# 8.5. Selection Sort¶

## 8.5.1. Selection Sort¶

Consider again the problem of sorting a pile of phone bills for the past year. Another intuitive approach might be to look through the pile until you find the bill for January, and pull that out. Then look through the remaining pile until you find the bill for February, and add that behind January. Proceed through the ever-shrinking pile of bills to select the next one in order until you are done. This is the inspiration for our last \(\Theta(n^2)\) sort, called Selection Sort. The \(i\)’th pass of Selection Sort “selects” the \(i\)’th largest key in the array, placing that record at the end of the array. In other words, Selection Sort first finds the largest key in an unsorted list, then the next largest, and so on. Its unique feature is that there are few record swaps. To find the next-largest key value requires searching through the entire unsorted portion of the array, but only one swap is required to put the record into place. Thus, the total number of swaps required will be \(n-1\) (we get the last record in place “for free”).

Here is an implementation for Selection Sort.

```
static <T extends Comparable<T>> void selsort(T[] A) {
for (int i=0; i<A.length-1; i++) { // Select i'th biggest record
int bigindex = 0; // Current biggest index
for (int j=1; j<A.length-i; j++) // Find the max value
if (A[j].compareTo(A[bigindex]) > 0) // Found something bigger
bigindex = j; // Remember bigger index
Swap.swap(A, bigindex, A.length-i-1); // Put it into place
}
}
```

```
static <T extends Comparable<T>> void selsort(T[] A) {
for (int i=0; i<A.length-1; i++) { // Select i'th biggest record
int bigindex = 0; // Current biggest index
for (int j=1; j<A.length-i; j++) { // Find the max value
if (A[j].compareTo(A[bigindex]) > 0) {// Found something bigger
bigindex = j; // Remember bigger index
}
}
swap(A, bigindex, A.length-i-1); // Put it into place
}
}
```

```
def selsort(A):
for i in range(len(A)): # Select i'th biggest record
bigindex = 0; # Current biggest index
for j in range (1, len(A) - i): # Find the max value
if (A[j] > A[bigindex]): # Found something bigger
bigindex = j; # Remember bigger index
swap(A, bigindex, len(A) - i - 1); # Put it into place
```

Consider the example of the following array.

Now we continue with the second pass. However, since the largest record is already at the right end, we will not need to look at it again.

Selection Sort continues in this way until the entire array is sorted.

The following visualization puts it all together.

Now try for yourself to see if you understand how Selection Sort works.

## 8.5.2. Selection Sort Analysis¶

Any algorithm can be written in slightly different ways. For example, we could have written Selection Sort to find the smallest record, the next smallest, and so on. We wrote this version of Selection Sort to mimic the behavior of our Bubble Sort implementation as closely as possible. This shows that Selection Sort is essentially a Bubble Sort except that rather than repeatedly swapping adjacent values to get the next-largest record into place, we instead remember the position of the record to be selected and do one swap at the end.

This visualization analyzes the number of comparisons and swaps required by Selection Sort.

There is another approach to keeping the cost of swapping records low, and it can be used by any sorting algorithm even when the records are large. This is to have each element of the array store a pointer to a record rather than store the record itself. In this implementation, a swap operation need only exchange the pointer values. The large records do not need to move. This technique is illustrated by the following visualization. Additional space is needed to store the pointers, but the return is a faster swap operation.

Here are some review questions to check how well you understand Selection Sort.