# Maximum number of given operations to remove the entire string

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: 4
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: 1

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the maximum number ` `// of given operations required to ` `// remove the given string entirely ` `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 = 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 code ` `int` `main() ` `{ ` `    ``string s = ``"abababab"``; ` ` `  `    ``cout << find(s); ` ` `  `    ``return` `0; ` `} `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the maximum number ` `# of given operations required to ` `# remove the given entirely ` `def` `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 code ` `s ``=` `"abababab"` `print``(find(s)) ` ` `  `# This code is contributed by Mohit Kumar `

Output:

```4
```

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Second year Department of Information Technology Jadavpur University

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