C++ Program to Count rotations required to sort given array in non-increasing order
Last Updated :
27 Jan, 2022
Given an array arr[] consisting of N integers, the task is to sort the array in non-increasing order by minimum number of anti-clockwise rotations. If it is not possible to sort the array, then print “-1”. Otherwise, print the count of rotations.
Examples:
Input: arr[] = {2, 1, 5, 4, 3}
Output: 2
Explanation: Two anti-clockwise rotations are required to sort the array in decreasing order, i.e. {5, 4, 3, 2, 1}
Input: arr[] = {2, 3, 1}
Output: -1
Approach: The idea is to traverse the given array arr[] and count the number of indices satisfying arr[i + 1] > arr[i]. Follow the steps below to solve the problem:
- Store the count of arr[i + 1] > arr[i] in a variable and also store the index when arr[i+1] > arr[i].
- If the value of count is N – 1, then the array is sorted in non-decreasing order. The required steps are exactly (N – 1).
- If the value of count is 0, then the array is already sorted in non-increasing order.
- If the value of count is 1 and arr[0] ? arr[N – 1], then the required number of rotations is equal to (index + 1), by performing shifting of all the numbers upto that index. Also, check if arr[0] ? arr[N – 1] to ensure if the sequence is non-increasing.
- Otherwise, it is not possible to sort the array in non-increasing order.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void minMovesToSort( int arr[], int N)
{
int count = 0;
int index;
for ( int i = 0; i < N - 1; i++) {
if (arr[i] < arr[i + 1]) {
count++;
index = i;
}
}
if (count == 0) {
cout << "0" ;
}
else if (count == N - 1) {
cout << N - 1;
}
else if (count == 1
&& arr[0] <= arr[N - 1]) {
cout << index + 1;
}
else {
cout << "-1" ;
}
}
int main()
{
int arr[] = { 2, 1, 5, 4, 2 };
int N = sizeof (arr)
/ sizeof (arr[0]);
minMovesToSort(arr, N);
return 0;
}
|
Time Complexity: O(N)
Auxiliary Space: O(1)
Please refer complete article on Count rotations required to sort given array in non-increasing order for more details!
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