# 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] 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}
```

## 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 at i is maximum of prefix arr[0..i], then we can make a partition ending with 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); ` `    ``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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