In selection sort,

- In each pass smallest/largest element is selected and placed in a sorted list.
- Th entire array is divided into two parts: sorted and unsorted
- In each pass, in the unsorted subarray, the smallest element is selected and exchanged with the first element.

**Algorithm**

1. Declare and initialize necessary variables such as array[], i, j, large, n etc.

2. for ( i = n - 1; i > 0; i--), repeat following steps

large = x[0];

index = 0;

2.1 for(j = i ; j <= i; j++)

if x[j] > large

large = x[i]

index = j

2.2 x[index] =x[i]

x[i] = large

3. Display the sorted array

**Source Code:**

#include<iostream>using namespace std;class SelectionSort{public:int no_of_elements;int elements[10];public:void getarray();void sortit(int [], int);void display();};void SelectionSort::getarray(){cout<<"How many elements? ";cin>>no_of_elements;cout<<"Insert array of element to sort: ";for(int i=0;i<no_of_elements;i++){cin>>elements[i];}}void SelectionSort::sortit(int x[], int n){int i, indx, j, large;for(i = n - 1; i > 0; i--){large = x[0];indx = 0;for(j = 1; j <= i; j++){if(x[j] > large){large = x[j];indx = j;}}x[indx] = x[i];x[i] = large;}}void SelectionSort::display(){cout<<"The sorted array is :\n";for(int i = 0 ; i < no_of_elements; i++){cout<<elements[i]<<" ";}cout<<endl;}int main(){SelectionSort SS;SS.getarray();SS.sortit(SS.elements,SS.no_of_elements);SS.display();return 0;}

**Output:**

How many elements? 6

Insert array of element to sort: 78 56 110 12 56 26

The sorted list is 12 26 56 56 78 110

**Efficiency of Straight selection sort**

In this first pass, it makes (n – 1) comparisons, in second pass it makes (n – 2) comparisons and so on, So total number of comparisons is n(n-1)/2 which is O(n^2)

thanks.

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