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

Given an array arr[] of size n such that elements of arr[] in range [0, 1, ..n-1]. 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}
```

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

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

 `// 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[0]); ` `    ``cout << maxPartitions(arr, n); ` `    ``return` `0; ` `} `

 `// 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 vt_m. `

 `# 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# 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. `

 ` `

Output:
```4
```

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