Selection Sort

Sorting Algorithm


About Selection Sort

In this technique we find the smallest element in each pass and place it in the appropriate position to get the elements in ascending or descending order.
In pass 1, smallest element is searched between a[0] to a[n-1] and swapped with a[0].
In pass 2, smallest element is searched between a[1] to a[n-1] and swapped with a[1].
In a similar way the process is carried out n-1 times.

  • Selection sort requires n – 1 pass to sort an array of n elements.
  • In each pass we search for the smallest element from the search range and swap it with appropriate place.
  • In each pass we have n – k comparisons, where n is the number of elements and k is the pass number.
    So, 1st pass requires n – 1 comparisons,
    kth pass requires n – k comparisons and
    the last pass requires 1 comparison.


a[0:n-1] is an array of n elements.
temp is a variable to facilitate exchange.
    for k = 1 to n-1 by 1 do	//this is for pass
        Set small = a[k-1];
        Set pos = k-1;
        for j = k to n-1 by 1 do	//this is for searching small element
            if(a[j] < small) then
                Set small = a[j];
                Set pos = j;
        if(pos != k-1) then	//swap value
            Set temp = a[k-1];
            Set a[k-1] = a[pos];
            Set a[pos] = temp;

Code in C

#include <stdio.h>

//function declaration
void selectionSort(int *a, int n);

int main(){
	//variable declaration
	int arr[5], i;
	for(i = 0; i < 5; i++)
		scanf("%d", &arr[i]);
	selectionSort(arr, 5);	//passing arr address and no. of elements
	for(i = 0; i < 5; i++)
		printf("%d\n", arr[i]);
	return 0;

//function definition
void selectionSort(int *a, int n){
	int k, j, pos, small, temp;
	for(k = 1; k <= n-1; k++){
		small = a[k-1];
		pos = k-1;
		for(j = k; j <= n-1; j++){
			if(a[j] < small){
				small = a[j];
				pos = j;
		if(pos != k-1){
			temp = a[k-1];
			a[k-1] = a[pos];
			a[pos] = temp;

Time complexity

1st pass requires (n-1) comparison
2nd pass required (n-2) comparison
... last pass requires 1 comparison
So, total comparison = (n-1) + (n-2) + ... + 1 = n(n-1)/2 = O(n2)