Minimize a string by removing all occurrences of another string

Given two strings S1 and S2 of length N and M respectively, consisting of lowercase letters, the task is to find the minimum length to which S1 can be reduced by removing all occurrences of the string S2 from the string S1.

Examples:

Input: S1 =”fffoxoxoxfxo”, S2 = “fox”;
Output: 3
Explanation:
By removing “fox” starting from index 2, the string modifies to “ffoxoxfxo”.
By removing “fox” starting from index 1, the string modifies to “foxfxo”.
By removing “fox” starting from index 0, the string modifies to “fxo”.
Therefore, the minimum length of string S1 after removing all occurrences of S2 is 3.

Input: S1 =”abcd”, S2 = “pqr”
Output: 4

Approach: The idea to solve this problem is to use Stack Data Structure. Follow the steps below to solve the given problem:

• Initialize a stack to store the characters of the string S1 in it.
• Traverse the given string S1 and perform the following operations:
  • Push the current character in the stack.
  • If the size of the stack is at least M, then check if the top M characters of the stack form the string S2 or not. If found to be true, then remove those characters.
• After completing the above steps, the remaining size of the stack is the required minimum length.

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 minimum length
// to which string str can be reduced to
// by removing all occurrences of string K
int minLength(string str, int N,
              string K, int M)
{

    // Initialize stack of characters
    stack<char> stackOfChar;

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

        // Push character into the stack
        stackOfChar.push(str[i]);

        // If stack size >= K.size()
        if (stackOfChar.size() >= M) {

            // Create empty string to
            // store characters of stack
            string l = "";

            // Traverse the string K in reverse
            for (int j = M - 1; j >= 0; j--) {

                // If any of the characters
                // differ, it means that K
                // is not present in the stack
                if (K[j] != stackOfChar.top()) {

                    // Push the elements
                    // back into the stack
                    int f = 0;
                    while (f != l.size()) {

                        stackOfChar.push(l[f]);
                        f++;
                    }

                    break;
                }

                // Store the string
                else {

                    l = stackOfChar.top()
                        + l;

                    // Remove top element
                    stackOfChar.pop();
                }
            }
        }
    }

    // Size of stack gives the
    // minimized length of str
    return stackOfChar.size();
}

// Driver Code
int main()
{
    string S1 = "fffoxoxoxfxo";
    string S2 = "fox";

    int N = S1.length();
    int M = S2.length();

    // Function Call
    cout << minLength(S1, N, S2, M);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG
{

// Function to find the minimum length
// to which String str can be reduced to
// by removing all occurrences of String K
static int minLength(String str, int N,
              String K, int M)
{

    // Initialize stack of characters
    Stack<Character> stackOfChar = new Stack<Character>();

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

        // Push character into the stack
        stackOfChar.add(str.charAt(i));

        // If stack size >= K.size()
        if (stackOfChar.size() >= M)
        {

            // Create empty String to
            // store characters of stack
            String l = "";

            // Traverse the String K in reverse
            for (int j = M - 1; j >= 0; j--)
            {

                // If any of the characters
                // differ, it means that K
                // is not present in the stack
                if (K.charAt(j) != stackOfChar.peek())
                {

                    // Push the elements
                    // back into the stack
                    int f = 0;
                    while (f != l.length())
                    {
                        stackOfChar.add(l.charAt(f));
                        f++;
                    }

                    break;
                }

                // Store the String
                else
                {
                    l = stackOfChar.peek()
                        + l;

                    // Remove top element
                    stackOfChar.pop();
                }
            }
        }
    }

    // Size of stack gives the
    // minimized length of str
    return stackOfChar.size();
}

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

    int N = S1.length();
    int M = S2.length();

    // Function Call
    System.out.print(minLength(S1, N, S2, M));
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach

# Function to find the minimum length
# to which string str can be reduced to
# by removing all occurrences of string K
def minLength(Str, N, K, M) :

    # Initialize stack of characters
    stackOfChar = []

    for i in range(N) :

        # Push character into the stack
        stackOfChar.append(Str[i])

        # If stack size >= K.size()
        if (len(stackOfChar) >= M) :

            # Create empty string to
            # store characters of stack
            l = ""

            # Traverse the string K in reverse
            for j in range(M - 1, -1, -1) :

                # If any of the characters
                # differ, it means that K
                # is not present in the stack
                if (K[j] != stackOfChar[-1]) :

                    # Push the elements
                    # back into the stack
                    f = 0
                    while (f != len(l)) :
                        stackOfChar.append(l[f])
                        f += 1

                    break

                # Store the string
                else :
                    l = stackOfChar[-1] + l

                    # Remove top element
                    stackOfChar.pop()

    # Size of stack gives the
    # minimized length of str
    return len(stackOfChar)

# Driver code
S1 = "fffoxoxoxfxo"
S2 = "fox"

N = len(S1)
M = len(S2)

# Function Call
print(minLength(S1, N, S2, M))

# This code is contributed by divyeshrabadiya07

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{

// Function to find the minimum length
// to which String str can be reduced to
// by removing all occurrences of String K
static int minLength(String str, int N,
              String K, int M)
{

    // Initialize stack of characters
    Stack<char> stackOfChar = new Stack<char>();
    for (int i = 0; i < N; i++)
    {

        // Push character into the stack
        stackOfChar.Push(str[i]);

        // If stack size >= K.Count
        if (stackOfChar.Count >= M)
        {

            // Create empty String to
            // store characters of stack
            String l = "";

            // Traverse the String K in reverse
            for (int j = M - 1; j >= 0; j--)
            {

                // If any of the characters
                // differ, it means that K
                // is not present in the stack
                if (K[j] != stackOfChar.Peek())
                {

                    // Push the elements
                    // back into the stack
                    int f = 0;
                    while (f != l.Length)
                    {
                        stackOfChar.Push(l[f]);
                        f++;
                    }
                    break;
                }

                // Store the String
                else
                {
                    l = stackOfChar.Peek()
                        + l;

                    // Remove top element
                    stackOfChar.Pop();
                }
            }
        }
    }

    // Size of stack gives the
    // minimized length of str
    return stackOfChar.Count;
}

// Driver Code
public static void Main(String[] args)
{
    String S1 = "fffoxoxoxfxo";
    String S2 = "fox";

    int N = S1.Length;
    int M = S2.Length;

    // Function Call
    Console.Write(minLength(S1, N, S2, M));
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// JavaScript program for the above approach

// Function to find the minimum length
// to which string str can be reduced to
// by removing all occurrences of string K
function minLength(str, N, K, M)
{

    // Initialize stack of characters
    var stackOfChar = [];

    for (var i = 0; i < N; i++) {

        // Push character into the stack
        stackOfChar.push(str[i]);

        // If stack size >= K.size()
        if (stackOfChar.length >= M) {

            // Create empty string to
            // store characters of stack
            var l = "";

            // Traverse the string K in reverse
            for (var j = M - 1; j >= 0; j--) {

                // If any of the characters
                // differ, it means that K
                // is not present in the stack
                if (K[j] != stackOfChar[stackOfChar.length-1]) {

                    // Push the elements
                    // back into the stack
                    var f = 0;
                    while (f != l.length) {

                        stackOfChar.push(l[f]);
                        f++;
                    }

                    break;
                }

                // Store the string
                else {

                    l = stackOfChar[stackOfChar.length-1]
                        + l;

                    // Remove top element
                    stackOfChar.pop();
                }
            }
        }
    }

    // Size of stack gives the
    // minimized length of str
    return stackOfChar.length;
}

// Driver Code
var S1 = "fffoxoxoxfxo";
var S2 = "fox";
var N = S1.length;
var M = S2.length;

// Function Call
document.write( minLength(S1, N, S2, M));

</script>

Output:

3

Time complexity: O(N*M), as we are using nested loops for traversing N*M times.
Auxiliary Space: O(N), as we are using extra space for stack.