Given a string str that represents a large number, the task is to find the minimum number of segments the given string can be divided such that each segment is a prime number in the range of 1 to 106.
Examples:
Input: str = “13499315”
Output: 3
Explanation:
The number can be segmented as [13499, 31, 5]Input: str = “43”
Output: 1
Explanation:
The number can be segmented as [43]
Naive Approach: The idea is to consider every prefix up to 6 digits( Since it is given that the primes are less than 106) and check if it is a prime number or not. If the prefix is a prime number, then recursively call the function to check the remaining string. If a non-negative number is returned, then it is considered as a possible arrangement. If none of the possible combinations returns a positive number, then -1 is printed.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <iostream>#include <string>using namespace std;// Function to check whether a string// is a prime number or notbool checkPrime(string number){ int num = stoi(number); for (int i = 2; i * i <= num; i++) if ((num % i) == 0) return false; return true;}// A recursive function to find the minimum// number of segments the given string can// be divided such that every segment is a primeint splitIntoPrimes(string number){ // If the number is null if (number.length() == 0) return 0; // checkPrime function is called to check if // the number is a prime or not. if (number.length() <= 6 and checkPrime(number)) return 1; else { int numLen = number.length(); // A very large number denoting maximum int ans = 1000000; // Consider a minimum of 6 and length // since the primes are less than 10 ^ 6 for (int i = 1; i <= 6 && i <= numLen; i++) { if (checkPrime(number.substr(0, i))) { // Recursively call the function // to check for the remaining string int val = splitIntoPrimes(number.substr(i)); if (val != -1) { // Evaluating minimum splits // into Primes for the suffix ans = min(ans, 1 + val); } } } // Checks if no combination found if (ans == 1000000) return -1; return ans; }}// Driver codeint main(){ cout << splitIntoPrimes("13499315") << "\n"; cout << splitIntoPrimes("43") << "\n"; return 0;} |
Java
// Java implementation of the above approachimport java.util.*;class GFG{ // Function to check whether a String// is a prime number or notstatic boolean checkPrime(String number){ int num = Integer.valueOf(number); for (int i = 2; i * i <= num; i++) if ((num % i) == 0) return false; return true;} // A recursive function to find the minimum// number of segments the given String can// be divided such that every segment is a primestatic int splitIntoPrimes(String number){ // If the number is null if (number.length() == 0) return 0; // checkPrime function is called to check if // the number is a prime or not. if (number.length() <= 6 && checkPrime(number)) return 1; else { int numLen = number.length(); // A very large number denoting maximum int ans = 1000000; // Consider a minimum of 6 and length // since the primes are less than 10 ^ 6 for (int i = 1; i <= 6 && i <= numLen; i++) { if (checkPrime(number.substring(0, i))) { // Recursively call the function // to check for the remaining String int val = splitIntoPrimes(number.substring(i)); if (val != -1) { // Evaluating minimum splits // into Primes for the suffix ans = Math.min(ans, 1 + val); } } } // Checks if no combination found if (ans == 1000000) return -1; return ans; }} // Driver codepublic static void main(String[] args){ System.out.print(splitIntoPrimes("13499315")+ "\n"); System.out.print(splitIntoPrimes("43")+ "\n");}}// This code is contributed by Rajput-Ji |
Python3
# Python3 implementation of the above approach# Function to check whether a string # is a prime number or notdef checkPrime(number) : num = int(number) for i in range(2, int(num**0.5)) : if((num % i) == 0) : return False return True# A recursive function to find the minimum# number of segments the given string can # be divided such that every segment is a primedef splitIntoPrimes(number) : # If the number is null if( number == '' ) : return 0 # checkPrime function is called to check if # the number is a prime or not. if( len(number)<= 6 and checkPrime(number) ) : return 1 else : numLen = len(number) # A very large number denoting maximum ans = 1000000 # Consider a minimum of 6 and length # since the primes are less than 10 ^ 6 for i in range( 1, (min( 6, numLen ) + 1) ) : if( checkPrime( number[:i] ) ) : # Recursively call the function # to check for the remaining string val = splitIntoPrimes( number[i:] ) if(val != -1) : # Evaluating minimum splits # into Primes for the suffix ans = min(ans, 1 + val) # Checks if no combination found if( ans == 1000000 ) : return -1 return ans # Driver codeprint(splitIntoPrimes("13499315"))print(splitIntoPrimes("43")) |
C#
// C# implementation of the above approachusing System;class GFG{ // Function to check whether a String// is a prime number or notstatic bool checkPrime(String number){ int num = Int32.Parse(number); for (int i = 2; i * i <= num; i++) if ((num % i) == 0) return false; return true;} // A recursive function to find the minimum// number of segments the given String can// be divided such that every segment is a primestatic int splitIntoPrimes(String number){ // If the number is null if (number.Length == 0) return 0; // checkPrime function is called to check if // the number is a prime or not. if (number.Length <= 6 && checkPrime(number)) return 1; else { int numLen = number.Length; // A very large number denoting maximum int ans = 1000000; // Consider a minimum of 6 and length // since the primes are less than 10 ^ 6 for (int i = 1; i <= 6 && i <= numLen; i++) { if (checkPrime(number.Substring(0, i))) { // Recursively call the function // to check for the remaining String int val = splitIntoPrimes(number.Substring(i)); if (val != -1) { // Evaluating minimum splits // into Primes for the suffix ans = Math.Min(ans, 1 + val); } } } // Checks if no combination found if (ans == 1000000) return -1; return ans; }} // Driver codepublic static void Main(String[] args){ Console.Write(splitIntoPrimes("13499315")+ "\n"); Console.Write(splitIntoPrimes("43")+ "\n");}}// This code is contributed by sapnasingh4991 |
Javascript
<script>// JavaScript implementation of the above approach// Function to check whether a string// is a prime number or notfunction checkPrime(number){ let num = String(number); for (let i = 2; i * i <= num; i++) if ((num % i) == 0) return false; return true;}// A recursive function to find the minimum// number of segments the given string can// be divided such that every segment is a primefunction splitIntoPrimes(number){ // If the number is null if (number.length == 0) return 0; // checkPrime function is called to check if // the number is a prime or not. if (number.length <= 6 && checkPrime(number)) return 1; else { let numLen = number.length; // A very large number denoting maximum let ans = 1000000; // Consider a minimum of 6 and length // since the primes are less than 10 ^ 6 for (let i = 1; i <= 6 && i <= numLen; i++) { if (checkPrime(number.substr(0, i))) { // Recursively call the function // to check for the remaining string let val = splitIntoPrimes(number.substr(i)); if (val != -1) { // Evaluating minimum splits // into Primes for the suffix ans = Math.min(ans, 1 + val); } } } // Checks if no combination found if (ans == 1000000) return -1; return ans; }}// Driver codedocument.write(splitIntoPrimes("13499315") + "<br>");document.write(splitIntoPrimes("43") + "<br>");// This code is contributed by gfgking</script> |
Output:
3 1
Time Complexity:
- The time complexity for the above approach would be of O(N5/2) where N is the length of the input string.
- The complexity to find all the possible combinations recursively is O(N2).
- For every combination, to check if the number is a prime number or not, an additional O(N0.5) time is used.
- This makes the time complexity O(N5/2).
Dynamic Programming Approach: The given problem is seen to exhibit an overlapping subproblem property. Therefore, dynamic programming can be used to efficiently solve this question.
A splitDP[] array is defined and used where splitDP[i] denotes the minimum number of splits required in the prefix string of length ‘i’ to break it into the prime subdivision.
The splitDP[] array is filled in the following way:
- A for loop is used to iterate through all the indices of the given string.
- For every index ‘i’ from the above loop, another loop is iterated from 1 to 6 to check if the substring from (i + j)th index forms a prime or not.
- If it forms a prime number, then the value at splitDP[] is updated as:
splitDP[i + j] = min(splitDP[i + j], 1 + splitDP[i]);
- After updating all the values of the array, the value at the last index is the minimum number of splits for the entire string.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// Function to check whether a string// is a prime number or notbool checkPrime(string number){ int num = stoi(number); for (int i = 2; i * i <= num; i++) if ((num % i) == 0) return false; return true;}// A function to find the minimum// number of segments the given string// can be divided such that every// segment is a primeint splitIntoPrimes(string number){ int numLen = number.length(); // Declare a splitdp[] array // and initialize to -1 int splitDP[numLen + 1]; memset(splitDP, -1, sizeof(splitDP)); // Build the DP table in // a bottom-up manner for (int i = 1; i <= numLen; i++) { // Initially Check if the entire prefix is Prime if (i <= 6 && checkPrime(number.substr(0, i))) splitDP[i] = 1; // If the Given Prefix can be split into Primes // then for the remaining string from i to j // Check if Prime. If yes calculate // the minimum split till j if (splitDP[i] != -1) { for (int j = 1; j <= 6 && i + j <= numLen; j++) { // To check if the substring from i to j // is a prime number or not if (checkPrime(number.substr(i, j))) { // If it is a prime, then update the dp array if (splitDP[i + j] == -1) splitDP[i + j] = 1 + splitDP[i]; else splitDP[i + j] = min(splitDP[i + j], 1 + splitDP[i]); } } } } // Return the minimum number of splits // for the entire string return splitDP[numLen];}// Driver codeint main(){ cout << splitIntoPrimes("13499315") << "\n"; cout << splitIntoPrimes("43") << "\n"; return 0;} |
Java
// Java implementation of the above approachimport java.util.*;class GFG{ // Function to check whether a String// is a prime number or notstatic boolean checkPrime(String number){ if(number.length()==0) return true; int num = Integer.parseInt(number); for (int i = 2; i * i <= num; i++) if ((num % i) == 0) return false; return true;} // A function to find the minimum// number of segments the given String// can be divided such that every// segment is a primestatic int splitIntoPrimes(String number){ int numLen = number.length(); // Declare a splitdp[] array // and initialize to -1 int []splitDP = new int[numLen + 1]; Arrays.fill(splitDP, -1); // Build the DP table in // a bottom-up manner for (int i = 1; i <= numLen; i++) { // Initially Check if the entire prefix is Prime if (i <= 6 && checkPrime(number.substring(0, i))) splitDP[i] = 1; // If the Given Prefix can be split into Primes // then for the remaining String from i to j // Check if Prime. If yes calculate // the minimum split till j if (splitDP[i] != -1) { for (int j = 1; j <= 6 && i + j <= numLen; j++) { // To check if the subString from i to j // is a prime number or not if (checkPrime(number.substring(i, i+j))) { // If it is a prime, then update the dp array if (splitDP[i + j] == -1) splitDP[i + j] = 1 + splitDP[i]; else splitDP[i + j] = Math.min(splitDP[i + j], 1 + splitDP[i]); } } } } // Return the minimum number of splits // for the entire String return splitDP[numLen];} // Driver codepublic static void main(String[] args){ System.out.print(splitIntoPrimes("13499315")+ "\n"); System.out.print(splitIntoPrimes("43")+ "\n");}}// This code contributed by Princi Singh |
Python3
# Python 3 implementation of the above approachfrom math import sqrt# Function to check whether a string# is a prime number or notdef checkPrime(number): if(len(number) == 0): return True num = int(number) for i in range(2,int(sqrt(num)) + 1, 1): if ((num % i) == 0): return False return True# A function to find the minimum# number of segments the given string# can be divided such that every# segment is a primedef splitIntoPrimes(number): numLen = len(number) # Declare a splitdp[] array # and initialize to -1 splitDP = [-1 for i in range(numLen + 1)] # Build the DP table in # a bottom-up manner for i in range(1, numLen + 1, 1): # Initially Check if the entire prefix is Prime if (i <= 6 and checkPrime(number[0:i])): splitDP[i] = 1 # If the Given Prefix can be split into Primes # then for the remaining string from i to j # Check if Prime. If yes calculate # the minimum split till j if (splitDP[i] != -1): j = 1 while(j <= 6 and i + j <= numLen): # To check if the substring from i to j # is a prime number or not if (checkPrime(number[i:i+j])): # If it is a prime, then update the dp array if (splitDP[i + j] == -1): splitDP[i + j] = 1 + splitDP[i] else: splitDP[i + j] = min(splitDP[i + j], 1 + splitDP[i]) j += 1 # Return the minimum number of splits # for the entire string return splitDP[numLen]# Driver codeif __name__ == '__main__': print(splitIntoPrimes("13499315")) print(splitIntoPrimes("43"))# This code is contributed by Surendra_Gangwar |
C#
// C# implementation of the above approachusing System;class GFG{ // Function to check whether a String// is a prime number or notstatic bool checkPrime(String number){ if(number.Length==0) return true; int num = Int32.Parse(number); for (int i = 2; i * i <= num; i++) if ((num % i) == 0) return false; return true;} // A function to find the minimum// number of segments the given String// can be divided such that every// segment is a primestatic int splitIntoPrimes(String number){ int numLen = number.Length; // Declare a splitdp[] array // and initialize to -1 int []splitDP = new int[numLen + 1]; for (int i = 0; i <= numLen; i++) splitDP[i] = -1; // Build the DP table in // a bottom-up manner for (int i = 1; i <= numLen; i++) { // Initially Check if the entire prefix is Prime if (i <= 6 && checkPrime(number.Substring(0, i))) splitDP[i] = 1; // If the Given Prefix can be split into Primes // then for the remaining String from i to j // Check if Prime. If yes calculate // the minimum split till j if (splitDP[i] != -1) { for (int j = 1; j <= 6 && i + j <= numLen; j++) { // To check if the subString from i to j // is a prime number or not if (checkPrime(number.Substring(i, j))) { // If it is a prime, then update the dp array if (splitDP[i + j] == -1) splitDP[i + j] = 1 + splitDP[i]; else splitDP[i + j] = Math.Min(splitDP[i + j], 1 + splitDP[i]); } } } } // Return the minimum number of splits // for the entire String return splitDP[numLen];} // Driver codepublic static void Main(String[] args){ Console.Write(splitIntoPrimes("13499315")+ "\n"); Console.Write(splitIntoPrimes("43")+ "\n");}}// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript implementation of the above approach// Function to check whether a String// is a prime number or notfunction checkPrime(number){ if (number.length == 0) return true; let num = parseInt(number); for(let i = 2; i * i <= num; i++) if ((num % i) == 0) return false; return true;}// A function to find the minimum// number of segments the given String// can be divided such that every// segment is a primefunction splitIntoPrimes(number){ let numLen = number.length; // Declare a splitdp[] array // and initialize to -1 let splitDP = new Array(numLen + 1); for(let i = 0; i < splitDP.length; i++) { splitDP[i] = -1; } // Build the DP table in // a bottom-up manner for(let i = 1; i <= numLen; i++) { // Initially Check if the entire prefix is Prime if (i <= 6 && checkPrime(number.substring(0, i))) splitDP[i] = 1; // If the Given Prefix can be split into Primes // then for the remaining String from i to j // Check if Prime. If yes calculate // the minimum split till j if (splitDP[i] != -1) { for(let j = 1; j <= 6 && i + j <= numLen; j++) { // To check if the subString from i to j // is a prime number or not if (checkPrime(number.substring(i, i + j))) { // If it is a prime, then update the // dp array if (splitDP[i + j] == -1) splitDP[i + j] = 1 + splitDP[i]; else splitDP[i + j] = Math.min( splitDP[i + j], 1 + splitDP[i]); } } } } // Return the minimum number of splits // for the entire String return splitDP[numLen];}// Driver codedocument.write(splitIntoPrimes("13499315") + "<br>");document.write(splitIntoPrimes("43") + "<br>");// This code is contributed by avanitrachhadiya2155</script> |
Output:
3 1
Time Complexity:
- The time complexity of the above approach is O(N3/2) where N is the length of the input string.
- The time to iterate through all the indices is O(N).
- Since the inner for loop runs a constant number of times for every index, it’s run time can be considered as constant.
- For every index, the time taken to check whether the number is a prime or not is of O(N0.5).
- Therefore, the overall time complexity is O(N3/2).
Optimized Dynamic Programming Approach: The above approach can further be optimized by using the concept Sieve of Eratosthenes to precompute and store whether a number is prime or not and reducing the time complexity to check for a number at every iteration.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// Function to precompute all the primes// upto 1000000 and store it in a set// using Sieve of Eratosthenesvoid getPrimesFromSeive(set<string>& primes){ bool prime[1000001]; memset(prime, true, sizeof(prime)); prime[0] = prime[1] = false; for (int i = 2; i * i <= 1000000; i++) { if (prime[i] == true) { for (int j = i * i; j <= 1000000; j += i) prime[j] = false; } } // Here to_string() is used // for converting int to string for (int i = 2; i <= 1000000; i++) { if (prime[i] == true) primes.insert(to_string(i)); }}// A function to find the minimum// number of segments the given string// can be divided such that every// segment is a primeint splitIntoPrimes(string number){ int numLen = number.length(); // Declare a splitdp[] array // and initialize to -1 int splitDP[numLen + 1]; memset(splitDP, -1, sizeof(splitDP)); // Call sieve function to store primes in // primes array set<string> primes; getPrimesFromSeive(primes); // Build the DP table in a bottom-up manner for (int i = 1; i <= numLen; i++) { // If the prefix is prime then the prefix // will be found in the prime set if (i <= 6 && (primes.find(number.substr(0, i)) != primes.end())) splitDP[i] = 1; // If the Given Prefix can be split into Primes // then for the remaining string from i to j // Check if Prime. If yes calculate // the minimum split till j if (splitDP[i] != -1) { for (int j = 1; j <= 6 && i + j <= numLen; j++) { // To check if the substring from i to j // is a prime number or not if (primes.find(number.substr(i, j)) != primes.end()) { // If it is a prime, then update the dp array if (splitDP[i + j] == -1) splitDP[i + j] = 1 + splitDP[i]; else splitDP[i + j] = min(splitDP[i + j], 1 + splitDP[i]); } } } } // Return the minimum number of splits // for the entire string return splitDP[numLen];}int main(){ cout << splitIntoPrimes("13499315") << "\n"; cout << splitIntoPrimes("43") << "\n"; return 0;} |
Java
// Java implementation of the above approachimport java.util.*;class GFG{ // Function to precompute all the primes// upto 1000000 and store it in a set// using Sieve of Eratosthenesstatic void getPrimesFromSeive(HashSet<String> primes){ boolean []prime = new boolean[1000001]; Arrays.fill(prime, true); prime[0] = prime[1] = false; for (int i = 2; i * i <= 1000000; i++) { if (prime[i] == true) { for (int j = i * i; j <= 1000000; j += i) prime[j] = false; } } // Here to_String() is used // for converting int to String for (int i = 2; i <= 1000000; i++) { if (prime[i] == true) primes.add(String.valueOf(i)); }} // A function to find the minimum// number of segments the given String// can be divided such that every// segment is a primestatic int splitIntoPrimes(String number){ int numLen = number.length(); // Declare a splitdp[] array // and initialize to -1 int []splitDP = new int[numLen + 1]; Arrays.fill(splitDP, -1); // Call sieve function to store primes in // primes array HashSet<String> primes = new HashSet<String>(); getPrimesFromSeive(primes); // Build the DP table in a bottom-up manner for (int i = 1; i <= numLen; i++) { // If the prefix is prime then the prefix // will be found in the prime set if (i <= 6 && (primes.contains(number.substring(0, i)))) splitDP[i] = 1; // If the Given Prefix can be split into Primes // then for the remaining String from i to j // Check if Prime. If yes calculate // the minimum split till j if (splitDP[i] != -1) { for (int j = 1; j <= 6 && i + j <= numLen; j++) { // To check if the subString from i to j // is a prime number or not if (primes.contains(number.substring(i, i+j))) { // If it is a prime, then update the dp array if (splitDP[i + j] == -1) splitDP[i + j] = 1 + splitDP[i]; else splitDP[i + j] = Math.min(splitDP[i + j], 1 + splitDP[i]); } } } } // Return the minimum number of splits // for the entire String return splitDP[numLen];} public static void main(String[] args){ System.out.print(splitIntoPrimes("13499315")+ "\n"); System.out.print(splitIntoPrimes("43")+ "\n");}}// This code contributed by Princi Singh |
Python3
# Python3 implementation of the above approach# Function to precompute all the primes# upto 1000000 and store it in a set# using Sieve of Eratosthenesdef getPrimesFromSeive(primes): prime = [True] * (1000001) prime[0], prime[1] = False, False i = 2 while (i * i <= 1000000): if (prime[i] == True): for j in range(i * i, 1000001, i): prime[j] = False i += 1 # Here str() is used for # converting int to string for i in range(2, 1000001): if (prime[i] == True): primes.append(str(i))# A function to find the minimum# number of segments the given string# can be divided such that every# segment is a primedef splitIntoPrimes(number): numLen = len(number) # Declare a splitdp[] array # and initialize to -1 splitDP = [-1] * (numLen + 1) # Call sieve function to store # primes in primes array primes = [] getPrimesFromSeive(primes) # Build the DP table in a bottom-up manner for i in range(1, numLen + 1): # If the prefix is prime then the prefix # will be found in the prime set if (i <= 6 and (number[0 : i] in primes)): splitDP[i] = 1 # If the Given Prefix can be split into Primes # then for the remaining string from i to j # Check if Prime. If yes calculate # the minimum split till j if (splitDP[i] != -1): j = 1 while (j <= 6 and (i + j <= numLen)): # To check if the substring from i to j # is a prime number or not if (number[i : i + j] in primes): # If it is a prime, then # update the dp array if (splitDP[i + j] == -1): splitDP[i + j] = 1 + splitDP[i] else: splitDP[i + j] = min(splitDP[i + j], 1 + splitDP[i]) j += 1 # Return the minimum number of # splits for the entire string return splitDP[numLen]# Driver codeprint(splitIntoPrimes("13499315"))print(splitIntoPrimes("43"))# This code is contributed by chitranayal |
C#
// C# implementation of the above approachusing System;using System.Collections.Generic;class GFG{// Function to precompute all the primes// upto 1000000 and store it in a set// using Sieve of Eratosthenesstatic void getPrimesFromSeive(HashSet<String> primes){ bool []prime = new bool[1000001]; for(int i = 0; i < 1000001; i++) prime[i] = true; prime[0] = prime[1] = false; for(int i = 2; i * i <= 1000000; i++) { if (prime[i] == true) { for(int j = i * i; j <= 1000000; j += i) prime[j] = false; } } // Converting int to String for(int i = 2; i <= 1000000; i++) { if (prime[i] == true) primes.Add(String.Join("", i)); }}// A function to find the minimum// number of segments the given String// can be divided such that every// segment is a primestatic int splitIntoPrimes(String number){ int numLen = number.Length; // Declare a splitdp[] array // and initialize to -1 int []splitDP = new int[numLen + 1]; for(int i = 0; i < numLen + 1; i++) splitDP[i] = -1; // Call sieve function to store primes // in primes array HashSet<String> primes = new HashSet<String>(); getPrimesFromSeive(primes); // Build the DP table in a bottom-up manner for(int i = 1; i <= numLen; i++) { // If the prefix is prime then the prefix // will be found in the prime set if (i <= 6 && (primes.Contains (number.Substring(0, i)))) splitDP[i] = 1; // If the given prefix can be split into // primes, then for the remaining String // from i to j check if prime. If yes // calculate the minimum split till j if (splitDP[i] != -1) { for(int j = 1; j <= 6 && i + j <= numLen; j++) { // To check if the subString from // i to j is a prime number or not if (primes.Contains(number.Substring(i, j))) { // If it is a prime, then update // the dp array if (splitDP[i + j] == -1) splitDP[i + j] = 1 + splitDP[i]; else splitDP[i + j] = Math.Min(splitDP[i + j], 1 + splitDP[i]); } } } } // Return the minimum number of // splits for the entire String return splitDP[numLen];}public static void Main(String[] args){ Console.Write(splitIntoPrimes("13499315") + "\n"); Console.Write(splitIntoPrimes("43") + "\n");}}// This code is contributed by sapnasingh4991 |
Javascript
<script>// Javascript implementation of the above approach// Function to precompute all the primes// upto 1000000 and store it in a set// using Sieve of Eratosthenesfunction getPrimesFromSeive(primes){ let prime = new Array(1000001); for(let i=0;i<prime.length;i++) { prime[i]=true; } prime[0] = prime[1] = false; for (let i = 2; i * i <= 1000000; i++) { if (prime[i] == true) { for (let j = i * i; j <= 1000000; j += i) prime[j] = false; } } // Here to_String() is used // for converting int to String for (let i = 2; i <= 1000000; i++) { if (prime[i] == true) primes.add((i).toString()); }}// A function to find the minimum// number of segments the given String// can be divided such that every// segment is a primefunction splitIntoPrimes(number){ let numLen = number.length; // Declare a splitdp[] array // and initialize to -1 let splitDP = new Array(numLen + 1); for(let i=0;i<splitDP.length;i++) { splitDP[i]=-1; } // Call sieve function to store primes in // primes array let primes = new Set(); getPrimesFromSeive(primes); // Build the DP table in a bottom-up manner for (let i = 1; i <= numLen; i++) { // If the prefix is prime then the prefix // will be found in the prime set if (i <= 6 && (primes.has(number.substring(0, i)))) splitDP[i] = 1; // If the Given Prefix can be split into Primes // then for the remaining String from i to j // Check if Prime. If yes calculate // the minimum split till j if (splitDP[i] != -1) { for (let j = 1; j <= 6 && i + j <= numLen; j++) { // To check if the subString from i to j // is a prime number or not if (primes.has(number.substring(i, i+j))) { // If it is a prime, then update the dp array if (splitDP[i + j] == -1) splitDP[i + j] = 1 + splitDP[i]; else splitDP[i + j] = Math.min(splitDP[i + j], 1 + splitDP[i]); } } } } // Return the minimum number of splits // for the entire String return splitDP[numLen];}document.write(splitIntoPrimes("13499315")+ "<br>");document.write(splitIntoPrimes("43")+ "<br>");// This code is contributed by rag2127</script> |
Output:
3 1
Time Complexity:
- This is the most efficient method as this runs in O(N) time complexity where N is the length of the input string.
- Since the sieve of Eratosthenes has a run time of O(N*log(log(N))) and the list of primes up to 106, the precomputation complexity can be calculated. However, since this is performed only once for any number of strings, it is not counted in calculating time complexity.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
