Check if reversing a sub array make the array sorted
Last Updated :
13 Sep, 2023
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
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;
}
// Driver 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 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;
}
// Driver 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 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# 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;
}
// Driver 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 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
?>
<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>
Time Complexity: O(n*log(n) ).
Auxiliary Space: O(n).
Method 3: Linear time solution (O(n)):
The idea to solve this problem is based on the observation that if we perform one rotation of any subarray in the sorted array (increasing order), then we there will be exactly one subarray which will be in decreasing order. So, we have to find that rotated subarray and perform one rotation on it. Finally check if the array becomes sorted or not.
- Initialize two variables x and y with -1.
- Iterate over the array.
- Find the first number for which a[i] > a[i+1] and store it into x.
- Similarly, Store index i+1 as well into y, As this will keep track of the ending of the subarray which is needed to reverse.
- Check if x == -1 then array is already sorted so return true.
- Otherwise, reverse the array from index x to index y.
- Traverse the array to check for every element is sorted or not.
- If not sorted, return false.
- Finally, return true.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
bool sortArr(int a[], int n)
{
int x = -1;
int y = -1;
for (int i = 0; i < n - 1; i++) {
if (a[i] > a[i + 1]) {
if (x == -1) {
x = i;
}
y = i + 1;
}
}
if (x != -1) {
reverse(a + x, a + y + 1);
for (int i = 0; i < n - 1; i++) {
if (a[i] > a[i + 1]) {
return false;
return 0;
}
}
}
return true;
}
// Driver Program
int main()
{
int arr[] = { 1, 2, 5, 4, 3 };
int n = sizeof(arr) / sizeof(arr[0]);
sortArr(arr, n) ? (cout << "Yes" << endl)
: (cout << "No" << endl);
return 0;
}
//This code is contributed by Shaurya Dixit (B19EE077)
public class GFG {
static void reverse(int[] a,int x,int y)
{
while(x<y)
{
int temp = a[x];
a[x] = a[y];
a[y] = temp;
x++;
y--;
}
}
static boolean sortArr(int[] a, int n)
{
int x = -1;
int y = -1;
for (int i = 0; i < n - 1; i++) {
if (a[i] > a[i + 1]) {
if (x == -1) {
x = i;
}
y = i + 1;
}
}
if (x != -1) {
reverse(a,x,y);
for (int i = 0; i < n - 1; i++) {
if (a[i] > a[i + 1]) {
return false;
}
}
}
return true;
}
// Driver Code
public static void main (String[] args)
{
int arr[] = { 1, 2, 5, 4, 3 };
int n = arr.length;
if(sortArr(arr, n))
{
System.out.println("Yes");
}
else
{
System.out.println("No");
}
}
}
// This code is contributed by aditya942003patil
def reverse(a, x, y):
while(x < y):
temp = a[x]
a[x] = a[y]
a[y] = temp
x += 1
y -= 1
def sortArr(a, n):
x, y = -1, -1
for i in range(n-1):
if(a[i] > a[i+1]):
if(x == -1):
x = i
y = i + 1
if(x != -1):
reverse(a, x, y)
for i in range(0, n-1):
if(a[i] > a[i+1]):
return False
return True
arr = [1, 2, 5, 4, 3]
n = len(arr)
if(sortArr(arr, n)):
print("Yes")
else:
print("No")
# This code is contributed by lokesh
// C# code to implement the above approach
using System;
public class GFG {
static void reverse(int[] a, int x, int y)
{
while (x < y) {
int temp = a[x];
a[x] = a[y];
a[y] = temp;
x++;
y--;
}
}
static bool sortArr(int[] a, int n)
{
int x = -1;
int y = -1;
for (int i = 0; i < n - 1; i++) {
if (a[i] > a[i + 1]) {
if (x == -1) {
x = i;
}
y = i + 1;
}
}
if (x != -1) {
reverse(a, x, y);
for (int i = 0; i < n - 1; i++) {
if (a[i] > a[i + 1]) {
return false;
}
}
}
return true;
}
static public void Main()
{
// Code
int[] arr = { 1, 2, 5, 4, 3 };
int n = arr.Length;
if (sortArr(arr, n)) {
Console.WriteLine("Yes");
}
else {
Console.WriteLine("No");
}
}
}
// This code is contributed by lokeshmvs21.
function sortArr(arr, n) {
let x=-1;
let y=-1;
for (let i = 0; i < n-1; i++) {
if (arr[i] > arr[i + 1]) {
if (x == -1) {
x = i;
}
y = i + 1;
}
}
if (x != -1) {
while(x<y)
{
let temp=arr[x];
arr[x]=arr[y];
arr[y]=temp;
x++;
y--;
}
for (let i = 0; i < n - 1; i++) {
if (arr[i] > arr[i + 1]) {
return false;
return 0;
}
}
}
return true;
}
// Driver Code
let arr = [1, 2, 5, 4, 3];
let n = arr.length;
if (sortArr(arr, n)) {
console.log("Yes");
} else {
console.log("No");
}
// This code is contributed by garg28harsh.
Time Complexity: O(n)
Auxiliary Space: O(1)
Method 4: Another 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;
}
// Driver 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 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;
}
// Driver 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 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# 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;
}
// Driver 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
<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>
Time Complexity: O(n).
Auxiliary Space: O(1).
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