Given a string. The task is to find the maximum number **P**, such that given **string** can be partitioned into P contiguous substrings such that any two adjacent substring must be different. More formally and .

**Examples:**

Input:str = “aabccd”

Output:4

Explanation:

We can divide the given string into four string, like “a”, “ab”, “c”, “cd”. We can not divide

it more than four parts, if we do then the condition will not

satisfy

Input:str = “aaaa”

Output:3

**Approach: **

- Here we only have to focus on the value of P, not upon finding those P substrings.
- We will solve it greedily, we always check the current string that we have with the previous string that has been taken already.
- If we found that both of them are same then we will go forward otherwise create a partition here and change the previous track of the string to current string, means we will treat this current string as the previous string for future comparison.

Below is the implementation of above Approach:

## C++

`// C++ implementation of the above approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Return the count of string ` `int` `maxPartition(string s) ` `{ ` ` ` `// P will store the answer ` ` ` `int` `n = s.length(), P = 0; ` ` ` ` ` `// Current will store current string ` ` ` `// Previous will store the previous ` ` ` `// string that has been taken already ` ` ` `string current = ` `""` `, previous = ` `""` `; ` ` ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` ` ` `// Add a character to current string ` ` ` `current += s[i]; ` ` ` ` ` `if` `(current != previous) { ` ` ` ` ` `// Here we will create a partition and ` ` ` `// update the previous string with ` ` ` `// current string ` ` ` `previous = current; ` ` ` ` ` `// Now we will clear the current string ` ` ` `current.clear(); ` ` ` ` ` `// Increment the count of partition. ` ` ` `P++; ` ` ` `} ` ` ` `} ` ` ` ` ` `return` `P; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` ` ` `string s = ` `"geeksforgeeks"` `; ` ` ` ` ` `int` `ans = maxPartition(s); ` ` ` ` ` `cout << ans << ` `"\n"` `; ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// Java implementation of the above approach ` `class` `GFG ` `{ ` `// Return the count of string ` `static` `int` `maxPartition(String s) ` `{ ` ` ` `// P will store the answer ` ` ` `int` `n = s.length(), P = ` `0` `; ` ` ` ` ` `// Current will store current string ` ` ` `// Previous will store the previous ` ` ` `// string that has been taken already ` ` ` `String current = ` `""` `, previous = ` `""` `; ` ` ` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) ` ` ` `{ ` ` ` ` ` `// Add a character to current string ` ` ` `current += s.charAt(i); ` ` ` ` ` `if` `(!current.equals(previous)) ` ` ` `{ ` ` ` ` ` `// Here we will create a partition and ` ` ` `// update the previous string with ` ` ` `// current string ` ` ` `previous = current; ` ` ` ` ` `// Now we will clear the current string ` ` ` `current = ` `""` `; ` ` ` ` ` `// Increment the count of partition. ` ` ` `P++; ` ` ` `} ` ` ` `} ` ` ` `return` `P; ` `} ` ` ` `// Driver code ` `public` `static` `void` `main (String[] args) ` `{ ` ` ` `String s = ` `"geeksforgeeks"` `; ` ` ` ` ` `int` `ans = maxPartition(s); ` ` ` ` ` `System.out.println(ans); ` `} ` `} ` ` ` `// This code is contributed by ihritik ` |

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## Python3

`# Python3 implementation of the above approach ` ` ` `# Return the count of string ` `def` `maxPartition(s): ` ` ` ` ` `# P will store the answer ` ` ` `n ` `=` `len` `(s) ` ` ` `P ` `=` `0` ` ` ` ` `# Current will store current string ` ` ` `# Previous will store the previous ` ` ` `# that has been taken already ` ` ` `current ` `=` `"" ` ` ` `previous ` `=` `"" ` ` ` ` ` `for` `i ` `in` `range` `(n): ` ` ` ` ` `# Add a character to current string ` ` ` `current ` `+` `=` `s[i] ` ` ` ` ` `if` `(current !` `=` `previous): ` ` ` ` ` `# Here we will create a partition and ` ` ` `# update the previous with ` ` ` `# current string ` ` ` `previous ` `=` `current ` ` ` ` ` `# Now we will clear the current string ` ` ` `current ` `=` `"" ` ` ` ` ` `# Increment the count of partition. ` ` ` `P ` `+` `=` `1` ` ` ` ` `return` `P ` ` ` `# Driver code ` `s ` `=` `"geeksforgeeks"` ` ` `ans ` `=` `maxPartition(s) ` ` ` `print` `(ans) ` ` ` `# This code is contributed by Mohit Kumar ` |

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## C#

`// C# implementation of the above approach ` `using` `System; ` `class` `GFG ` `{ ` `// Return the count of string ` `static` `int` `maxPartition(` `string` `s) ` `{ ` ` ` `// P will store the answer ` ` ` `int` `n = s.Length, P = 0; ` ` ` ` ` `// Current will store current string ` ` ` `// Previous will store the previous ` ` ` `// string that has been taken already ` ` ` `string` `current = ` `""` `, previous = ` `""` `; ` ` ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `{ ` ` ` ` ` `// Add a character to current string ` ` ` `current += s[i]; ` ` ` ` ` `if` `(!current.Equals(previous)) ` ` ` `{ ` ` ` ` ` `// Here we will create a partition and ` ` ` `// update the previous string with ` ` ` `// current string ` ` ` `previous = current; ` ` ` ` ` `// Now we will clear the current string ` ` ` `current = ` `""` `; ` ` ` ` ` `// Increment the count of partition. ` ` ` `P++; ` ` ` `} ` ` ` `} ` ` ` `return` `P; ` `} ` ` ` `// Driver code ` `public` `static` `void` `Main () ` `{ ` ` ` `string` `s = ` `"geeksforgeeks"` `; ` ` ` ` ` `int` `ans = maxPartition(s); ` ` ` ` ` `Console.WriteLine(ans); ` `} ` `} ` ` ` `// This code is contributed by ihritik ` |

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**Output:**

11

**Time Complexity:** O(N), where N is the length of the string.

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