Sort an Array which contain 1 to N values in O(N) using Cycle Sort

Prerequisite: Cycle Sort

Given an array arr[] of elements from 1 to N, the task is to sort the given array in O(N) time.

Examples:



Input: arr[] = { 2, 1, 5, 4, 3}
Output: 1 2 3 4 5
Explanation:
Since arr[0] = 2 is not at correct position, then swap arr[0] with arr[arr[0] – 1]
Now array becomes: arr[] = {1, 2, 5, 4, 3}

Now arr[2] = 5 is not at correct position, then swap arr[3] with arr[arr[3] – 1]
Now the array becomes: arr[] = {1, 2, 3, 4, 5}
Now the array is sorted.

Input: arr[] = {1, 2, 3, 4, 5, 6}
Output: 1 2 3 4 5 6
Explanation:
The array is already sorted.

Approach: This problem can be solved using Greedy Approach. Below are the steps:

  1. Traverse the given array arr[].
  2. If the current element is not at the correct position i.e., arr[i] is not equal to i+1 then, swap the current element with the element with it’s correct position.
    For Example:

    Let arr[] = {2, 1, 4, 5, 3}
    Since, arr[0] = 2, which is not at it’s correct position 1.
    Then swap this element with it’s correct position, i.e., swap(arr[0], arr[2-1])
    Then array becomes: arr[] = {1, 2, 4, 5, 3}

  3. If the current element is at the correct position then check for the next element.
  4. Repeat the above steps till we reach the end of the array.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
  
#include "bits/stdc++.h"
using namespace std;
  
// Function to swap two a & b value
void swap(int* a, int* b)
{
    int temp = *a;
    *a = *b;
    *b = temp;
}
  
// Function to print array element
void printArray(int arr[], int N)
{
  
    // Traverse the array
    for (int i = 0; i < N; i++) {
        cout << arr[i] << ' ';
    }
}
  
// Function to sort the array in O(N)
void sortArray(int arr[], int N)
{
  
    // Traverse the array
    for (int i = 0; i < N;) {
  
        // If the current element is
        // at correct position
        if (arr[i] == i + 1) {
            i++;
        }
  
        // Else swap the current element
        // with it's correct position
        else {
            swap(&arr[i], &arr[arr[i] - 1]);
        }
    }
}
  
// Driver Code
int main()
{
  
    int arr[] = { 2, 1, 5, 3, 4 };
    int N = sizeof(arr) / sizeof(arr[0]);
  
    // Function call to sort the array
    sortArray(arr, N);
  
    // Function call to print the array
    printArray(arr, N);
    return 0;
}

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Java

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// Java program for the above approach 
class Main{
      
// Function to print array element 
public static void printArray(int arr[], int N) 
      
    // Traverse the array 
    for(int i = 0; i < N; i++)
    
       System.out.print(arr[i] + " ");
    
      
// Function to sort the array in O(N) 
public static void sortArray(int arr[], int N) 
  
    // Traverse the array 
    for(int i = 0; i < N;)
    
  
       // If the current element is 
       // at correct position 
       if (arr[i] == i + 1)
       
           i++; 
       
         
       // Else swap the current element 
       // with it's correct position 
       else 
       
           // Swap the value of 
           // arr[i] and arr[arr[i]-1]
           int temp1 = arr[i]; 
           int temp2 = arr[arr[i] - 1];
           arr[i] = temp2;
           arr[temp1 - 1] = temp1;
       
    
  
// Driver Code    
public static void main(String[] args)
{
    int arr[] = { 2, 1, 5, 3, 4 }; 
    int N = arr.length; 
  
    // Function call to sort the array 
    sortArray(arr, N); 
  
    // Function call to print the array 
    printArray(arr, N); 
}
}
  
// This code is contributed by divyeshrabadiya07

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Python3

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# Python3 program for the above approach 
  
# Function to print array element 
def printArray(arr, N):
      
    # Traverse the array 
    for i in range(N):
        print(arr[i], end = ' ')
          
# Function to sort the array in O(N) 
def sortArray(arr, N):
      
    i = 0
      
    # Traverse the array 
    while i < N:
          
        # If the current element is 
        # at correct position
        if arr[i] == i + 1:
            i += 1
          
        # Else swap the current element 
        # with it's correct position
        else:
              
            # Swap the value of 
            # arr[i] and arr[arr[i]-1] 
            temp1 = arr[i]
            temp2 = arr[arr[i] - 1]
            arr[i] = temp2
            arr[temp1 - 1] = temp1
      
# Driver code 
if __name__=='__main__':
      
    arr = [ 2, 1, 5, 3, 4 ]
    N = len(arr)
      
    # Function call to sort the array
    sortArray(arr, N)
      
    # Function call to print the array
    printArray(arr, N)
  
# This code is contributed by rutvik_56    

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Output:

1 2 3 4 5

Time Complexity: O(N), where N is the length of the array.
Auxiliary Space: O(1)

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