Check if an Array can be Sorted by picking only the corner Array elements

Last Updated : 29 Apr, 2021

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++

// 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; }

Java

// 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

Python3

# 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

// 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

JavaScript

<script> // Javascript Program to implement // the above approach  // Function to check if an array can // be sorted using given operations function check(arr, n) {     var 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 var arr = [ 2, 3, 4, 10, 4, 3, 1 ]; var n = arr.length; if (check(arr, n) == true)     document.write( "Yes"+             "<br>"); else     document.write( "No"          + "<br>");  // This code is contributed by noob2000. </script>

Output:

Yes

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

Check if array can be sorted by swapping adjacent elements of opposite parity

varuntak22

Improve

Article Tags :

Practice Tags :

Check if an Array can be Sorted by picking only the corner Array elements

Similar Reads