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C++ Program to Modify given array to a non-decreasing array by rotation

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  • Last Updated : 27 Jan, 2022

Given an array arr[] of size N (consisting of duplicates), the task is to check if the given array can be converted to a non-decreasing array by rotating it. If it’s not possible to do so, then print “No“. Otherwise, print “Yes“.

Examples:

Input: arr[] = {3, 4, 5, 1, 2}
Output: Yes
Explanation: After 2 right rotations, the array arr[] modifies to {1, 2, 3, 4, 5}

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

Approach: The idea is based on the fact that a maximum of N distinct arrays can be obtained by rotating the given array and check for each individual rotated array, whether it is non-decreasing or not. Follow the steps below to solve the problem:

  • Initialize a vector, say v, and copy all the elements of the original array into it.
  • Sort the vector v.
  • Traverse the original array and perform the following steps:
    • Rotate by 1 in each iteration.
    • If the array becomes equal to vector v, print “Yes“. Otherwise, print “No“.

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 a
// non-decreasing array can be obtained
// by rotating the original array
void rotateArray(vector<int>& arr, int N)
{
    // Stores copy of original array
    vector<int> v = arr;
  
    // Sort the given vector
    sort(v.begin(), v.end());
  
    // Traverse the array
    for (int i = 1; i <= N; ++i) {
  
        // Rotate the array by 1
        rotate(arr.begin(),
               arr.begin() + 1, arr.end());
  
        // If array is sorted
        if (arr == v) {
  
            cout << "YES" << endl;
            return;
        }
    }
  
    // If it is not possible to
    // sort the array
    cout << "NO" << endl;
}
  
// Driver Code
int main()
{
    // Given array
    vector<int> arr = { 3, 4, 5, 1, 2 };
  
    // Size of the array
    int N = arr.size();
  
    // Function call to check if it is possible
    // to make array non-decreasing by rotating
    rotateArray(arr, N);
}

Output

YES

Time Complexity: O(N2)
Auxiliary Space: O(N)

Please refer complete article on Modify given array to a non-decreasing array by rotation for more details!


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