Check if reversing a sub array make the array sorted
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++
- Java
- Python3
- C#
- PHP
- Javascript
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
Python3
C#
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Yes
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.
- Traverse the array to check for every element is sorted or not.
- Finally, return true.
Below is the implementation of the above approach:
- C++
- Java
- Python3
- C#
- Javascript
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) |
Java
Python3
C#
Javascript
Yes
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++
- Java
- Python3
- C#
- Javascript
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
Python3
C#
Javascript
Yes
Time Complexity: O(n).
Auxiliary Space: O(1).
