Skip to content
Related Articles

Related Articles

Improve Article
Minimize a string by removing all occurrences of another string
  • Difficulty Level : Medium
  • Last Updated : 31 May, 2021

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:



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 occurences 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 occurences 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 occurences 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 occurences 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 occurences 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)
Auxiliary Space: O(N)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live 




My Personal Notes arrow_drop_up
Recommended Articles
Page :