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++ 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 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 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# 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 |
<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> |
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.