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Check if an Array can be Sorted by picking only the corner Array elements
  • Difficulty Level : Easy
  • Last Updated : 18 Aug, 2020

Given an array arr[] consisting of N elements, the task is to check if the given array can be sorted by picking only corner elements i.e., elements either from left or right side of the array can be chosen.

Examples:

Input: arr[] = {2, 3, 4, 10, 4, 3, 1} 
Output: Yes 
Explanation: 
The order of picking elements from the array and placing in the sorted array are as follows: 
{2, 3, 4, 10, 4, 3, 1} -> {1} 
{2, 3, 4, 10, 4, 3} -> {1, 2} 
{3, 4, 10, 4, 3} -> {1, 2, 3} 
{4, 10, 4, 3} -> {1, 2, 3, 3} 
{4, 10, 4} -> {1, 2, 3, 3, 4} 
{10, 4} -> {1, 2, 3, 3, 4, 4} 
{10} -> {1, 2, 3, 3, 4, 4, 10}
Input: a[] = {2, 4, 2, 3} 
Output: No 
 

Approach: To solve the problem, we need to use a concept similar to Bitonic Sequence.Follow the below steps to solve the problem:

  • Traverse the array and check if the sequence of array elements is decreasing, i.e. if the next element is smaller than previous element, then all the remaining elements should be decreasing or equal as well.
  • That is, if the sequence is non-increasing, non-decreasing or non-decreasing followed by non-increasing, only then the array can be sorted by the given operations.

Below is implementation of above approach:



C++

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// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to check if an array can
// be sorted using given operations
bool check(int arr[], int n)
{
    int i, g;
    g = 0;
    for (i = 1; i < n; i++) {
  
        // If sequence becomes increasing
        // after an already non-decreasing to
        // non-increasing pattern
        if (arr[i] - arr[i - 1] > 0 && g == 1)
            return false;
  
        // If a decreasing pattern is observed
        if (arr[i] - arr[i - 1] < 0)
            g = 1;
    }
    return true;
}
  
// Driver Code
int main()
{
  
    int arr[] = { 2, 3, 4, 10, 4, 3, 1 };
    int n = sizeof(arr) / sizeof(int);
    if (check(arr, n) == true)
        cout << "Yes"
                "\n";
    else
        cout << "No"
             << "\n";
  
    return 0;
}

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Java

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// Java program to implement
// the above approach
class GFG{
  
// Function to check if an array can
// be sorted using given operations
static boolean check(int arr[], int n) 
{
    int i, g;
    g = 0;
  
    for(i = 1; i < n; i++)
    {
          
        // If sequence becomes increasing
        // after an already non-decreasing to
        // non-increasing pattern
        if (arr[i] - arr[i - 1] > 0 && g == 1)
            return false;
  
        // If a decreasing pattern is observed
        if (arr[i] - arr[i - 1] < 0)
            g = 1;
    }
    return true;
}
  
// Driver Code
public static void main(String[] args) 
{
    int arr[] = { 2, 3, 4, 10, 4, 3, 1 };
    int n = arr.length;
      
    if (check(arr, n) == true)
    {
        System.out.println("Yes");
    } else
    {
        System.out.println("No");
    }
}
}
  
// This code is contributed by rutvik_56

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Python3

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# Python3 program to implement
# the above approach
  
# Function to check if an array can 
# be sorted using given operations
def check(arr, n):
  
    g = 0
      
    for i in range(1, n):
  
        # If sequence becomes increasing
        # after an already non-decreasing to
        # non-increasing pattern
        if(arr[i] - arr[i - 1] > 0 and g == 1):
            return False
  
        # If a decreasing pattern is observed
        if(arr[i] - arr[i] < 0):
            g = 1
  
    return True
  
# Driver Code
arr = [ 2, 3, 4, 10, 4, 3, 1 ]
n = len(arr)
  
if(check(arr, n) == True):
    print("Yes")
else:
    print("No")
  
# This code is contributed by Shivam Singh

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C#

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// C# program to implement
// the above approach
using System;
class GFG{
  
// Function to check if an array can
// be sorted using given operations
static bool check(int []arr, int n) 
{
    int i, g;
    g = 0;
  
    for(i = 1; i < n; i++)
    {
          
        // If sequence becomes increasing
        // after an already non-decreasing to
        // non-increasing pattern
        if (arr[i] - arr[i - 1] > 0 && g == 1)
            return false;
  
        // If a decreasing pattern is observed
        if (arr[i] - arr[i - 1] < 0)
            g = 1;
    }
    return true;
}
  
// Driver Code
public static void Main(String[] args) 
{
    int []arr = { 2, 3, 4, 10, 4, 3, 1 };
    int n = arr.Length;
      
    if (check(arr, n) == true)
    {
        Console.WriteLine("Yes");
    } else
    {
        Console.WriteLine("No");
    }
}
}
  
// This code is contributed by sapnasingh4991

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Output: 

Yes


 

 

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

 

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