Selection Sort

Problem Description

Given an unsorted array of size N, use selection sort to sort arr[] in increasing order.

Examples

Example 1:

Input:
N = 5
arr[] = {4, 1, 3, 9, 7}
Output:
1 3 4 7 9
Explanation:
Maintain sorted (in bold) and unsorted subarrays.
Select 1. Array becomes 1 4 3 9 7.
Select 3. Array becomes 1 3 4 9 7.
Select 4. Array becomes 1 3 4 9 7.
Select 7. Array becomes 1 3 4 7 9.
Select 9. Array becomes 1 3 4 7 9.

Example 2:

Input:
N = 10
arr[] = {10, 9, 8, 7, 6, 5, 4, 3, 2, 1}
Output:
1 2 3 4 5 6 7 8 9 10

Your Task

You dont need to read input or print anything. Complete the functions select() and selectionSort() ,where select() takes arr[] and starting point of unsorted array i as input parameters and returns the selected element for each iteration in selection sort, and selectionSort() takes the array and it's size and sorts the array.

Expected Time Complexity: O(N^2)

Expected Auxiliary Space: O(1)

Constraints

• 1 ≤ N ≤ 10^3

Problem Explanation

The task is to traverse the array and sort the array using selection sort.

Code Implementation

C++ Solution

// C++ program for implementation of
// selection sort
#include <bits/stdc++.h>
using namespace std;

// Function for Selection sort
void selectionSort(int arr[], int n)
{
    int i, j, min_idx;

    // One by one move boundary of
    // unsorted subarray
    for (i = 0; i < n - 1; i++) {

        // Find the minimum element in
        // unsorted array
        min_idx = i;
        for (j = i + 1; j < n; j++) {
            if (arr[j] < arr[min_idx])
                min_idx = j;
        }

        // Swap the found minimum element
        // with the first element
        if (min_idx != i)
            swap(arr[min_idx], arr[i]);
    }
}

// Function to print an array
void printArray(int arr[], int size)
{
    int i;
    for (i = 0; i < size; i++) {
        cout << arr[i] << " ";
        cout << endl;
    }
}

// Driver program
int main()
{
    int arr[] = { 64, 25, 12, 22, 11 };
    int n = sizeof(arr) / sizeof(arr[0]);

    // Function Call
    selectionSort(arr, n);
    cout << "Sorted array: \n";
    printArray(arr, n);
    return 0;
}

// Java program for implementation of Selection Sort
import java.io.*;
public class SelectionSort
{
    void sort(int arr[])
    {
        int n = arr.length;

        // One by one move boundary of unsorted subarray
        for (int i = 0; i < n-1; i++)
        {
            // Find the minimum element in unsorted array
            int min_idx = i;
            for (int j = i+1; j < n; j++)
                if (arr[j] < arr[min_idx])
                    min_idx = j;

            // Swap the found minimum element with the first
            // element
            int temp = arr[min_idx];
            arr[min_idx] = arr[i];
            arr[i] = temp;
        }
    }

    // Prints the array
    void printArray(int arr[])
    {
        int n = arr.length;
        for (int i=0; i<n; ++i)
            System.out.print(arr[i]+" ");
        System.out.println();
    }

    // Driver code to test above
    public static void main(String args[])
    {
        SelectionSort ob = new SelectionSort();
        int arr[] = {64,25,12,22,11};
        ob.sort(arr);
        System.out.println("Sorted array");
        ob.printArray(arr);
    }
}

# Python program for implementation of Selection
# Sort
A = [64, 25, 12, 22, 11]

# Traverse through all array elements
for i in range(len(A)-1):
    
    # Find the minimum element in remaining 
    # unsorted array
    min_idx = i
    for j in range(i+1, len(A)):
        if A[min_idx] > A[j]:
            min_idx = j
            
    # Swap the found minimum element with 
    # the first element        
    A[i], A[min_idx] = A[min_idx], A[i]

# Driver code to test above
print ("Sorted array")
for i in range(len(A)):
    print(A[i],end=" ") 

<script>
// Javascript program for implementation of selection sort
function swap(arr,xp, yp)
{
    var temp = arr[xp];
    arr[xp] = arr[yp];
    arr[yp] = temp;
}

function selectionSort(arr,  n)
{
    var i, j, min_idx;

    // One by one move boundary of unsorted subarray
    for (i = 0; i < n-1; i++)
    {
        // Find the minimum element in unsorted array
        min_idx = i;
        for (j = i + 1; j < n; j++)
        if (arr[j] < arr[min_idx])
            min_idx = j;

        // Swap the found minimum element with the first element
        swap(arr,min_idx, i);
    }
}

function printArray( arr,  size)
{
    var i;
    for (i = 0; i < size; i++)
        document.write(arr[i] + " ");
    document.write(" <br>");
}

var arr = [64, 25, 12, 22, 11];
    var n = 5;
    selectionSort(arr, n);
    document.write("Sorted array: <br>");
    printArray(arr, n);

// This code is contributed by akshitsaxenaa09.
</script>

Solution Logic:

1. For the first position in the sorted array, the whole array is traversed from index 0 to 4 sequentially. The first position where 64 is stored presently, after traversing whole array it is clear that 11 is the lowest value.

2. Thus, replace 64 with 11. After one iteration 11, which happens to be the least value in the array, tends to appear in the first position of the sorted list.

3. For the second position, where 25 is present, again traverse the rest of the array in a sequential manner.

4. After traversing, we found that 12 is the second lowest value in the array and it should appear at the second place in the array, thus swap these values.

5. Now, for third place, where 25 is present again traverse the rest of the array and find the third least value present in the array.

6. While traversing, 22 came out to be the third least value and it should appear at the third place in the array, thus swap 22 with element present at third position.

7. Similarly, for fourth position traverse the rest of the array and find the fourth least element in the array

8. As 25 is the 4th lowest value hence, it will place at the fourth position.

9. At last the largest value present in the array automatically get placed at the last position in the array

10. The resulted array is the sorted array.

Time Complexity

• The time complexity is $O(N^2)$ as there are two nested loops

Space Complexity

• The auxiliary space complexity is $O(1)$ due to the only extra memory used is for temporary variables while swapping two values in Array. The selection sort never makes more than $O(N)$ swaps and can be useful when memory writing is costly.