# Split the given string into Primes : Digit DP

• Difficulty Level : Hard
• Last Updated : 02 Jun, 2021

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:
Explanation:
The number can be segmented as [13499, 31, 5]

Input: str = “43”
Output:
Explanation:
The number can be segmented as 

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 ``#include ``using` `namespace` `std;` `// Function to check whether a string``// is a prime number or not``bool` `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 prime``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 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 code``int` `main()``{``    ``cout << splitIntoPrimes(``"13499315"``) << ``"\n"``;``    ``cout << splitIntoPrimes(``"43"``) << ``"\n"``;``    ``return` `0;``}`

## Java

 `// Java implementation of the above approach``import` `java.util.*;` `class` `GFG{`` ` `// Function to check whether a String``// is a prime number or not``static` `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 prime``static` `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 code``public` `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 not``def` `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 prime``def` `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 code``print``(splitIntoPrimes(``"13499315"``))``print``(splitIntoPrimes(``"43"``))`

## C#

 `// C# implementation of the above approach``using` `System;` `class` `GFG{``  ` `// Function to check whether a String``// is a prime number or not``static` `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 prime``static` `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 code``public` `static` `void` `Main(String[] args)``{``    ``Console.Write(splitIntoPrimes(``"13499315"``)+ ``"\n"``);``    ``Console.Write(splitIntoPrimes(``"43"``)+ ``"\n"``);``}``}` `// This code is contributed by sapnasingh4991`

## Javascript

 ``

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 ``using` `namespace` `std;` `// Function to check whether a string``// is a prime number or not``bool` `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 prime``int` `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 code``int` `main()``{``    ``cout << splitIntoPrimes(``"13499315"``) << ``"\n"``;``    ``cout << splitIntoPrimes(``"43"``) << ``"\n"``;``    ``return` `0;``}`

## Java

 `// Java implementation of the above approach` `import` `java.util.*;` `class` `GFG{`` ` `// Function to check whether a String``// is a prime number or not``static` `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 prime``static` `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 code``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

 `# Python 3 implementation of the above approach``from` `math ``import` `sqrt` `# Function to check whether a string``# is a prime number or not``def` `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 prime``def` `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 code``if` `__name__ ``=``=` `'__main__'``:``    ``print``(splitIntoPrimes(``"13499315"``))``    ``print``(splitIntoPrimes(``"43"``))` `# This code is contributed by Surendra_Gangwar`

## C#

 `// C# implementation of the above approach``using` `System;` `class` `GFG{``  ` `// Function to check whether a String``// is a prime number or not``static` `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 prime``static` `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 code``public` `static` `void` `Main(String[] args)``{``    ``Console.Write(splitIntoPrimes(``"13499315"``)+ ``"\n"``);``    ``Console.Write(splitIntoPrimes(``"43"``)+ ``"\n"``);``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``

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 ``using` `namespace` `std;` `// Function to precompute all the primes``// upto 1000000 and store it in a set``// using Sieve of Eratosthenes``void` `getPrimesFromSeive(set& primes)``{``    ``bool` `prime;``    ``memset``(prime, ``true``, ``sizeof``(prime));``    ``prime = prime = ``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 prime``int` `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 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 approach` `import` `java.util.*;` `class` `GFG{`` ` `// Function to precompute all the primes``// upto 1000000 and store it in a set``// using Sieve of Eratosthenes``static` `void` `getPrimesFromSeive(HashSet 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 prime``static` `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 primes = ``new` `HashSet();``    ``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 Eratosthenes``def` `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 prime``def` `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 code``print``(splitIntoPrimes(``"13499315"``))``print``(splitIntoPrimes(``"43"``))` `# This code is contributed by chitranayal`

## C#

 `// C# implementation of the above approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG{` `// Function to precompute all the primes``// upto 1000000 and store it in a set``// using Sieve of Eratosthenes``static` `void` `getPrimesFromSeive(HashSet primes)``{``    ``bool` `[]prime = ``new` `bool``;``    ` `    ``for``(``int` `i = 0; i < 1000001; i++)``       ``prime[i] = ``true``;``    ``prime = prime = ``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 prime``static` `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 primes = ``new` `HashSet();``    ``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

 ``

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.

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