Searching Arrays

COMP 1010 Home > Selection Search Algorithm

Introduction A common programming scenario involves searching an array for a particular value or key. (Placeholder)

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Selection Searching

Selection searches are often used in instances where the kth smallest item must be found. The easiest way to do this is to first sort the array into ascending (or descending) order, and then iterate through each item until the conditions have been satisfied.

The idea behind a selection sort is very simple: iterate through each element in the array, and for each element look at all remaining elements to see if there is something smaller that should be in this position. If there is, exchange the two values.

Example: sort this array

|30|10|20|40|60|

1) Outer loop: iterate through each element

i = 0 (30)

2) Inner loop: iterate through the remaining elements to find a smaller element that should be in this position

j = 1 (10)

3) Since the element at i is greater than the element at j, swap them - the array will now be:

|10|30|20|40|60|

4) Again with the inner loop, we will move to the next element and compare:

i = 0 (10), j = 2 (20)

Since all the other elements in the list are greater than the 0th (10), nothing else will happen for this outer loop.

Loop 2: In the outer loop, we will move to the next element (1)

Here is an example of a very simple selection sorting algorithm:

// Selection sort algorithm

public static void selectionSort(int[] haystack) {
    for (int i = 0; i < haystack.length - 1; i++) {
        for (int j = i + 1; j < haystack.length; j++) {
            if (haystack[i] > haystack[j]) {
                //... Exchange elements
                int temp = haystack[i];
                haystack[i] = haystack[j];
                haystack[j] = temp;
            }
        }
    }
}

After the list has been sorted, you can then iterate through the list simply using a for loop.

// selectionSearch - returns the index at which the value was found, or -1 if it was not found

public static int selectionSearch(int needle, int[] haystack) {
    for (int i = 0; i < haystack.length; i++) {
        if (needle == haystack[i]) {
            return i;  // if the value was found, return its index
        }
    }
    return -1;  // if the value has not been found return -1
}

Binary Search

To help understand how the binary search works, consider this scenario:

You are using a phone book to look up a friend with the last name “McAdams”. If you used a sequential search method, you would open the phone book to the first page, then flip to the second page, continuing page by page until you have reached the M’s, and finally find McAdams.

Is that realistic? No! You would open the middle of the phone book, check if M's are on the right, or the left, and continue cutting the phone book until you find the right page.

This is the idea of the binary search. If we are searching for an element in a sorted array, we can inspect a single item and eliminate half of the array each inspection.

Here is an example of how the binary search works. Example: search for “40” in this array:

           |10|20|30|40|50|  

1) Inspect the middle element (30).
2) Is "40" this middle element, or to the left, or right of the middle element? It's on the right.
3) Eliminate the left portion of the array.

           |30|40|50|

4) Inspect the new middle element (40).
5) Is "40" this middle element, or to the left, or right of the middle element? It's the middle element.
6) Binary search is done - we have found the "40" in element 3.

Now that you understand how the binary search works, let's look at the

Binary Search Algorithm:

public static int binarySearch(int[] list, int searchValue)
{
	int left;		//left element index
	int right;		//right element index
	int middle;		//middle element index
	int result;		//
	int middleElement;	//

	result = -1;
	left = 0;
	right = list.length-1;

	while ((left <= right) && (result == -1))
	{
		middle = (left + right) / 2;
		middleElement = list(middle);
		if (middleElement == searchValue)
		{
			result = middle;
		}
		else if (middleElement < searchValue)
		{
			left = middle + 1;
		}
		else if (middleElement > searchValue)
		{
			right = middle – 1;
		}
	}//end while
	return result;
}//end binarySearch()