# Explain Selection Sort in detail?

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 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);
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)