Skip to content
Related Articles

Related Articles

Check if reversing a sub array make the array sorted

View Discussion
Improve Article
Save Article
  • Difficulty Level : Medium
  • Last Updated : 22 Jun, 2022
View Discussion
Improve Article
Save Article

Given an array of n distinct integers. The task is to check whether reversing any one sub-array can make the array sorted or not. If the array is already sorted or can be made sorted by reversing any one subarray, print “Yes“, else print “No“.
Examples: 

Input : arr [] = {1, 2, 5, 4, 3}
Output : Yes
By reversing the subarray {5, 4, 3}, 
the array will be sorted.

Input : arr [] = { 1, 2, 4, 5, 3 }
Output : No

Method 1: Brute force (O(n3))
Consider every subarray and check if reversing the subarray makes the whole array sorted. If yes, return True. If reversing any of the subarrays doesn’t make the array sorted, then return False. Considering every subarray will take O(n2), and for each subarray, checking whether the whole array will get sorted after reversing the subarray in consideration will take O(n). Thus overall complexity would be O(n3).

Method 2: Sorting ( O(n*log(n) )) 
The idea is to compare the given array with its sorted version. Make a copy of the given array and sort it. Now, find the first index and last index in the given array which does not match with the sorted array. If no such indices are found (given array was already sorted), return True. Else check if the elements between the found indices are in decreasing order, if Yes then return True else return False

if Below is the implementation of the above approach:

C++




// C++ program to check whether reversing a
// sub array make the array sorted or not
#include<bits/stdc++.h>
using namespace std;
 
// Return true, if reversing the subarray will
// sort the array, else return false.
bool checkReverse(int arr[], int n)
{
    // Copying the array.
    int temp[n];
    for (int i = 0; i < n; i++)
        temp[i] = arr[i];
 
    // Sort the copied array.
    sort(temp, temp + n);
 
    // Finding the first mismatch.
    int front;
    for (front = 0; front < n; front++)
        if (temp[front] != arr[front])
            break;
 
    // Finding the last mismatch.
    int back;
    for (back = n - 1; back >= 0; back--)
        if (temp[back] != arr[back])
            break;
 
    // If whole array is sorted
    if (front >= back)
        return true;
 
    // Checking subarray is decreasing or not.
    do
    {
        front++;
        if (arr[front - 1] < arr[front])
            return false;
    } while (front != back);
 
    return true;
}
 
// Driven Program
int main()
{
    int arr[] = { 1, 2, 5, 4, 3 };
    int n = sizeof(arr)/sizeof(arr[0]);
 
    checkReverse(arr, n)? (cout << "Yes" << endl):
                          (cout << "No" << endl);
    return 0;
}

Java




// Java program to check whether reversing a
// sub array make the array sorted or not
 
import java.util.Arrays;
 
class GFG {
 
// Return true, if reversing the subarray will
// sort the array, else return false.
    static boolean checkReverse(int arr[], int n) {
        // Copying the array.
        int temp[] = new int[n];
        for (int i = 0; i < n; i++) {
            temp[i] = arr[i];
        }
 
        // Sort the copied array.
        Arrays.sort(temp);
 
        // Finding the first mismatch.
        int front;
        for (front = 0; front < n; front++) {
            if (temp[front] != arr[front]) {
                break;
            }
        }
 
        // Finding the last mismatch.
        int back;
        for (back = n - 1; back >= 0; back--) {
            if (temp[back] != arr[back]) {
                break;
            }
        }
 
        // If whole array is sorted
        if (front >= back) {
            return true;
        }
 
        // Checking subarray is decreasing or not.
        do {
            front++;
            if (arr[front - 1] < arr[front]) {
                return false;
            }
        } while (front != back);
 
        return true;
    }
 
// Driven Program
    public static void main(String[] args) {
 
        int arr[] = {1, 2, 5, 4, 3};
        int n = arr.length;
 
        if (checkReverse(arr, n)) {
            System.out.print("Yes");
        } else {
            System.out.print("No");
        }
    }
 
}
//This code contributed by 29AjayKumar

Python3




# Python3 program to check whether
# reversing a sub array make the
# array sorted or not
 
# Return true, if reversing the
# subarray will sort the array,
# else return false.
def checkReverse(arr, n):
 
    # Copying the array
    temp = [0] * n
    for i in range(n):
        temp[i] = arr[i]
 
    # Sort the copied array.
    temp.sort()
 
    # Finding the first mismatch.
    for front in range(n):
        if temp[front] != arr[front]:
            break
 
    # Finding the last mismatch.
    for back in range(n - 1, -1, -1):
        if temp[back] != arr[back]:
            break
 
    #If whole array is sorted
    if front >= back:
        return True
    while front != back:
        front += 1
        if arr[front - 1] < arr[front]:
            return False
    return True
 
# Driver code
arr = [1, 2, 5, 4, 3]
n = len(arr)
if checkReverse(arr, n) == True:
    print("Yes")
else:
    print("No")
 
# This code is contributed
# by Shrikant13

C#




// C# program to check whether reversing a
// sub array make the array sorted or not
using System;
 
class GFG
{
 
// Return true, if reversing the
// subarray will sort the array,
// else return false.
static bool checkReverse(int []arr, int n)
{
    // Copying the array.
    int []temp = new int[n];
    for (int i = 0; i < n; i++)
    {
        temp[i] = arr[i];
    }
 
    // Sort the copied array.
    Array.Sort(temp);
 
    // Finding the first mismatch.
    int front;
    for (front = 0; front < n; front++)
    {
        if (temp[front] != arr[front])
        {
            break;
        }
    }
 
    // Finding the last mismatch.
    int back;
    for (back = n - 1; back >= 0; back--)
    {
        if (temp[back] != arr[back])
        {
            break;
        }
    }
 
    // If whole array is sorted
    if (front >= back)
    {
        return true;
    }
 
    // Checking subarray is decreasing
    // or not.
    do
    {
        front++;
        if (arr[front - 1] < arr[front])
        {
            return false;
        }
    } while (front != back);
 
    return true;
}
 
// Driven Program
public static void Main()
{
    int []arr = {1, 2, 5, 4, 3};
    int n = arr.Length;
 
    if (checkReverse(arr, n))
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
}
 
// This code is contributed
// by PrinciRaj

PHP




<?php
// PHP program to check whether reversing a
// sub array make the array sorted or not
 
// Return true, if reversing the subarray
// will sort the array, else return false.
function checkReverse($arr, $n)
{
    // Copying the array.
    $temp[$n] = array();
    for ($i = 0; $i < $n; $i++)
        $temp[$i] = $arr[$i];
 
    // Sort the copied array.
    sort($temp, 0);
 
    // Finding the first mismatch.
    $front;
    for ($front = 0; $front < $n; $front++)
        if ($temp[$front] != $arr[$front])
            break;
 
    // Finding the last mismatch.
    $back;
    for ($back = $n - 1; $back >= 0; $back--)
        if ($temp[$back] != $arr[$back])
            break;
 
    // If whole array is sorted
    if ($front >= $back)
        return true;
 
    // Checking subarray is decreasing or not.
    do
    {
        $front++;
        if ($arr[$front - 1] < $arr[$front])
            return false;
    } while ($front != $back);
 
    return true;
}
 
// Driver Code
$arr = array( 1, 2, 5, 4, 3 );
$n = sizeof($arr);
 
if(checkReverse($arr, $n))
    echo "Yes" . "\n";
else
    echo "No" . "\n";
 
// This code is contributed
// by Akanksha Rai
?>

Javascript




<script>
 
// Javascript program to check whether reversing a
// sub array make the array sorted or not
 
// Return true, if reversing the subarray will
// sort the array, else return false.
    function checkReverse(arr, n) {
        // Copying the array.
        let temp = [];
        for (let i = 0; i < n; i++) {
            temp[i] = arr[i];
        }
 
        // Sort the copied array.
        temp.sort();
 
        // Finding the first mismatch.
        let front;
        for (front = 0; front < n; front++) {
            if (temp[front] != arr[front]) {
                break;
            }
        }
 
        // Finding the last mismatch.
        let back;
        for (back = n - 1; back >= 0; back--) {
            if (temp[back] != arr[back]) {
                break;
            }
        }
 
        // If whole array is sorted
        if (front >= back) {
            return true;
        }
 
        // Checking subarray is decreasing or not.
        do {
            front++;
            if (arr[front - 1] < arr[front]) {
                return false;
            }
        } while (front != back);
 
        return true;
    }
 
// Driver Code
 
    let arr = [1, 2, 5, 4, 3];
    let n = arr.length;
 
    if (checkReverse(arr, n)) {
        document.write("Yes");
    } else {
        document.write("No");
    }
 
</script>

Output

Yes

Time Complexity: O(n*log(n) ).
Auxiliary Space: O(n).

Method 3: Linear time solution (O(n)) 
Observe, that the answer will be True when the array is already sorted or when the array consists of three parts. The first part is increasing subarray, then decreasing subarray, and then again increasing subarray. So, we need to check that array contains increasing elements then some decreasing elements, and then increasing elements if this is the case the answer will be True. In all other cases, the answer will be False.

Note: Simply finding the three parts does not guarantee the answer to be True eg consider

 arr [] = {10,20,30,40,4,3,2,50,60,70} 

The answer would be False in this case although we are able to find three parts. We will be handling the validity of the three parts in the code below.

Below is the implementation of this approach: 

C++




// C++ program to check whether reversing a sub array
// make the array sorted or not
#include<bits/stdc++.h>
using namespace std;
 
// Return true, if reversing the subarray will sort t
// he array, else return false.
bool checkReverse(int arr[], int n)
{
    if (n == 1)
        return true;
 
    // Find first increasing part
    int i;
    for (i=1; i < n && arr[i-1] < arr[i]; i++);
    if (i == n)
        return true;
 
    // Find reversed part
    int j = i;
    while (j < n && arr[j] < arr[j-1])
    {
        if (i > 1 && arr[j] < arr[i-2])
            return false;
        j++;
    }
 
    if (j == n)
        return true;
 
    // Find last increasing part
    int k = j;
 
    // To handle cases like {1,2,3,4,20,9,16,17}
    if (arr[k] < arr[i-1])
       return false;
 
    while (k > 1 && k < n)
    {
        if (arr[k] < arr[k-1])
            return false;
        k++;
    }
    return true;
}
 
// Driven Program
int main()
{
    int arr[] = {1, 3, 4, 10, 9, 8};
    int n = sizeof(arr)/sizeof(arr[0]);
    checkReverse(arr, n)? cout << "Yes" : cout << "No";
    return 0;
}

Java




// Java program to check whether reversing a sub array
// make the array sorted or not
 
class GFG {
 
// Return true, if reversing the subarray will sort t
// he array, else return false.
    static boolean checkReverse(int arr[], int n) {
        if (n == 1) {
            return true;
        }
 
        // Find first increasing part
        int i;
        for (i = 1; arr[i - 1] < arr[i] && i < n; i++);
        if (i == n) {
            return true;
        }
 
        // Find reversed part
        int j = i;
        while (j < n && arr[j] < arr[j - 1]) {
            if (i > 1 && arr[j] < arr[i - 2]) {
                return false;
            }
            j++;
        }
 
        if (j == n) {
            return true;
        }
 
        // Find last increasing part
        int k = j;
 
        // To handle cases like {1,2,3,4,20,9,16,17}
        if (arr[k] < arr[i - 1]) {
            return false;
        }
 
        while (k > 1 && k < n) {
            if (arr[k] < arr[k - 1]) {
                return false;
            }
            k++;
        }
        return true;
    }
 
// Driven Program
    public static void main(String[] args) {
 
        int arr[] = {1, 3, 4, 10, 9, 8};
        int n = arr.length;
 
        if (checkReverse(arr, n)) {
            System.out.print("Yes");
        } else {
            System.out.print("No");
        }
    }
 
}
 
// This code is contributed
// by Rajput-Ji

Python3




# Python3 program to check whether reversing
# a sub array make the array sorted or not
import math as mt
 
# Return True, if reversing the subarray
# will sort the array, else return False.
def checkReverse(arr, n):
 
    if (n == 1):
        return True
 
    # Find first increasing part
    i = 1
    for i in range(1, n):
        if arr[i - 1] < arr[i] :
            if (i == n):
                return True
          
        else:
            break
 
    # Find reversed part
    j = i
    while (j < n and arr[j] < arr[j - 1]):
      
        if (i > 1 and arr[j] < arr[i - 2]):
            return False
        j += 1
 
    if (j == n):
        return True
 
    # Find last increasing part
    k = j
 
    # To handle cases like 1,2,3,4,20,9,16,17
    if (arr[k] < arr[i - 1]):
        return False
 
    while (k > 1 and k < n):
     
        if (arr[k] < arr[k - 1]):
            return False
        k += 1
     
    return True
 
# Driver Code
arr = [ 1, 3, 4, 10, 9, 8]
n = len(arr)
if checkReverse(arr, n):
    print("Yes")
else:
    print("No")
         
# This code is contributed by
# Mohit kumar 29

C#




// C# program to check whether reversing a
// sub array make the array sorted or not
  
using System;
public class GFG{
 
// Return true, if reversing the subarray will sort t
// he array, else return false.
    static bool checkReverse(int []arr, int n) {
        if (n == 1) {
            return true;
        }
 
        // Find first increasing part
        int i;
        for (i = 1; arr[i - 1] < arr[i] && i < n; i++);
        if (i == n) {
            return true;
        }
 
        // Find reversed part
        int j = i;
        while (j < n && arr[j] < arr[j - 1]) {
            if (i > 1 && arr[j] < arr[i - 2]) {
                return false;
            }
            j++;
        }
 
        if (j == n) {
            return true;
        }
 
        // Find last increasing part
        int k = j;
 
        // To handle cases like {1,2,3,4,20,9,16,17}
        if (arr[k] < arr[i - 1]) {
            return false;
        }
 
        while (k > 1 && k < n) {
            if (arr[k] < arr[k - 1]) {
                return false;
            }
            k++;
        }
        return true;
    }
 
 
// Driven Program
    public static void Main() {
 
        int []arr = {1, 3, 4, 10, 9, 8};
        int n = arr.Length;
 
        if (checkReverse(arr, n)) {
            Console.Write("Yes");
        } else {
            Console.Write("No");
        }
    }
}
// This code is contributed
// by 29AjayKumar

Javascript




<script>
 
// Javascript program to check whether reversing a sub array
// make the array sorted or not
 
// Return true, if reversing the subarray will sort t
// he array, else return false.
function checkReverse( arr, n)
{
    if (n == 1)
        return true;
 
    // Find first increasing part
    let i;
    for (i=1; i < n && arr[i-1] < arr[i]; i++);
    if (i == n)
        return true;
 
    // Find reversed part
    let j = i;
    while (j < n && arr[j] < arr[j-1])
    {
        if (i > 1 && arr[j] < arr[i-2])
            return false;
        j++;
    }
 
    if (j == n)
        return true;
 
    // Find last increasing part
    let k = j;
 
    // To handle cases like {1,2,3,4,20,9,16,17}
    if (arr[k] < arr[i-1])
    return false;
 
    while (k > 1 && k < n)
    {
        if (arr[k] < arr[k-1])
            return false;
        k++;
    }
    return true;
}
 
     
    // Driver program
     
        let arr = [1, 3, 4, 10, 9, 8];
        let n = arr.length;
  
        if (checkReverse(arr, n)) {
            document.write("Yes");
        } else {
            document.write("No");
        }
     
     
</script>

Output

Yes

Time Complexity: O(n).
Auxiliary Space: O(1).

This article is contributed by Anuj Chauhan and improved by Nakshatra Chhillar. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!