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

• Difficulty Level : Medium
• Last Updated : 30 Dec, 2021

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}

Recommended: Please solve it on “PRACTICE” 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.

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 codeint main(){    int arr[] = { 1, 0, 2, 3, 4 };    int n = sizeof(arr) / sizeof(arr);    cout << maxPartitions(arr, n);    return 0;}

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 vt_m.

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 codearr = [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 arrayusing 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.



Javascript


Output:
4

Time Complexity: O(N)

Space Complexity: O(1)

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