Explain Selection Sort in detail?

Software Development
#1

The selection sort algorithm sorts an array by repeatedly finding the minimum element (considering ascending order) from the unsorted part and putting it at the beginning.

The algorithm maintains two subarrays in a given array.

The subarray which already sorted.
The remaining subarray was unsorted.

Below is the implementation of the above approach:

#include <bits/stdc++.h>
using namespace std;
 
//Swap function
void swap(int *xp, int *yp)
{
    int temp = *xp;
    *xp = *yp;
    *yp = temp;
}
 
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 to test above functions

int main()
{
    int arr[] = {64, 25, 12, 22, 11};
    int n = sizeof(arr)/sizeof(arr[0]);
    selectionSort(arr, n);
    cout << "Sorted array: \n";
    printArray(arr, n);
    return 0;
}

Output

Sorted array: 
11 12 22 25 64

Complexity Analysis of Selection Sort:

Time Complexity: The time complexity of Selection Sort is O(N2) as there are two nested loops:

One loop to select an element of Array one by one = O(N)
Another loop to compare that element with every other Array element = O(N)
Therefore overall complexity = O(N) * O(N) = O(N*N) = O(N2)