Maximum number of given operations to remove the entire string

• Last Updated : 07 May, 2021

Given a string str containing lowercase English characters, we can perform the following two operations on the given string:

1. Remove the entire string.
2. Remove a prefix of the string str[0…i] only if it is equal to the sub-string str[(i + 1)…(2 * i + 1)].

The task is to find the maximum number of operations required to delete the entire string.

Examples:

Input: str = “abababab”
Output:
Operation 1: Delete prefix “ab” and the string becomes “ababab”.
Operation 2: Delete prefix “ab” and the string becomes “abab”.
Operation 3: Delete prefix “ab”, str = “ab”.
Operation 4: Delete the entire string.

Input: s = “abc”
Output:

Approach: In order to maximize the number of operations, the prefix that is to be deleted must be of minimum length i.e. starting from a single character and appending the successive characters one by one find the minimum length prefix that satisfies the given condition. One operation will be required to delete this prefix say str[0…i] then recursively call the same function to find the result for the substring str[i+1…2*i+1]. If there is no such prefix that could be deleted then return 1 as the only possible operation is to delete the entire string.

Below is the implementation of the above approach:

C++

 // C++ implementation of the approach#includeusing namespace std; // Function to return the maximum number// of given operations required to// remove the given string entirelyint find(string s){     // If length of the string is zero    if (s.length() == 0)        return 0;     // Single operation can delete the entire string    int c = 1;     // To store the prefix of the string    // which is to be deleted    string d = "";     for (int i = 0; i < s.length(); i++) {         // Prefix s[0..i]        d += s[i];         // To store the substring s[i+1...2*i+1]        string s2 = s.substr(i + 1, d.length());         // If the prefix s[0...i] can be deleted        if (s2 == d) {             // 1 operation to remove the current prefix            // and then recursively find the count of            // operations for the substring s[i+1...n-1]            c = 1 + find(s.substr(i + 1));            break;        }    }     // Entire string has to be deleted    return c;} // Driver codeint main(){    string s = "abababab";     cout << find(s);     return 0;}

Java

 // Java implementation of the approachimport java.util.*; class GFG{   // Function to return the maximum number// of given operations required to// remove the given string entirelystatic int find(String s){         // If length of the string is zero    if (s.length() == 0)        return 0;       // Single operation can delete    // the entire string    int c = 1;       // To store the prefix of the string    // which is to be deleted    String d = "";       for(int i = 0; i < s.length(); i++)    {                 // Prefix s[0..i]        d += s.charAt(i);                 // To store the substring s[i+1...2*i+1]        String s2 = "";                 if (i + d.length() < s.length())        {            s2 = s.substring(i + 1,               d.length() + (i + 1));        }                 // If the prefix s[0...i] can be deleted        if (s2.equals(d))        {                         // 1 operation to remove the current prefix            // and then recursively find the count of            // operations for the substring s[i+1...n-1]            c = 1 + find(s.substring(i + 1));            break;        }    }       // Entire string has to be deleted    return c;}   // Driver codepublic static void main(String[] args){    String s = "abababab";         System.out.print(find(s));}} // This code is contributed by pratham76

Python3

 # Python3 implementation of the approach # Function to return the maximum number# of given operations required to# remove the given entirelydef find(s):     # If length of the is zero    if (len(s) == 0):        return 0     # Single operation can delete    # the entire string    c = 1     # To store the prefix of the string    # which is to be deleted    d = ""     for i in range(len(s)):         # Prefix s[0..i]        d += s[i]         # To store the subs[i+1...2*i+1]        s2 = s[i + 1:i + 1 + len(d)]         # If the prefix s[0...i] can be deleted        if (s2 == d):             # 1 operation to remove the current prefix            # and then recursively find the count of            # operations for the subs[i+1...n-1]            c = 1 + find(s[i + 1:])            break     # Entire has to be deleted    return c # Driver codes = "abababab"print(find(s)) # This code is contributed by Mohit Kumar

C#

 // C# implementation of the approachusing System;class GFG{  // Function to return the maximum number// of given operations required to// remove the given string entirelystatic int find(string s){      // If length of the string is zero    if (s.Length == 0)        return 0;      // Single operation can delete the entire string    int c = 1;      // To store the prefix of the string    // which is to be deleted    string d = "";      for (int i = 0; i < s.Length; i++)    {          // Prefix s[0..i]        d += s[i];          // To store the substring s[i+1...2*i+1]       string s2 = "";       if(i + d.Length < s.Length)       {           s2 = s.Substring(i + 1, d.Length);       }               // If the prefix s[0...i] can be deleted       if (s2 == d)       {            // 1 operation to remove the current prefix            // and then recursively find the count of            // operations for the substring s[i+1...n-1]            c = 1 + find(s.Substring(i + 1));            break;       }    }      // Entire string has to be deleted    return c;}      // Driver code    public static void Main(string[] args)    {       string s = "abababab";       Console.Write(find(s));    }} // This code is contributed by rutvik

Javascript


Output:
4

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