Split the string into minimum parts such that each part is in the another string

Given two strings A and B, the task is to split the string A into the minimum number of substrings such that each substring is in the string B.

Note: If there is no way to split the string, then print -1 

Examples:

Input: A = “abcdab”, B = “dabc” 
Output: 2 
Explanation:
The two substrings of A which is also present in B are – 
{“abc”, “dab”}

Input: A = “abcde”, B = “edcb” 
Output: -1 
Explanation:
There is no way to split the string A into substring 
Such that each string is also present in the B.

Approach:

• Construct a trie of every substring of B.
• After that, we’ll use Dynamic programming to find the minimum number of parts to break the string A such that every part is a substring of B.

Recurrence Relation in Dynamic Programming:

dp[i] = minimum number of parts to break the string A up to ith prefix.

dp[0] = 0
for i in {0, S1.length}
    for j in {i, S1.length}
        if(S1[i, ... j] is found in trie
            dp[j] = min(dp[j], dp[i] + 1);
        else
            break;

dp[S1.length] is the
minimum number of parts.

Below is the implementation of the above approach:

C++

// C++ implementation to split the
// string into minimum number of
// parts such that each part is also
// present in the another string

#include <bits/stdc++.h>
using namespace std;

const int INF = 1e9 + 9;

// Node of Trie
struct TrieNode {
    TrieNode* child[26] = { NULL };
};

// Function to insert a node in the
// Trie Data Structure
void insert(int idx, string& s,
            TrieNode* root)
{
    TrieNode* temp = root;
    for (int i = idx; i < s.length(); i++) {

        // Inserting every character from idx
        // till end to string into trie
        if (temp->child[s[i] - 'a'] == NULL)

            // if there is no edge corresponding
            // to the ith character,
            // then make a new node
            temp->child[s[i] - 'a'] = new TrieNode;

        temp = temp->child[s[i] - 'a'];
    }
}

// Function to find the minimum
// number of parts such that each
// part is present into another string
int minCuts(string S1, string S2)
{
    int n1 = S1.length();
    int n2 = S2.length();

    // Making a new trie
    TrieNode* root = new TrieNode;

    for (int i = 0; i < n2; i++) {

        // Inserting every substring
        // of S2 in trie
        insert(i, S2, root);
    }

    // Creating dp array and
    // init it with infinity
    vector<int> dp(n1 + 1, INF);

    // Base Case
    dp[0] = 0;
    for (int i = 0; i < n1; i++) {

        // Starting the cut from ith character
        // taking temporary node pointer
        // for checking whether the substring
        // [i, j) is present in trie of not
        TrieNode* temp = root;
        for (int j = i + 1; j <= n1; j++) {
            if (temp->child[S1[j - 1] - 'a'] == NULL)

                // if the jth character is not in trie
                // we'll break
                break;

            // Updating the ending of
            // jth character with dp[i] + 1
            dp[j] = min(dp[j], dp[i] + 1);

            // Descending the trie pointer
            temp = temp->child[S1[j - 1] - 'a'];
        }
    }

    // Answer not possible
    if (dp[n1] >= INF)
        return -1;
    else
        return dp[n1];
}

// Driver Code
int main()
{
    string S1 = "abcdab";
    string S2 = "dabc";

    cout << minCuts(S1, S2);
}

Java

// Java implementation to split the
// String into minimum number of
// parts such that each part is also
// present in the another String
import java.util.*;

class GFG{

static int INF = (int)(1e9 + 9);

// Node of Trie
static class TrieNode
{
    TrieNode []child = new TrieNode[26];
};

// Function to insert a node in the
// Trie Data Structure
static void insert(int idx, String s,
                   TrieNode root)
{
    TrieNode temp = root;
    for(int i = idx; i < s.length(); i++)
    {

        // Inserting every character from idx
        // till end to String into trie
        if (temp.child[s.charAt(i) - 'a'] == null)

            // If there is no edge corresponding
            // to the ith character,
            // then make a new node
            temp.child[s.charAt(i) - 'a'] = new TrieNode();

        temp = temp.child[s.charAt(i) - 'a'];
    }
}

// Function to find the minimum
// number of parts such that each
// part is present into another String
static int minCuts(String S1, String S2)
{
    int n1 = S1.length();
    int n2 = S2.length();

    // Making a new trie
    TrieNode root = new TrieNode();

    for(int i = 0; i < n2; i++)
    {

        // Inserting every subString
        // of S2 in trie
        insert(i, S2, root);
    }

    // Creating dp array and
    // init it with infinity
    int []dp = new int[n1 + 1];
    Arrays.fill(dp, INF);

    // Base Case
    dp[0] = 0;

    for(int i = 0; i < n1; i++)
    {

        // Starting the cut from ith character
        // taking temporary node pointer
        // for checking whether the subString
        // [i, j) is present in trie of not
        TrieNode temp = root;
        for(int j = i + 1; j <= n1; j++)
        {
            if (temp.child[S1.charAt(j - 1) - 'a'] == null)

                // If the jth character is not in trie
                // we'll break
                break;

            // Updating the ending of
            // jth character with dp[i] + 1
            dp[j] = Math.min(dp[j], dp[i] + 1);

            // Descending the trie pointer
            temp = temp.child[S1.charAt(j - 1) - 'a'];
        }
    }

    // Answer not possible
    if (dp[n1] >= INF)
        return -1;
    else
        return dp[n1];
}

// Driver Code
public static void main(String[] args)
{
    String S1 = "abcdab";
    String S2 = "dabc";

    System.out.print(minCuts(S1, S2));
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 implementation to split the
# string into minimum number of
# parts such that each part is also
# present in the another string
INF = 1e9 + 9

# Node of Trie
class TrieNode():
    def __init__(self):

        self.child = [None] * 26

# Function to insert a node in the
# Trie Data Structure
def insert(idx, s, root):

    temp = root

    for i in range(idx, len(s)):

        # Inserting every character from idx
        # till end to string into trie
        if temp.child[ord(s[i]) -
                      ord('a')] == None:

            # If there is no edge corresponding
            # to the ith character,
            # then make a new node
            temp.child[ord(s[i]) -
                       ord('a')] = TrieNode()

        temp = temp.child[ord(s[i]) - ord('a')]

# Function to find the minimum
# number of parts such that each
# part is present into another string
def minCuts(S1, S2):

    n1 = len(S1)
    n2 = len(S2)

    # Making a new trie
    root = TrieNode()

    for i in range(n2):

        # Inserting every substring
        # of S2 in trie
        insert(i, S2, root)

    # Creating dp array and
    # init it with infinity
    dp = [INF] * (n1 + 1)

    # Base Case
    dp[0] = 0

    for i in range(n1):

        # Starting the cut from ith character
        # taking temporary node pointer
        # for checking whether the substring
        # [i, j) is present in trie of not
        temp = root
        for j in range(i + 1, n1 + 1):
            if temp.child[ord(S1[j - 1]) -
                          ord('a')] == None:

                # If the jth character is not
                # in trie we'll break
                break

            # Updating the ending of
            # jth character with dp[i] + 1
            dp[j] = min(dp[j], dp[i] + 1)

            # Descending the trie pointer
            temp = temp.child[ord(S1[j - 1]) -
                              ord('a')]

    # Answer not possible
    if dp[n1] >= INF:
        return -1
    else:
        return dp[n1]

# Driver Code
S1 = "abcdab"
S2 = "dabc"

print(minCuts(S1, S2))

# This code is contributed by Shivam Singh

C#

// C# implementation to split the
// String into minimum number of
// parts such that each part is also
// present in the another String
using System;
class GFG{

static int INF = (int)(1e9 + 9);

// Node of Trie
class TrieNode
{
    public TrieNode []child =
                    new TrieNode[26];
};

// Function to insert a node in the
// Trie Data Structure
static void insert(int idx, String s,
                   TrieNode root)
{
    TrieNode temp = root;
    for(int i = idx; i < s.Length; i++)
    {
        // Inserting every character from idx
        // till end to String into trie
        if (temp.child[s[i] - 'a'] == null)

            // If there is no edge corresponding
            // to the ith character,
            // then make a new node
            temp.child[s[i] - 'a'] =
                       new TrieNode();

        temp = temp.child[s[i] - 'a'];
    }
}

// Function to find the minimum
// number of parts such that each
// part is present into another String
static int minCuts(String S1, String S2)
{
    int n1 = S1.Length;
    int n2 = S2.Length;

    // Making a new trie
    TrieNode root = new TrieNode();

    for(int i = 0; i < n2; i++)
    {
        // Inserting every subString
        // of S2 in trie
        insert(i, S2, root);
    }

    // Creating dp array and
    // init it with infinity
    int []dp = new int[n1 + 1];
    for(int i = 0; i <= n1; i++)
        dp[i] = INF;

    // Base Case
    dp[0] = 0;

    for(int i = 0; i < n1; i++)
    {
        // Starting the cut from ith character
        // taking temporary node pointer
        // for checking whether the subString
        // [i, j) is present in trie of not
        TrieNode temp = root;
        for(int j = i + 1; j <= n1; j++)
        {
            if (temp.child[S1[j-1] - 'a'] == null)

                // If the jth character is not in trie
                // we'll break
                break;

            // Updating the ending of
            // jth character with dp[i] + 1
            dp[j] = Math.Min(dp[j], dp[i] + 1);

            // Descending the trie pointer
            temp = temp.child[S1[j - 1] - 'a'];
        }
    }

    // Answer not possible
    if (dp[n1] >= INF)
        return -1;
    else
        return dp[n1];
}

// Driver Code
public static void Main(String[] args)
{
    String S1 = "abcdab";
    String S2 = "dabc";
    Console.Write(minCuts(S1, S2));
}
}
// This code is contributed by shikhasingrajput

Javascript

<script>
    // Javascript implementation to split the
    // String into minimum number of
    // parts such that each part is also
    // present in the another String

    let INF = (1e9 + 9);

    // Node of Trie
    class Node
    {
        constructor()
        {

        }
    }

    function TrieNode()
    {
        let temp = new Node();
        temp.child = new Node(26);
        for(let i = 0; i < 26; i++)
        {
            temp.child[i] = null;
        }
        return temp;
    }

    // Function to insert a node in the
    // Trie Data Structure
    function insert(idx, s, root)
    {
        let temp = root;
        for(let i = idx; i < s.length; i++)
        {

            // Inserting every character from idx
            // till end to String into trie
            if (temp.child[s[i].charCodeAt() - 'a'.charCodeAt()] == null)

                // If there is no edge corresponding
                // to the ith character,
                // then make a new node
                temp.child[s[i].charCodeAt() - 'a'.charCodeAt()] = new TrieNode();

            temp = temp.child[s[i].charCodeAt() - 'a'.charCodeAt()];
        }
    }

    // Function to find the minimum
    // number of parts such that each
    // part is present into another String
    function minCuts(S1, S2)
    {
        let n1 = S1.length;
        let n2 = S2.length;

        // Making a new trie
        let root = new TrieNode();

        for(let i = 0; i < n2; i++)
        {

            // Inserting every subString
            // of S2 in trie
            insert(i, S2, root);
        }

        // Creating dp array and
        // init it with infinity
        let dp = new Array(n1 + 1);
        dp.fill(INF);

        // Base Case
        dp[0] = 0;

        for(let i = 0; i < n1; i++)
        {

            // Starting the cut from ith character
            // taking temporary node pointer
            // for checking whether the subString
            // [i, j) is present in trie of not
            let temp = root;
            for(let j = i + 1; j <= n1; j++)
            {
                if (temp.child[S1[j - 1].charCodeAt() - 'a'.charCodeAt()] == null)

                    // If the jth character is not in trie
                    // we'll break
                    break;

                // Updating the ending of
                // jth character with dp[i] + 1
                dp[j] = Math.min(dp[j], dp[i] + 1);

                // Descending the trie pointer
                temp = temp.child[S1[j - 1].charCodeAt() - 'a'.charCodeAt()];
            }
        }

        // Answer not possible
        if (dp[n1] >= INF)
            return -1;
        else
            return dp[n1];
    }

    let S1 = "abcdab";
    let S2 = "dabc";

    document.write(minCuts(S1, S2));

    // This code is contributed by mukesh07.
</script>

Output:

2

Time complexity: O(|S1||S2|^2) where |S1| and |S2| are the lengths of S1 and S2, respectively.
Auxiliary Space: The space complexity of the program is dominated by the size of the Trie data structure. In the worst case, the size of the Trie will be proportional to the size of S2. So, the space complexity of the program is O(|S2|)