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Sort Array by splitting into subarrays where each element belongs to only subarray

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  • Difficulty Level : Easy
  • Last Updated : 15 Feb, 2022
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Given an array arr[] of N distinct integers, the task is to check if it is possible to sort the array in increasing order by performing the following operations in order exactly once:

  • Split the array arr[] into exactly Y(1 <= Y <= N) non-empty subarrays such that each element belongs to exactly one subarray.
  • Reorder the subarrays in any arbitrary order.
  • Merge the reordered subarrays.

Examples:

Input: arr[ ] = {6, 3, 4, 2, 1}, Y = 4
Output: Yes
Explanation:
The operations can be performed as:

  • Split the array into exactly 4 non-empty subarrays: {6, 3, 4, 2, 1} -> {6}, {3, 4}, {2}, {1}
  • Reorder the subarrays: {6}, {3, 4}, {2}, {1} -> {1}, {2}, {3, 4}, {6}
  • Merging the subarrays: {1}, {2}, {3, 4}, {6} -> {1, 2, 3, 4, 6} (sorted)

Input: arr[ ] = {1, -4, 0, -2}, Y = 2
Output: No

Approach: The main idea is if the minimum number of splits required to sort the given array arr[] is less than or equal to Y, then it is always possible to sort the array. Also, the required number of splits must be equal to the count of the minimum number of splits required to divide the array into the minimum number of non-decreasing subarrays.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Initialization of pair
pair<int, int> a[100001];
 
// Function to check if it is possible
// to sort the array by performing the
// operations exactly once in order
void checkForSortedSubarray(int n, int y, int arr[])
{
 
    // Traversing the array
    for (int i = 1; i <= n; i++) {
        // Storing the array element
        // in the first part of pair
        a[i].first = arr[i];
 
        // Storing the index as second
        // element in the pair
        a[i].second = i;
    }
 
    // Initialize Count
    int cnt = 1;
 
    // Sorting the array
    sort(a + 1, a + n + 1);
 
    for (int i = 1; i <= n - 1; i++) {
 
        // Checking if the index lies
        // in order
        if (a[i].second != a[i + 1].second - 1)
            cnt++;
    }
 
    // If minimum splits required is
    // greater than y
    if (cnt > y)
        cout << "No";
    else
        cout << "Yes";
}
 
// Driver Code
int main()
 
{
    int Y = 4;
    int arr[] = { 6, 3, 4, 2, 1 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    checkForSortedSubarray(N, Y, arr);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Initialization of pair
static pair []a = new pair[100001];
static class pair
{
    int first, second;
    public pair(int first, int second) 
    {
        this.first = first;
        this.second = second;
    }   
}
// Function to check if it is possible
// to sort the array by performing the
// operations exactly once in order
static void checkForSortedSubarray(int n, int y, int arr[])
{
 
    // Traversing the array
    for (int i = 1; i <= n; i++) {
        // Storing the array element
        // in the first part of pair
        a[i].first = arr[i];
 
        // Storing the index as second
        // element in the pair
        a[i].second = i;
    }
 
    // Initialize Count
    int cnt = 1;
 
    // Sorting the array
    Arrays.sort(a,(a, b) -> a.first - b.first);
 
    for (int i = 1; i <= n - 1; i++) {
 
        // Checking if the index lies
        // in order
        if (a[i].second != a[i + 1].second - 1)
            cnt++;
    }
 
    // If minimum splits required is
    // greater than y
    if (cnt > y)
        System.out.print("No");
    else
        System.out.print("Yes");
}
 
// Driver Code
public static void main(String[] args)
 
{
    int Y = 4;
    int arr[] = { 6, 3, 4, 2, 1 };
    int N = arr.length;
    for(int i = 0;i<a.length;i++) {
        a[i] = new pair(0, 0);
    }
    checkForSortedSubarray(N-1, Y, arr);
 
}
}
 
// This code is contributed by Princi Singh

Python3




# Python 3 program for the above approach
 
# Initialization of pair
 
# Function to check if it is possible
# to sort the array by performing the
# operations exactly once in order
def checkForSortedSubarray(n, y, arr, a):
   
    # Traversing the array
    for i in range(0, n, 1):
       
        # Storing the array element
        # in the first part of pair
        a[i][0] = arr[i]
 
        # Storing the index as second
        # element in the pair
        a[i][1] = i
 
    # Initialize Count
    cnt = 1
 
    # Sorting the array
    a.sort()
 
    for i in range(0,n,1):
        # Checking if the index lies
        # in order
        if (a[i][1] != a[i + 1][1] - 1):
            cnt += 1
 
    # If minimum splits required is
    # greater than y
    if (cnt > y):
        print("Yes")
    else:
        print("No")
 
# Driver Code
if __name__ == '__main__':
    Y = 4
    a = [[0,0] for i in range(100001)]
    arr = [6, 3, 4, 2, 1]
    N = len(arr)
    checkForSortedSubarray(N, Y, arr,a)
     
    # This code is contributed by SURENDRA_GANGWAR.

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
public class GFG {
 
  // Initialization of pair
  static List<pair> a = new List<pair>();
 
  public class pair {
    public int first, second;
 
    public pair(int first, int second) {
      this.first = first;
      this.second = second;
    }
  }
 
  // Function to check if it is possible
  // to sort the array by performing the
  // operations exactly once in order
  static void checkForSortedSubarray(int n, int y, int []arr) {
 
    // Traversing the array
    for (int i = 1; i <= n; i++)
    {
 
      // Storing the array element
      // in the first part of pair
      a[i].first = arr[i];
 
      // Storing the index as second
      // element in the pair
      a[i].second = i;
    }
 
    // Initialize Count
    int cnt = 1;
 
    // Sorting the array
    a.Sort((c, b) => c.first - b.first);
 
    for (int i = 1; i <= n - 1; i++) {
 
      // Checking if the index lies
      // in order
      if (a[i].second != a[i + 1].second - 1)
        cnt++;
    }
 
    // If minimum splits required is
    // greater than y
    if (cnt > y)
      Console.Write("No");
    else
      Console.Write("Yes");
  }
 
  // Driver Code
  public static void Main(String[] args)
 
  {
    int Y = 4;
    int []arr = { 6, 3, 4, 2, 1 };
    int N = arr.Length;
    for (int i = 0; i < 100001; i++) {
      a.Add(new pair(0, 0));
    }
    checkForSortedSubarray(N - 1, Y, arr);
 
  }
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
        // JavaScript Program to implement
        // the above approach
 
        // Function to check if it is possible
        // to sort the array by performing the
        // operations exactly once in order
         
        // Initialization of pair
        var a = [];
        function checkForSortedSubarray(n, y, arr)
        {
 
            // Traversing the array
            for (let i = 0; i < n; i++)
            {
             
                // Storing the array element
                // in the first part of pair
 
                // Storing the index as second
                // element in the pair
                a.push({
                    first: arr[i],
                    second: i
                })
            }
 
            // Initialize Count
            let cnt = 1;
 
            // Sorting the array
            a.sort(function (a, b) { return a.first - b.first; })
 
            for (let i = 0; i < n - 1; i++) {
 
                // Checking if the index lies
                // in order
                if (a[i].second != a[i + 1].second - 1)
                    cnt++;
            }
 
            // If minimum splits required is
            // greater than y
            if (cnt > y)
                document.write("No");
            else
                document.write("Yes");
        }
 
        // Driver Code
        let Y = 4;
        let arr = [6, 3, 4, 2, 1];
        let N = arr.length;
 
        checkForSortedSubarray(N, Y, arr);
 
// This code is contributed by Potta Lokesh
 
    </script>

Output

Yes

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


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