# Maximum number of partitions that can be sorted individually to make sorted

• Difficulty Level : Medium
• Last Updated : 28 Apr, 2022

Given an array arr[] of size n such that elements of arr[] in range [0, 1, ..n-1] where every number is present at most once. Our task is to divide the array into maximum number of partitions that can be sorted individually, then concatenated to make the whole array sorted.
Examples :

```Input : arr[] = [2, 1, 0, 3]
Output : 2
If divide arr[] into two partitions
{2, 1, 0} and {3}, sort then and concatenate
then, we get the whole array sorted.

Input : arr[] = [2, 1, 0, 3, 4, 5]
Output : 4
The maximum number of partitions are four, we
get these partitions as {2, 1, 0}, {3}, {4}
and {5}

Input : arr[] = [0, 1, 2, 3, 4, 5]
Output : 6
The maximum number of partitions are six, we
get these partitions as {0}, {1}, {2}, {3}, {4}
and {5}```

The idea is based on the fact that if an element at i is maximum of prefix arr[0..i], then we can make a partition ending with i.

## C++

 `// CPP program to find Maximum number of partitions such``// that we can get a sorted array.``#include ``using` `namespace` `std;` `// Function to find maximum partitions.``int` `maxPartitions(``int` `arr[], ``int` `n)``{``    ``int` `ans = 0, max_so_far = 0;``    ``for` `(``int` `i = 0; i < n; ++i) {` `        ``// Find maximum in prefix arr[0..i]``        ``max_so_far = max(max_so_far, arr[i]);` `        ``// If maximum so far is equal to index, we can make``        ``// a new partition ending at index i.``        ``if` `(max_so_far == i)``            ``ans++;``    ``}``    ``return` `ans;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 1, 0, 2, 3, 4 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``cout << maxPartitions(arr, n);``    ``return` `0;``}` `// This code is contributed by Aditya Kumar (adityakumar129)`

## C

 `// C program to find Maximum number of partitions such that``// we can get a sorted array.``#include ` `// Find maximum between two numbers.``int` `max(``int` `num1, ``int` `num2)``{``    ``return` `(num1 > num2) ? num1 : num2;``}` `// Function to find maximum partitions.``int` `maxPartitions(``int` `arr[], ``int` `n)``{``    ``int` `ans = 0, max_so_far = 0;``    ``for` `(``int` `i = 0; i < n; ++i) {` `        ``// Find maximum in prefix arr[0..i]``        ``max_so_far = max(max_so_far, arr[i]);` `        ``// If maximum so far is equal to index, we can make``        ``// a new partition ending at index i.``        ``if` `(max_so_far == i)``            ``ans++;``    ``}``    ``return` `ans;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 1, 0, 2, 3, 4 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``printf``(``"%d"``, maxPartitions(arr, n));``    ``return` `0;``}` `// This code is contributed by Aditya Kumar (adityakumar129)`

## Java

 `// java program to find Maximum number of partitions such``// that we can get a sorted array` `import` `java.io.*;` `class` `GFG {``    ``// Function to find maximum partitions.``    ``static` `int` `maxPartitions(``int` `arr[], ``int` `n)``    ``{``        ``int` `ans = ``0``, max_so_far = ``0``;``        ``for` `(``int` `i = ``0``; i < n; ++i) {` `            ``// Find maximum in prefix arr[0..i]``            ``max_so_far = Math.max(max_so_far, arr[i]);` `            ``// If maximum so far is equal to index, we can``            ``// make a new partition ending at index i.``            ``if` `(max_so_far == i)``                ``ans++;``        ``}``        ``return` `ans;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``1``, ``0``, ``2``, ``3``, ``4` `};``        ``int` `n = arr.length;``        ``System.out.println(maxPartitions(arr, n));``    ``}``}` `// This code is contributed by Aditya Kumar (adityakumar129)`

## Python3

 `# Python3 program to find Maximum``# number of partitions such that``# we can get a sorted array.` `# Function to find maximum partitions.``def` `maxPartitions(arr, n):` `    ``ans ``=` `0``; max_so_far ``=` `0``    ``for` `i ``in` `range``(``0``, n):` `        ``# Find maximum in prefix arr[0..i]``        ``max_so_far ``=` `max``(max_so_far, arr[i])` `        ``# If maximum so far is equal to``        ``# index, we can make a new partition``        ``# ending at index i.``        ``if` `(max_so_far ``=``=` `i):``            ``ans ``+``=` `1``    ` `    ``return` `ans` `# Driver code``arr ``=` `[``1``, ``0``, ``2``, ``3``, ``4``]``n ``=` `len``(arr)``print``(maxPartitions(arr, n))` `# This code is contributed by Smitha Dinesh Semwal.`

## C#

 `// C# program to find Maximum number of partitions``// such that we can get a sorted array``using` `System;` `class` `GFG``{``    ``// Function to find maximum partitions.``    ``static` `int` `maxPartitions(``int` `[]arr, ``int` `n)``    ``{``        ``int` `ans = 0, max_so_far = 0;``        ``for` `(``int` `i = 0; i < n; ++i)``        ``{``    ` `            ``// Find maximum in prefix arr[0..i]``            ``max_so_far = Math.Max(max_so_far, arr[i]);``    ` `            ``// If maximum so far is equal to index,``            ``// we can make a new partition ending at``            ``// index i.``            ``if` `(max_so_far == i)``                ``ans++;``        ``}``        ``return` `ans;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main ()``    ``{``        ``int` `[]arr = { 1, 0, 2, 3, 4 };``        ``int` `n = arr.Length;``        ``Console.Write (maxPartitions(arr, n));``            ` `    ``}``}` `// This code is contributed by nitin mittal.`

## PHP

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## Javascript

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Output:

`4`

Time Complexity: O(N)

Space Complexity: O(1)

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