Maximize partitions that if sorted individually makes the whole Array sorted

Given an array arr[]. The task is to divide arr[] into the maximum number of partitions, such that, those partitions if sorted individually make the whole array sorted.

Examples:

Input: arr[] = { 28, 9, 18, 32, 60, 50, 75, 70 }
Output: 4
Explanation: Following are the partitions in which the array is divided. 
If we divide arr[] into four partitions {28, 9, 18}, {32}, { 60, 50}, and {75, 70}, sort them and concatenate. 
Sorting all of them indivudually makes the whole array sorted. 
Hence, 4 is the answer.

Input: arr[] = { 2, 1, 0, 3, 4, 5 }
Output: 4

Approach: This problem is implementation-based. Follow the steps below to solve the given problem.  

• Create a maximum array that calculates the maximum element to the left till that index of the array.
• Create a minimum array that calculates the minimum element to the right till that index of the array.
• Iterate through the array, each time all elements to the leftMax[] are smaller (or equal) to all elements to the rightMax[], that means there is a new chunk, so increment the count by 1.
• Return count+1 as the final answer.

Below is the implementation of the above approach. 

C++

// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find maximum partitions.
int maxPartitions(vector<int>& arr)
{
    int N = arr.size();

    // To keep track of max
    // and min elements at every index
    vector<int> leftMax(arr.size());
    vector<int> rightMin(arr.size());

    leftMax[0] = arr[0];

    for (int i = 1; i < N; i++) {
        leftMax[i] = max(leftMax[i - 1],
                         arr[i]);
    }

    rightMin[N - 1] = arr[N - 1];

    for (int i = N - 2; i >= 0; i--) {
        rightMin[i] = min(rightMin[i + 1],
                          arr[i]);
    }

    int count = 0;

    for (int i = 0; i < N - 1; i++) {
        if (leftMax[i] <= rightMin[i + 1]) {
            count++;
        }
    }

    // Return count + 1 as the final answer
    return count + 1;
}

// Driver code
int main()
{
    vector<int> arr{ 10, 0, 21, 32, 68 };

    cout << maxPartitions(arr);
    return 0;
}

Java

// Java implementation of the above approach
import java.util.*;
public class GFG {

  // Function to find maximum partitions.
  static int maxPartitions(int[] arr)
  {
    int N = arr.length;

    // To keep track of max
    // and min elements at every index
    int[] leftMax = new int[arr.length];
    int[] rightMin = new int[arr.length];

    leftMax[0] = arr[0];

    for (int i = 1; i < N; i++) {
      leftMax[i] = Math.max(leftMax[i - 1], arr[i]);
    }

    rightMin[N - 1] = arr[N - 1];

    for (int i = N - 2; i >= 0; i--) {
      rightMin[i] = Math.min(rightMin[i + 1], arr[i]);
    }

    int count = 0;

    for (int i = 0; i < N - 1; i++) {
      if (leftMax[i] <= rightMin[i + 1]) {
        count++;
      }
    }

    // Return count + 1 as the final answer
    return count + 1;
  }

  // Driver code
  public static void main(String args[])
  {
    int[] arr = { 10, 0, 21, 32, 68 };

    System.out.println(maxPartitions(arr));
  }
}

// This code is contributed by Samim Hossain Mondal.

Python

# Python program for above approach

# Function to find maximum partitions.
def maxPartitions(arr):
    N = len(arr)

    # To keep track of max
    # and min elements at every index
    leftMax = []
    rightMin = []

    leftMax.append(arr[0])

    for i in range(1, N):
        leftMax.append(max(leftMax[i - 1], arr[i]))

    rightMin.append(arr[N - 1])

    for i in range(1, N):
        rightMin.append(min(rightMin[i - 1], arr[N - i - 1]))
    rightMin.reverse()
    count = 0

    for i in range(0, N - 1):
        if (leftMax[i] <= rightMin[i + 1]):
            count = count + 1

    # Return count + 1 as the final answer
    return count + 1

# Driver code
arr = [10, 0, 21, 32, 68]
print(maxPartitions(arr))

# This code is contributed by Samim Hossain Mondal.

C#

// C# program for above approach
using System;
class GFG {

  // Function to find maximum partitions.
  static int maxPartitions(int[] arr)
  {
    int N = arr.Length;

    // To keep track of max
    // and min elements at every index
    int[] leftMax = new int[arr.Length];
    int[] rightMin = new int[arr.Length];

    leftMax[0] = arr[0];

    for (int i = 1; i < N; i++) {
      leftMax[i] = Math.Max(leftMax[i - 1], arr[i]);
    }

    rightMin[N - 1] = arr[N - 1];

    for (int i = N - 2; i >= 0; i--) {
      rightMin[i] = Math.Min(rightMin[i + 1], arr[i]);
    }

    int count = 0;

    for (int i = 0; i < N - 1; i++) {
      if (leftMax[i] <= rightMin[i + 1]) {
        count++;
      }
    }

    // Return count + 1 as the final answer
    return count + 1;
  }

  // Driver code
  public static void Main()
  {
    int[] arr = { 10, 0, 21, 32, 68 };

    Console.WriteLine(maxPartitions(arr));
  }
}

// This code is contributed by ukasp.

Javascript

<script>
       // JavaScript code for the above approach

       // Function to find maximum partitions.
       function maxPartitions(arr)
       {
           let N = arr.length;

           // To keep track of max
           // and min elements at every index
           let leftMax = new Array(arr.length);
           let rightMin = new Array(arr.length);

           leftMax[0] = arr[0];

           for (let i = 1; i < N; i++) {
               leftMax[i] = Math.max(leftMax[i - 1],
                   arr[i]);
           }

           rightMin[N - 1] = arr[N - 1];

           for (let i = N - 2; i >= 0; i--) {
               rightMin[i] = Math.min(rightMin[i + 1],
                   arr[i]);
           }

           let count = 0;

           for (let i = 0; i < N - 1; i++) {
               if (leftMax[i] <= rightMin[i + 1]) {
                   count++;
               }
           }

           // Return count + 1 as the final answer
           return count + 1;
       }

       // Driver code
       let arr = [10, 0, 21, 32, 68];
       document.write(maxPartitions(arr));

        // This code is contributed by Potta Lokesh
   </script>

Output

4

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