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Check if removal of a subsequence of non-adjacent elements makes the array sorted

  • Last Updated : 01 Oct, 2021

Given a binary array arr[] of size N, the task is to check if the array arr[] can be made sorted by removing any subsequence of non-adjacent array elements. If the array can be made sorted, then print “Yes”. Otherwise, print “No”.

Examples:

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Input: arr[] = {1, 0, 1, 0, 1, 1, 0}
Output: Yes
Explanation: 
Remove the element at indices {1, 3, 6} modifies the given array to {1, 1, 1, 1}, which is sorted. Therefore, print Yes.



Input: arr[] = {0, 1, 1, 0, 0}
Output: No

Naive Approach: The simplest approach to solve the given problem is to generate all possible subsequences of non-adjacent elements and if there exists any subsequence whose removal sorts the given array, then print “Yes”. Otherwise, print “No”.

Time Complexity: O(2N)
Auxiliary Space: O(1)

Efficient Approach: The above approach can also be optimized based on the observation that the array can’t be sorted if and only if when there exists two consecutive 1s are present at adjacent indexes i and i + 1 and two consecutive 0s are present at adjacent indexes j and j + 1 such that (j > i). Follow the steps below to solve the problem:

  • Initialize a variable, say idx as -1 that stores the index if there are two consecutive 1s in the array.
  • Traverse the given array and if there exists any two consecutive 1s are present in the array then store that index in the variable idx and break out of the loop. Otherwise, removing all the 1s from the array and make the array sorted, and print “Yes”.
  • If the value of idx is “-1”, then print “Yes” as always removing all the 1s from the array makes the array sorted.
  • Otherwise, traverse the array again from the index idx and if there exist two consecutive 0s, then print and break out of the loop. Otherwise, removing all the 0s from the array and make the array sorted, and print “Yes”.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if it is possible
// to sort the array or not
void isPossibleToSort(int arr[], int N)
{
    // Stores the index if there are two
    // consecutive 1's in the array
    int idx = -1;
 
    // Traverse the given array
    for (int i = 1; i < N; i++) {
 
        // Check adjacent same elements
        // having values 1s
        if (arr[i] == 1
            && arr[i - 1] == 1) {
            idx = i;
            break;
        }
    }
 
    // If there are no two consecutive
    // 1s, then always remove all the
    // 1s from array & make it sorted
    if (idx == -1) {
        cout << "YES";
        return;
    }
 
    for (int i = idx + 1; i < N; i++) {
 
        // If two consecutive 0's are
        // present after two consecutive
        // 1's then array can't be sorted
        if (arr[i] == 0 && arr[i - 1] == 0) {
            cout << "NO";
            return;
        }
    }
 
    // Otherwise, print Yes
    cout << "YES";
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 0, 1, 0, 1, 1, 0 };
    int N = sizeof(arr) / sizeof(arr[0]);
    isPossibleToSort(arr, N);
 
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
public class GFG {
 
    // Function to check if it is possible
    // to sort the array or not
    static void isPossibleToSort(int arr[], int N)
    {
        // Stores the index if there are two
        // consecutive 1's in the array
        int idx = -1;
 
        // Traverse the given array
        for (int i = 1; i < N; i++) {
 
            // Check adjacent same elements
            // having values 1s
            if (arr[i] == 1 && arr[i - 1] == 1) {
                idx = i;
                break;
            }
        }
 
        // If there are no two consecutive
        // 1s, then always remove all the
        // 1s from array & make it sorted
        if (idx == -1) {
            System.out.println("YES");
            return;
        }
 
        for (int i = idx + 1; i < N; i++) {
 
            // If two consecutive 0's are
            // present after two consecutive
            // 1's then array can't be sorted
            if (arr[i] == 0 && arr[i - 1] == 0) {
                System.out.println("NO");
                return;
            }
        }
 
        // Otherwise, print Yes
        System.out.println("YES");
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { 1, 0, 1, 0, 1, 1, 0 };
        int N = arr.length;
        isPossibleToSort(arr, N);
    }
}
 
// This code is contributed by Kingash.

Python3




# Python3 program for the above approach
 
# Function to check if it is possible
# to sort the array or not
def isPossibleToSort(arr, N):
     
    # Stores the index if there are two
    # consecutive 1's in the array
    idx = -1
 
    # Traverse the given array
    for i in range(1, N):
         
        # Check adjacent same elements
        # having values 1s
        if (arr[i] == 1 and arr[i - 1] == 1):
            idx = i
            break
 
    # If there are no two consecutive
    # 1s, then always remove all the
    # 1s from array & make it sorted
    if (idx == -1):
        print("YES")
        return
 
    for i in range(idx + 1, N, 1):
         
        # If two consecutive 0's are
        # present after two consecutive
        # 1's then array can't be sorted
        if (arr[i] == 0 and arr[i - 1] == 0):
            print("NO")
            return
 
    # Otherwise, print Yes
    print("YES")
 
# Driver Code
if __name__ == '__main__':
     
    arr = [ 1, 0, 1, 0, 1, 1, 0 ]
    N = len(arr)
     
    isPossibleToSort(arr, N)
 
# This code is contributed by SURENDRA_GANGWAR

C#




// C# for the above approach
using System.IO;
using System;
 
class GFG{
     
// Function to check if it is possible
// to sort the array or not
static void isPossibleToSort(int[] arr, int N)
{
     
    // Stores the index if there are two
    // consecutive 1's in the array
    int idx = -1;
 
    // Traverse the given array
    for(int i = 1; i < N; i++)
    {
         
        // Check adjacent same elements
        // having values 1s
        if (arr[i] == 1 && arr[i - 1] == 1)
        {
            idx = i;
            break;
        }
    }
 
    // If there are no two consecutive
    // 1s, then always remove all the
    // 1s from array & make it sorted
    if (idx == -1)
    {
        Console.WriteLine("YES");
        return;
    }
 
    for(int i = idx + 1; i < N; i++)
    {
         
        // If two consecutive 0's are
        // present after two consecutive
        // 1's then array can't be sorted
        if (arr[i] == 0 && arr[i - 1] == 0)
        {
            Console.WriteLine("NO");
            return;
        }
    }
 
    // Otherwise, print Yes
    Console.WriteLine("YES");
}
 
// Driver code
static void Main()
{
    int[] arr = { 1, 0, 1, 0, 1, 1, 0 };
    int N = arr.Length;
     
    isPossibleToSort(arr, N);
}
}
 
// This code is contributed by abhinavjain194

Javascript




<script>
 
// JavaScript program for the above approach
 
// Function to check if it is possible
// to sort the array or not
function isPossibleToSort(arr, N)
{
    // Stores the index if there are two
    // consecutive 1's in the array
    var idx = -1;
     
    var i;
    // Traverse the given array
    for (i = 1; i < N; i++) {
 
        // Check adjacent same elements
        // having values 1s
        if (arr[i] == 1
            && arr[i - 1] == 1) {
            idx = i;
            break;
        }
    }
 
    // If there are no two consecutive
    // 1s, then always remove all the
    // 1s from array & make it sorted
    if (idx == -1) {
        document.write("YES");
        return;
    }
 
    for (i = idx + 1; i < N; i++) {
 
        // If two consecutive 0's are
        // present after two consecutive
        // 1's then array can't be sorted
        if (arr[i] == 0 && arr[i - 1] == 0) {
            document.write("NO");
            return;
        }
    }
 
    // Otherwise, print Yes
    document.write("YES");
}
 
// Driver Code
    var arr = [1, 0, 1, 0, 1, 1, 0];
    var N = arr.length;
    isPossibleToSort(arr, N);
 
</script>
Output: 
YES

 

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




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