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

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++
C
Java
Python3
C#
PHP
Javascript
// 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 codeint 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 codeint 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 codearr = [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 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.
<?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)
Space Complexity: O(1)
nitin mittal
MuraliSuresh
mv15
be1398
sakshamdoda
divyesh072019
prasanna1995
adityakumar129
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Sorting
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