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Length of Smallest Non Prime Subsequence in given numeric String

  • Last Updated : 16 Sep, 2021

Given a string S of size N consisting of digits [1, 9], the task is to find the length of the smallest subsequence in the string such that it is not a prime number.

Examples:

Input: S = “237”
Output: 2
Explanation:
There are 7 non empty subsequence {“2”, “3”, “7”, “23”, “27”, “37”, “237”}. Among these subsequence there are two subsequence that are not prime, i.e., 27 and 237. Thus as 27 has a length of 2. So print 2.

Input: S = “44444”
Output: 1

Approach: The idea to solve this problem is based on the observation that if on deleting j or more than j characters from the string S the string is a prime number, then the answer should be greater than j. Based on this fact, form all the strings such that deleting element from that string gives a prime number. All the possible strings from the above intuition are {2, 3, 5, 7, 23, 37, 53, 73} Thus the maximum possible size of the sub-sequence is 3. Therefore, the idea is to traverse over all the subsequences of size 1 and size 2 and if the subsequences are found to contain at least 1 element which is not present in the list then the size might be either 1 or 2. Otherwise, the size will be 3. Follow the steps below to solve the problem:

  • Initialize the boolean variable flag as false.
  • Initialize an empty string dummy.
  • Iterate over the range [0, N) using the variable j and perform the following tasks:
    • If the character at j-th position is not equal to 2 or 3 or 5 or 7, then print the answer as 1, set the value of flag as true and break.
  • If flag is true, then return otherwise perform the following tasks.
  • Iterate over the range [0, N) using the variable j and perform the following tasks:
    • Iterate over the range [j+1, N) using the variable j1 and perform the following tasks:
      • Build a dummy string with characters at j and j1 position.
      • If the dummy string is not equal to 2, 3, 5, 7, 23, 37, 53, and 73 then print the answer as 2 and set the value of flag as true and break.
  • If the flag is false and the length of the string S is greater than equal to 3, then print 3 as the answer else print -1.

Below is the implementation of the above approach: 

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the smallest
// length of resultant subsequence
void findMinimumSubsequence(
    string S)
{
    bool flag = false;
    string dummy;
 
    // Check for a subsequence of size 1
    for (int j = 0; j < S.length(); j++)
    {
        if (S[j] != '2' && S[j] != '3' && S[j] != '5' && S[j] != '7')
        {
            cout << 1;
            flag = true;
            break;
        }
    }
 
    // Check for a subsequence of size 2
 
    if (!flag)
    {
 
        for (int j = 0;
             j < S.length() - 1; j++)
        {
 
            for (int j1 = j + 1;
                 j1 < S.length(); j1++)
            {
 
                dummy = S[j] + S[j1];
 
                if (dummy != "23" && dummy != "37" && dummy != "53" && dummy != "73")
                {
                    cout << 2;
                }
                if (flag = true)
                    break;
            }
            if (flag = true)
                break;
        }
    }
 
    // If none of the above check is
    // successful then subsequence
    // must be of size 3
  
    if (!flag)
    {
        if (S.length() >= 3)
        {
 
            // Never executed
            cout << 3;
        }
        else
        {
            cout << -1;
        }
    }
}
 
// Driver Code
int main()
{
    string S = "237";
    findMinimumSubsequence(S);
    return 0;
}
 
// This code is contributed by Potta Lokesh

Java




// Java program for the above approach
 
import java.io.*;
 
class GFG {
 
    // Function to find the smallest
    // length of resultant subsequence
    public static void findMinimumSubsequence(
        String S)
    {
        boolean flag = false;
        StringBuilder dummy = new StringBuilder();
 
        // Check for a subsequence of size 1
        for (int j = 0; j < S.length(); j++) {
            if (S.charAt(j) != '2' && S.charAt(j) != '3'
                && S.charAt(j) != '5'
                && S.charAt(j) != '7') {
                System.out.println(1);
                flag = true;
                break;
            }
        }
 
        // Check for a subsequence of size 2
        if (!flag) {
        loop:
            for (int j = 0;
                 j < S.length() - 1; j++) {
 
                for (int j1 = j + 1;
                     j1 < S.length(); j1++) {
 
                    dummy = new StringBuilder(
                        Character.toString(S.charAt(j)));
                    dummy.append(S.charAt(j1));
 
                    if (!dummy.toString().equals("23")
                        && !dummy.toString().equals("37")
                        && !dummy.toString().equals("53")
                        && !dummy.toString().equals("73")) {
                        System.out.println(2);
                        flag = true;
                        break loop;
                    }
                }
            }
        }
 
        // If none of the above check is
        // successful then subsequence
        // must be of size 3
        if (!flag) {
            if (S.length() >= 3) {
 
                // Never executed
                System.out.println(3);
            }
            else {
                System.out.println(-1);
            }
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        String S = "237";
        findMinimumSubsequence(S);
    }
}

C#




// C# program for the above approach
using System;
class GFG
{
   
    // Function to find the smallest
    // length of resultant subsequence
    static void findMinimumSubsequence(string S)
    {
        bool flag = false;
        string dummy = "";
 
        // Check for a subsequence of size 1
        for (int j = 0; j < S.Length; j++) {
            if (S[j] != '2' && S[j] != '3' && S[j] != '5'
                && S[j] != '7') {
                Console.WriteLine(1);
                flag = true;
                break;
            }
        }
 
        // Check for a subsequence of size 2
        if (!flag) {
 
            for (int j = 0; j < S.Length - 1; j++) {
 
                for (int j1 = j + 1; j1 < S.Length; j1++) {
 
                    dummy = S[j].ToString()
                            + S[j1].ToString();
 
                    if (dummy != "23" && dummy != "37"
                        && dummy != "53" && dummy != "73") {
                        Console.WriteLine(2);
                    }
                    if (flag == true)
                        break;
                    else
                        flag = true;
                }
                if (flag == true)
                    break;
            }
        }
 
        // If none of the above check is
        // successful then subsequence
        // must be of size 3
 
        if (flag == false) {
            if (S.Length >= 3) {
 
                // Never executed
                Console.WriteLine(3);
            }
            else {
                Console.WriteLine(-1);
            }
        }
    }
 
    // Driver Code
    public static void Main()
    {
        string S = "237";
        findMinimumSubsequence(S);
    }
}
 
// This code is contributed by ukasp.

Python3




# Python 3 program for the above approach
 
# Function to find the smallest
# length of resultant subsequence
def findMinimumSubsequence(S):
    flag = False
    dummy = ''
 
    # Check for a subsequence of size 1
    for j in range(len(S)):
        if (S[j] != '2' and S[j] != '3' and S[j] != '5' and S[j] != '7'):
            print(1)
            flag = True
            break
 
    # Check for a subsequence of size 2
    if (flag == False):
        for j in range(len(S)):
            for j1 in range(j + 1,len(S),1):
                dummy = S[j] + S[j1]
 
                if (dummy != "23" and dummy != "37" and dummy != "53" and dummy != "73"):
                    print(2)
                if (flag == True):
                    break
                else:
                    flag = True
                     
            if (flag == True):
                break
 
    # If none of the above check is
    # successful then subsequence
    # must be of size 3
    if (flag == False):
        if (len(S) >= 3):
            # Never executed
            print(3)
        else:
            print(-1)
     
# Driver Code
if __name__ == '__main__':
    S = "237"
    findMinimumSubsequence(S)
     
    # This code is contributed by ipg2016107.

Javascript




<script>   
// JavaScript program for the above approach
 
// Function to find the smallest
// length of resultant subsequence
function findMinimumSubsequence(S)
{
    let flag = false;
     
    // Check for a subsequence of size 1
    for (let j = 0; j < S.length; j++)
    {
        if (S[j] != '2' && S[j] != '3' && S[j] != '5' && S[j] != '7')
        {
            document.write(1);
            flag = true;
            break;
        }
    }
 
    // Check for a subsequence of size 2
    if (flag == false)
    {
        for (let j = 0; j < S.length; j++)
        {
            for (let j1 = j + 1; j1 < S.length; j1++)
            {
                let dummy = S[j] + S[j1];
                if (dummy != "23" && dummy != "37" && dummy != "53" && dummy != "73")
                {
                    document.write(2);
                }
                if (flag == true)
                    break;
                else
                    flag = true;
            }
            if (flag == true)
                break;
        }
    }
 
    // If none of the above check is
    // successful then subsequence
    // must be of size 3
    if (flag == false)
    {
        if (S.length >= 3)
        {
 
            // Never executed
           document.write(3);
        }
        else
        {
            document.write(-1);
        }
    }
}
 
// Driver Code
    let S = "237";
    findMinimumSubsequence(S);
 
// This code is contributed by AnkThon
</script>
Output: 
2

 

Time Complexity: O(N2)
Auxiliary Space: O(1)


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