bit 
    for(int i = 0 ; i < bt ; i++)
    {
        temp = toBinary(i) + "0";
        int j = 0, x = n - temp.size(), y;
        while(j < x)
        {
            temp = "0" + temp;
            j++;
        }
        j = 0; 
        x = 0;
        y = -1;
          
        string sp = "", tp = "";
        bool flag = 0;
          
        while(j < n)
        {
            sp += str[j];
            if(temp[j] == '1')
            {
                tp += sp + ',';
                y = stoi(sp);
                  
                // Pruning step
                if(!primes[y])
                {
                    flag = 1;
                    break;
                }
                sp = "";
            }
            j++;
        }
        tp += sp;
        if(sp != "")
        {
            y = stoi(sp);
            if(!primes[y])
            flag = 1;
        }
        if(!flag)
        ans.push_back(tp);
    }
    if(ans.size() == 0)
    {
        cout << -1 << endl;
    }
    for(auto i:ans)
    {
        cout << i << endl;
    }
}
  
// Driver code
int main()
{
    string str = "11373";
    sieve();
      
    PrimeSplit(str);
      
    return 0;
}

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Output:

113,73
113,7,3
11,373
11,37,3
11,3,73
11,3,7,3

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