# 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 = 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:

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 = 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}

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++

 `// 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     `

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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