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

Try it on GfG Practice

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 them 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}

Approach:

The given code finds the maximum number of partitions that can be made from an array of integers. The code defines a function "maxPartitions" that takes an array and its length as input. The function then uses a loop to iterate through each element of the array and find the maximum element in the prefix ending at the current index. If the maximum element is equal to the current index, the function increments the count of partitions as a new partition can be made ending at that index. Finally, the function returns the count of partitions. The main function initializes an array with values and calls the "maxPartitions" function with the array and its length as inputs. The result of the function is then printed to the console.

Algorithm to find maximum partitions:

1. Initialize a variable "ans" to store the count of partitions and a variable "max_so_far" to store the maximum element in the prefix of the array.
2. Loop through each element in the array.
   a. Set "max_so_far" to the maximum of "max_so_far" and the current element.
   b. If "max_so_far" is equal to the current index, increment the value of "ans" by 1.
3. Return the value of "ans" as the result.

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 <bits/stdc++.h>
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;
}

// This code is contributed by Aditya Kumar (adityakumar129)

// C program to find Maximum number of partitions such that
// we can get a sorted array.
#include <stdio.h>

// 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[0]);
    printf("%d", maxPartitions(arr, n));
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

// 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 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.

 

<?php
// PHP program to find Maximum
// number of partitions such 
// that we can get a sorted array.

// Function to find maximum partitions.
function maxPartitions($arr, $n)
{
    $ans = 0; 
    $max_so_far = 0;
    for ($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
{
    $arr = array(1, 0, 2, 3, 4);
    $n = sizeof($arr) / sizeof($arr[0]);
    echo maxPartitions($arr, $n);
    return 0;
}

// This code is contributed by nitin mittal
?>

<script>
    // Javascript program to find Maximum number of partitions
    // such that we can get a sorted array.
    
    // Function to find maximum partitions.
    function maxPartitions(arr, n)
    {
        let ans = 0, max_so_far = 0;
        for (let 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;
    }
    
    let arr = [ 1, 0, 2, 3, 4 ];
    let n = arr.length;
    document.write(maxPartitions(arr, n));
    
    // This code is contributed by divyesh072019.
</script>

4

Time Complexity: O(N)
Auxiliary Space: O(1)