Nothing special here. It’s just a blog post for summarising my algorithm learning course. Although this was already taught in the University, I didn’t even know that it can be used for Selection Problem

# Selection Problem

Given an array of N items, find a k^{th} smallest item. For example, an array A =
`[5, 2, 9, 4, 10, 7]`

, the 3rd smallest item is `5`

, the 2nd smallest item is `4`

and the smallest
item is `2`

# The idea

- Based on Quick Sort
- Partition the array so that
- Entry
`a[j]`

is in place. - No larger entry to the left of
`j`

- No smaller entry to the right of
`j`

- Entry
- Repeat in one sub-array, depending on
`j`

, finished when`j`

equals`k`

# Java Implementation

```
public static Comparable select(Comparable[] a, int k) {
StdRandom.shuffle(a);
int lo = 0, hi = a.length - 1;
while (hi > lo)
{
int j = partition(a, lo, hi);
if (j < k) lo = j + 1;
else if (j > k) hi = j - 1;
else return a[k];
}
return a[k];
}
```

The idea is that, after each partition, the item `a[j]`

will be moved to the right place. So it will
become the **j ^{th}** smallest item. Repeat that process until we can find

`j = k`

.# Performance

**Quick-select** takes **linear** time on average. **Quick-select** uses **~½ N ^{2}**
compares in the worst case, but (as with quicksort) the random shuffle provides a probabilistic
guarantee.