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# Minimize a string by removing all occurrences of another string

• Difficulty Level : Medium
• Last Updated : 27 May, 2022

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 ` `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 stackOfChar = ``new` `Stack();` `    ``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

 ``

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.

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