Given a number in the form of a string s, the task is to calculate and display minimum splits required such that the segments formed are Prime or print Not Possible otherwise.
Examples:
Input: s = “2351”
Output : 0
Explanation: Given number is already prime.
Input: s = “2352”
Output: 2
Explanation: Resultant prime segments are 23,5,2
Input: s = “2375672”
Output : 2
Explanation: Resultant prime segments are 2,37567,2
Approach:
This problem is a variation of Matrix Chain Multiplication and can be solved using Dynamic programming.
Try all possible splits recursively and at each split, check whether the segments formed are prime or not. Consider a 2D array dp where dp[i][j] shows the minimum splits from index i to j and returns dp[0][n] where n is the length of the string.
Recurrence :
dp[i][j] = min(1 + solve(i, k) + solve(k + 1, j)) where i <= k <= j
Actually, in the exact recurrence written above, the left and right segments both are non-prime, then 1 + INT_MAX + INT_MAX will be negative which leads to an incorrect answer.
So, separate calculations for the left and right segments are required. If any segment is found to be non-prime, no need to proceed further. Return min(1+left+right) otherwise.
Base cases considered are :
- If the number is prime, return 0
- If i==j and the number is prime, return 0
- If i==j and the number is not prime, return INT_MAX
Below code is the implementation of above approach:
C++
#include <bits/stdc++.h>using namespace std;int dp[1000][1000] = { 0 };// Checking for primebool isprime(long long num){ if (num <= 1) return false; for (int i = 2; i * i <= num; i++) { if (num % i == 0) { return false; } } return true;}// Conversion of string to intlong long convert(string s, int i, int j){ long long temp = 0; for (int k = i; k <= j; k++) { temp = temp * 10 + (s[k] - '0'); } return temp;}// Function to get the minimum splitsint solve(string s, int i, int j){ // Convert the segment to integer or long long long long num = convert(s, i, j); // Number is prime if (isprime(num)) { return 0; } // If a single digit is prime if (i == j && isprime(num)) return 0; // If single digit is not prime if (i == j && isprime(num) == false) return INT_MAX; if (dp[i][j]) return dp[i][j]; int ans = INT_MAX; for (int k = i; k < j; k++) { // Recur for left segment int left = solve(s, i, k); if (left == INT_MAX) { continue; } // Recur for right segment int right = solve(s, k + 1, j); if (right == INT_MAX) { continue; } // Minimum from left and right segment ans = min(ans, 1 + left + right); } return dp[i][j] = ans;}int main(){ string s = "2352"; int n = s.length(); int cuts = solve(s, 0, n - 1); if (cuts != INT_MAX) { cout << cuts; } else { cout << "Not Possible"; }} |
Java
import java.util.*; class GFG { static int dp[][] = new int[1000][1000]; // Checking for prime static boolean isprime(long num){ int i; if (num <= 1) return false; for (i = 2; i * i <= num; i++) { if (num % i == 0) { return false; } } return true; } // Conversion of string to int static long convert(String s, int i, int j) { long temp = 0; int k; for (k = i; k <= j; k++) { temp = temp * 10 + (s.charAt(k) - '0'); } return temp; } // Function to get the minimum splits static int solve(String s, int i, int j) { int k; // Convert the segment to integer or long long long num = convert(s, i, j); // Number is prime if (isprime(num)) { return 0; } // If a single digit is prime if (i == j && isprime(num)) return 0; // If single digit is not prime if (i == j && isprime(num) == false) return Integer.MAX_VALUE; if (dp[i][j] != 0) return dp[i][j]; int ans = Integer.MAX_VALUE; for (k = i; k < j; k++) { // Recur for left segment int left = solve(s, i, k); if (left == Integer.MAX_VALUE) { continue; } // Recur for right segment int right = solve(s, k + 1, j); if (right == Integer.MAX_VALUE) { continue; } // Minimum from left and right segment ans = Math.min(ans, 1 + left + right); } return dp[i][j] = ans; } public static void main (String []args) { String s = "2352"; int n = s.length(); int cuts = solve(s, 0, n - 1); if (cuts != Integer.MAX_VALUE) { System.out.print(cuts); } else { System.out.print("Not Possible"); } } }// This code is contributed by chitranayal |
Python3
# Python3 Implementation of the above approachimport numpy as np;import sysdp = np.zeros((1000,1000)) ; INT_MAX = sys.maxsize;# Checking for prime def isprime(num) : if (num <= 1) : return False; for i in range(2, int(num ** (1/2)) + 1) : if (num % i == 0) : return False; return True; # Conversion of string to int def convert(s, i, j) : temp = 0; for k in range(i, j + 1) : temp = temp * 10 + (ord(s[k]) - ord('0')); return temp; # Function to get the minimum splits def solve(s, i, j) : # Convert the segment to integer or long long num = convert(s, i, j); # Number is prime if (isprime(num)) : return 0; # If a single digit is prime if (i == j and isprime(num)) : return 0; # If single digit is not prime if (i == j and isprime(num) == False) : return INT_MAX; if (dp[i][j]) : return dp[i][j]; ans = INT_MAX; for k in range(i, j) : # Recur for left segment left = solve(s, i, k); if (left == INT_MAX) : continue; # Recur for right segment right = solve(s, k + 1, j); if (right == INT_MAX) : continue; # Minimum from left and right segment ans = min(ans, 1 + left + right); dp[i][j] = ans; return ans;# Driver code if __name__ == "__main__" : s = "2352"; n = len(s); cuts = solve(s, 0, n - 1); if (cuts != INT_MAX) : print(cuts); else : print("Not Possible"); # This code is converted by Yash_R |
C#
using System;class GFG { static int [,]dp = new int[1000,1000]; // Checking for prime static bool isprime(long num){ int i; if (num <= 1) return false; for (i = 2; i * i <= num; i++) { if (num % i == 0) { return false; } } return true; } // Conversion of string to int static long convert(String s, int i, int j) { long temp = 0; int k; for (k = i; k <= j; k++) { temp = temp * 10 + (s[k] - '0'); } return temp; } // Function to get the minimum splits static int solve(String s, int i, int j) { int k; // Convert the segment to integer or long long long num = convert(s, i, j); // Number is prime if (isprime(num)) { return 0; } // If a single digit is prime if (i == j && isprime(num)) return 0; // If single digit is not prime if (i == j && isprime(num) == false) return int.MaxValue; if (dp[i,j] != 0) return dp[i, j]; int ans = int.MaxValue; for (k = i; k < j; k++) { // Recur for left segment int left = solve(s, i, k); if (left == int.MaxValue) { continue; } // Recur for right segment int right = solve(s, k + 1, j); if (right == int.MaxValue) { continue; } // Minimum from left and right segment ans = Math.Min(ans, 1 + left + right); } return dp[i,j] = ans; } public static void Main(String []args) { String s = "2352"; int n = s.Length; int cuts = solve(s, 0, n - 1); if (cuts != int.MaxValue) { Console.Write(cuts); } else { Console.Write("Not Possible"); } } }// This code is contributed by PrinciRaj1992 |
Javascript
<script>let dp = new Array(1000);for(let i = 0; i < 1000; i++){ dp[i] = new Array(1000)}// Checking for primefunction isprime(num){ if (num <= 1) return false; for (let i = 2; i * i <= num; i++) { if (num % i == 0) { return false; } } return true;}// Conversion of string to intfunction convert(s, i, j){ let temp = 0; for (let k = i; k <= j; k++) { temp = temp * 10 + (s[k] - '0'); } return temp;}// Function to get the minimum splitsfunction solve(s, i, j){ // Convert the segment to integer or long long let num = convert(s, i, j); // Number is prime if (isprime(num)) { return 0; } // If a single digit is prime if (i == j && isprime(num)) return 0; // If single digit is not prime if (i == j && isprime(num) == false) return Number.MAX_SAFE_INTEGER; if (dp[i][j]) return dp[i][j]; let ans = Number.MAX_SAFE_INTEGER; for (let k = i; k < j; k++) { // Recur for left segment let left = solve(s, i, k); if (left == Number.MAX_SAFE_INTEGER) { continue; } // Recur for right segment let right = solve(s, k + 1, j); if (right == Number.MAX_SAFE_INTEGER) { continue; } // Minimum from left and right segment ans = Math.min(ans, 1 + left + right); } return dp[i][j] = ans;} let s = "2352"; let n = s.length; let cuts = solve(s, 0, n - 1); if (cuts != Number.MAX_SAFE_INTEGER) { document.write(cuts); } else { document.write("Not Possible"); } // This code is contributed by _saurabh_jaiswal</script> |
Output
2
Time Complexity: O(n3/2)
Auxiliary Space: O(106)
Another approach : Using DP Tabulation method ( Iterative approach )
In this approach we create a 2D DP to compute and store subproblems this method is called DP tabulation because we find answer just by iterating the DP and calculate subproblems without the help of recursion.
Steps to solve this problem :
- Define a 2D array dp of size n x n, where n is the length of the input string.
- Calculate the ending index j as i + len – 1.
- Convert the substring from index i to j to a long long integer and check if it is prime or not. Store the result in dp[i][j].
- Iterate over all possible values of k such that i <= k < j.
- Calculate the number of cuts required for the left substring from i to k, and the right substring from k+1 to j.
- Add 1 to the sum of the cuts and store the minimum value in dp[i][j].
- Return dp[0][n-1] if it is not INT_MAX, else return -1, indicating that it is not possible to split the string into prime numbers.
Implementation :
C++
#include <bits/stdc++.h>using namespace std;long long dp[1000][1000] = { 0 };// Checking for primebool isprime(long long num){ if (num <= 1) return false; for (long long i = 2; i * i <= num; i++) { if (num % i == 0) { return false; } } return true;}// Conversion of string to intlong long convert(string s, int i, int j){ long long temp = 0; for (int k = i; k <= j; k++) { temp = temp * 10 + (s[k] - '0'); } return temp;}// Function to get the minimum splitslong long solve(string s){ int n = s.length(); // base cases for (int i = 0; i < n; i++) { dp[i][i] = (isprime(s[i] - '0')) ? 0 : -1; } // fill dp table for (int len = 2; len <= n; len++) { for (int i = 0; i <= n - len; i++) { int j = i + len - 1; long long num = convert(s, i, j); dp[i][j] = (isprime(num)) ? 0 : -1; // check for all possible splits for (int k = i; k < j; k++) { long long left = dp[i][k]; long long right = dp[k + 1][j]; // if either side has no valid split, skip // this split if (left == -1 || right == -1) { continue; } // update the minimum splits if (dp[i][j] == -1 || dp[i][j] > left + right + 1) { dp[i][j] = left + right + 1; } } } } // return the minimum splits return dp[0][n - 1];}// Driver codeint main(){ string s = "2375672"; long long cuts = solve(s); if (cuts != -1) { cout << cuts << endl; } else { cout << "Not Possible" << endl; } return 0;} |
Java
import java.util.*;public class Main { static long[][] dp = new long[1000][1000]; // Checking for prime static boolean isprime(long num) { if (num <= 1) return false; for (long i = 2; i * i <= num; i++) { if (num % i == 0) { return false; } } return true; } // Conversion of string to int static long convert(String s, int i, int j) { long temp = 0; for (int k = i; k <= j; k++) { temp = temp * 10 + (s.charAt(k) - '0'); } return temp; } // Function to get the minimum splits static long solve(String s) { int n = s.length(); // base cases for (int i = 0; i < n; i++) { dp[i][i] = (isprime(s.charAt(i) - '0')) ? 0 : -1; } // fill dp table for (int len = 2; len <= n; len++) { for (int i = 0; i <= n - len; i++) { int j = i + len - 1; long num = convert(s, i, j); dp[i][j] = (isprime(num)) ? 0 : -1; // check for all possible splits for (int k = i; k < j; k++) { long left = dp[i][k]; long right = dp[k + 1][j]; // if either side has no valid split, // skip this split if (left == -1 || right == -1) { continue; } // update the minimum splits if (dp[i][j] == -1 || dp[i][j] > left + right + 1) { dp[i][j] = left + right + 1; } } } } // return the minimum splits return dp[0][n - 1]; } // Driver code public static void main(String[] args) { String s = "2375672"; long cuts = solve(s); if (cuts != -1) { System.out.println(cuts); } else { System.out.println("Not Possible"); } }} |
Python3
# Python program for the above approachimport sys# Checking for primedef isprime(num): if num <= 1: return False for i in range(2, int(num**0.5)+1): if num % i == 0: return False return True# Conversion of string to intdef convert(s, i, j): temp = 0 for k in range(i, j+1): temp = temp * 10 + int(s[k]) return temp# Function to get the minimum splitsdef solve(s): n = len(s) global dp dp = [[0 for j in range(n)] for i in range(n)] # base cases for i in range(n): dp[i][i] = 0 if isprime(int(s[i])) else -1 # fill dp table for length in range(2, n+1): for i in range(n-length+1): j = i + length - 1 num = convert(s, i, j) dp[i][j] = 0 if isprime(num) else -1 # check for all possible splits for k in range(i, j): left = dp[i][k] right = dp[k+1][j] # if either side has no valid split, # skip this split if left == -1 or right == -1: continue # update the minimum splits if dp[i][j] == -1 or dp[i][j] > left + right + 1: dp[i][j] = left + right + 1 # return the minimum splits return dp[0][n-1]# Driver codeif __name__ == '__main__': s = "2375672" cuts = solve(s) if cuts != -1: print(cuts) else: print("Not Possible") |
C#
using System;public class Program{ static long[,] dp = new long[1000, 1000]; // Checking for prime static bool IsPrime(long num) { if (num <= 1) return false; for (long i = 2; i * i <= num; i++) { if (num % i == 0) { return false; } } return true; } // Conversion of string to int static long Convert(string s, int i, int j) { long temp = 0; for (int k = i; k <= j; k++) { temp = temp * 10 + (s[k] - '0'); } return temp; } // Function to get the minimum splits static long Solve(string s) { int n = s.Length; // base cases for (int i = 0; i < n; i++) { dp[i, i] = (IsPrime(s[i] - '0')) ? 0 : -1; } // fill dp table for (int len = 2; len <= n; len++) { for (int i = 0; i <= n - len; i++) { int j = i + len - 1; long num = Convert(s, i, j); dp[i, j] = (IsPrime(num)) ? 0 : -1; // check for all possible splits for (int k = i; k < j; k++) { long left = dp[i, k]; long right = dp[k + 1, j]; // if either side has no valid split, // skip this split if (left == -1 || right == -1) { continue; } // update the minimum splits if (dp[i, j] == -1 || dp[i, j] > left + right + 1) { dp[i, j] = left + right + 1; } } } } // return the minimum splits return dp[0, n - 1]; } // Driver code public static void Main(string[] args) { string s = "2375672"; long cuts = Solve(s); if (cuts != -1) { Console.WriteLine(cuts); } else { Console.WriteLine("Not Possible"); } }} |
Javascript
// JavaScript code to get the minimum splitslet dp = new Array(1000).fill(0).map(() => new Array(1000).fill(0));// Checking for primefunction isprime(num) { if (num <= 1) { return false; } for (let i = 2; i * i <= num; i++) { if (num % i == 0) { return false; } } return true;}// Conversion of string to intfunction convert(s, i, j) { let temp = 0; for (let k = i; k <= j; k++) { temp = temp * 10 + (s[k] - '0'); } return temp;}// Function to get the minimum splitsfunction solve(s) { let n = s.length; // base cases for (let i = 0; i < n; i++) { dp[i][i] = (isprime(s[i] - '0')) ? 0 : -1; } // fill dp table for (let len = 2; len <= n; len++) { for (let i = 0; i <= n - len; i++) { let j = i + len - 1; let num = convert(s, i, j); dp[i][j] = (isprime(num)) ? 0 : -1; // check for all possible splits for (let k = i; k < j; k++) { let left = dp[i][k]; let right = dp[k + 1][j]; // if either side has no valid split, skip this split if (left == -1 || right == -1) { continue; } // update the minimum splits if (dp[i][j] == -1 || dp[i][j] > left + right + 1) { dp[i][j] = left + right + 1; } } } } // return the minimum splits return dp[0][n - 1];}// Driver codelet s = "2375672";let cuts = solve(s);if (cuts != -1) { console.log(cuts);}else { console.log("Not Possible");} |
Output:
2
Time Complexity: O(n^3)
Auxiliary Space: O(N^2)
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