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]. 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}
The idea is based on the fact that if an element arr[i] is maximum of prefix arr[0..i], then we can make a partition ending with arr[i].
C++
// 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; } |
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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. |
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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 code arr = [1, 0, 2, 3, 4] n = len(arr) print(maxPartitions(arr, n)) # This code is contributed by Smitha Dinesh Semwal. |
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C#
// 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. |
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PHP
<?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 ?> |
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Output:
4
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