Given an array arr[] containing N integers, with duplicates. The task is to sort the array in increasing order using at most N cyclic shift on any sub-array.
Cyclic shift on any sub-array means cut out any subarray from the given array, use cyclic shift (rotate) in it by any offset, and put it back into the same place of the array.
Print the number of such shifting required to sort the array. There can be multiple possibilities.
Examples:
Input: arr[] = [1, 3, 2, 8, 5]
Output: 2
Explanation: Consider segment from index = 1 to index = 2. [1, 3, 2, 8, 5]. Now rotate this segment by 1 offset. The new array becomes, [1, 2, 3, 8, 5].
Then consider segment from index = 3 to index = 4, [1, 2, 3, 8, 5]. Rotate it by 1 offset, the new array becomes, [1, 2, 3, 5, 8].Input: arr[] = [2, 4, 1, 3]
Output: 2
Explanation: From index = 0 to index = 2, [2, 4, 1, 3]. Rotate this segment by 2 offset on left, the new array becomes, [1, 2, 4, 3].
Taking second segment from index = 2 to index = 3 of offset 1, rotate it the new array becomes, [1, 2, 4, 3].
Approach: There can be two cases:
- Case when the array is already sorted: Then we do not need to perform any shifting operation
- Case when the array is not sorted: For that follow the steps mentioned below:
- Create a variable count = 0 to store the total number of counts.
- Iterate from i = 0 to i = N-2. And for each iteration do the following operations:
- Create a variable minVal to store the minimum value found in this iteration and initiate it with the value of arr[i].
- Start iterating from i+1 to N-1. If the current value is less than minVal, update minVal.
- Mark that position right to perform cyclic shift from i to right by offset 1.
- If no such right value exists then simply move to the next iteration.
- Otherwise, rotate the array from i to right by 1 position and increment count by 1.
- Return the value of count as your answer.
Below is the C++ implementation for the above approach:
C++
// C++ code to implement the above approach#include <bits/stdc++.h>using namespace std;// Function to Sort given Array using// at most N cyclic shift on any subarrayint ShiftingSort(vector<int>& arr, int n){ vector<int> temp(arr.begin(), arr.end()); sort(temp.begin(), temp.end()); // Variable to store count of // shifting operations int count = 0; // If the vector arr is already sorted // the 0 operations shift required if (arr == temp) { return 0; } else { // Run a loop from 0 to n-2 index for (int index1 = 0; index1 < n - 1; index1++) { int minval = arr[index1]; int left = index1; int right = -1; // Loop from i+1 to n-1 // to find the minimum value for (int index2 = index1 + 1; index2 < n; index2++) { if (minval > arr[index2]) { minval = arr[index2]; right = index2; } } // Check if the shifting is required if (right != -1) { // Increment count operations count++; int index = right; int tempval = arr[right]; // Loop for shifting the array arr // from index = left to index = right while (index > left) { arr[index] = arr[index - 1]; index--; } arr[index] = tempval; } } } // Return the total operations return count;}// Driver codeint main(){ vector<int> arr{ 1, 3, 2, 8, 5 }; int N = 5; cout << ShiftingSort(arr, N) << "\n"; return 0;} |
Java
// Java code to implement the above approachimport java.util.*;public class GFG{ // Function to Sort given Array using// at most N cyclic shift on any subarraystatic int ShiftingSort(ArrayList<Integer> arr, int n){ ArrayList<Integer> temp = new ArrayList<Integer>(); for(int i = 0; i < arr.size(); i++) { temp.add(arr.get(i)); } Collections.sort(temp); // Variable to store count of // shifting operations int count = 0; // If the vector arr is already sorted // the 0 operations shift required if (arr.equals(temp)) { return 0; } else { // Run a loop from 0 to n-2 index for (int index1 = 0; index1 < n - 1; index1++) { int minval = arr.get(index1); int left = index1; int right = -1; // Loop from i+1 to n-1 // to find the minimum value for (int index2 = index1 + 1; index2 < n; index2++) { if (minval > arr.get(index2)) { minval = arr.get(index2); right = index2; } } // Check if the shifting is required if (right != -1) { // Increment count operations count++; int index = right; int tempval = arr.get(right); // Loop for shifting the array arr // from index = left to index = right while (index > left) { arr.set(index, arr.get(index - 1)); index--; } arr.set(index, tempval); } } } // Return the total operations return count;}// Driver codepublic static void main(String args[]){ ArrayList<Integer> arr = new ArrayList<Integer>(); arr.add(1); arr.add(3); arr.add(2); arr.add(8); arr.add(5); int N = 5; System.out.println(ShiftingSort(arr, N));}}// This code is contributed by Samim Hossain Mondal. |
Python3
# Python Program to implement# the above approach# Function to Sort given Array using# at most N cyclic shift on any subarraydef ShiftingSort(arr, n): temp = arr.copy() temp.sort() # Variable to store count of # shifting operations count = 0 # If the vector arr is already sorted # the 0 operations shift required if (arr == temp): return 0 else: # Run a loop from 0 to n-2 index for index1 in range(n - 1): minval = arr[index1] left = index1 right = -1 # Loop from i+1 to n-1 # to find the minimum value for index2 in range(index1 + 1, n): if (minval > arr[index2]): minval = arr[index2] right = index2 # Check if the shifting is required if (right != -1): # Increment count operations count += 1 index = right tempval = arr[right] # Loop for shifting the array arr # from index = left to index = right while (index > left): arr[index] = arr[index - 1] index -= 1 arr[index] = tempval # Return the total operations return count# Driver codearr = [1, 3, 2, 8, 5]N = 5print(ShiftingSort(arr, N))# This code is contributed by gfgking |
C#
// C# code to implement the above approachusing System;using System.Collections;using System.Collections.Generic;class GFG{ // Function to Sort given Array using// at most N cyclic shift on any subarraystatic int ShiftingSort(ArrayList arr, int n){ ArrayList temp = new ArrayList(); for(int i = 0; i < arr.Count; i++) { temp.Add(arr[i]); } temp.Sort(); // Variable to store count of // shifting operations int count = 0; // If the vector arr is already sorted // the 0 operations shift required if (arr.Equals(temp)) { return 0; } else { // Run a loop from 0 to n-2 index for (int index1 = 0; index1 < n - 1; index1++) { int minval = (int)arr[index1]; int left = index1; int right = -1; // Loop from i+1 to n-1 // to find the minimum value for (int index2 = index1 + 1; index2 < n; index2++) { if (minval > (int)arr[index2]) { minval = (int)arr[index2]; right = index2; } } // Check if the shifting is required if (right != -1) { // Increment count operations count++; int index = right; int tempval = (int)arr[right]; // Loop for shifting the array arr // from index = left to index = right while (index > left) { arr[index] = arr[index - 1]; index--; } arr[index] = tempval; } } } // Return the total operations return count;}// Driver codepublic static void Main(){ ArrayList arr = new ArrayList(); arr.Add(1); arr.Add(3); arr.Add(2); arr.Add(8); arr.Add(5); int N = 5; Console.Write(ShiftingSort(arr, N));}}// This code is contributed by Samim Hossain Mondal. |
Javascript
<script> // JavaScript Program to implement // the above approach // Function to Sort given Array using // at most N cyclic shift on any subarray function ShiftingSort(arr, n) { let temp = [...arr]; temp.sort(function (a, b) { return a - b }) // Variable to store count of // shifting operations let count = 0; // If the vector arr is already sorted // the 0 operations shift required if (arr == temp) { return 0; } else { // Run a loop from 0 to n-2 index for (let index1 = 0; index1 < n - 1; index1++) { let minval = arr[index1]; let left = index1; let right = -1; // Loop from i+1 to n-1 // to find the minimum value for (let index2 = index1 + 1; index2 < n; index2++) { if (minval > arr[index2]) { minval = arr[index2]; right = index2; } } // Check if the shifting is required if (right != -1) { // Increment count operations count++; let index = right; let tempval = arr[right]; // Loop for shifting the array arr // from index = left to index = right while (index > left) { arr[index] = arr[index - 1]; index--; } arr[index] = tempval; } } } // Return the total operations return count; } // Driver code let arr = [1, 3, 2, 8, 5]; let N = 5; document.write(ShiftingSort(arr, N) + '<br>'); // This code is contributed by Potta Lokesh </script> |
Output
2
Time Complexity: O(N2)
Space Complexity: O(1)
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