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

• Difficulty Level : Easy
• Last Updated : 19 May, 2021

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

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

• Traverse the given array arr[].
• 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 its correct position.
For Example:

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

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

Below is the implementation of the above approach:

## C++

 `// 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);` `    ``// Function call to sort the array``    ``sortArray(arr, N);` `    ``// Function call to print the array``    ``printArray(arr, N);``    ``return` `0;``}`

## Java

 `// 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`

## Python3

 `# 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   `

## C#

 `// C# program for the above approach``using` `System;``class` `GFG{``    ` `// Function to print array element``public` `static` `void` `printArray(``int` `[]arr, ``int` `N)``{   ``    ``// Traverse the array``    ``for``(``int` `i = 0; i < N; i++)``    ``{``       ``Console.Write(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 shikhasingrajput`

## Javascript

 ``
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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