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Sorting Algorithms · Sorting

Quicksort

Quicksort partitions around a pivot and recursively sorts the sides, often fast in practice but sensitive to pivot choices.

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Student Focus

Students learn why average-case speed and worst-case risk can coexist.

Guided Lesson Notes

Understanding Quicksort

Quicksort focuses on ordering data so later operations become simpler, faster, or easier to prove correct. Quicksort partitions around a pivot and recursively sorts the sides, often fast in practice but sensitive to pivot choices.

Model: view the array as unsorted, partially sorted, and final regions; the algorithm is the rule for growing the final order.

Invariant: after each pass or recursive call, a known part of the data must be correctly ordered or correctly partitioned.

Implementation: comparisons, swaps, shifts, partitions, merges, counts, buckets, or digit passes depending on the algorithm.

Use Quicksort when the problem becomes easier after values are ordered, grouped, compared with neighbors, or processed by rank.

Visual Model

A small picture for Quicksort

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3
9
4
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5

Key Ideas

  • Pivot selection
  • Partitioning
  • Average versus worst case

Practice Prompts

  • Partition a list by hand around a pivot.
  • Compare quicksort behavior on random and already sorted input.

Vocabulary

Terms students should be able to say clearly

Comparison sort

A sorting method that learns order by comparing pairs of items.

Stable sort

A sort that preserves the relative order of equal keys.

Partition

Splitting values around a pivot or rule.

Merge

Combining sorted parts into a larger sorted result.

In-place

Using only a small amount of extra storage.

Distribution

How values are spread across the input range.

Worked Example

Worked example: tracing Quicksort

Trace a small input, focusing on pivot selection.

  1. 1Write a small input where Quicksort is clearly useful.
  2. 2Label the part of the input related to Pivot selection.
  3. 3Perform one operation and explain which invariant is still true afterward.
  4. 4Run a second operation that touches an edge case, such as an empty side, duplicate value, boundary index, disconnected vertex, or tie.
  5. 5Finish by saying which operation dominates the runtime and why.

Complexity Check

Costs students should be able to explain

OperationTypical CostReason
Simple comparison sortsO(n^2)Selection, insertion, and bubble sort compare many pairs on larger inputs.
Divide-and-conquer sortsO(n log n)Merge sort and average quicksort reduce the problem by splitting.
Non-comparison sortsO(n + k)Counting-style methods can be faster when the value range is controlled.

Common Mistakes

What to watch while practicing

  • Coding before drawing the structure or state changes.
  • Forgetting the invariant that makes the algorithm correct.
  • Testing only the sample input and skipping boundary cases.
  • Giving Big-O without explaining which operation dominates the work.

Interview-Style Coding Problem

Quicksort interview problem: Merge Overlapping Intervals

Interview medium

Problem

Given a list of intervals, merge all overlapping intervals and print the condensed list.

Input

The first line contains n. The next n lines contain start and end values.

Output

Print the number of merged intervals, followed by one interval per line.

Sample Input

4
1 3
2 6
8 10
15 18

Sample Output

3
1 6
8 10
15 18

Why the sample works

Intervals [1, 3] and [2, 6] overlap, so they become [1, 6].

Approach

  1. 1Sort intervals by starting point.
  2. 2Keep the current merged interval.
  3. 3If the next interval starts before the current end, extend the end.
  4. 4Otherwise save the current interval and start a new one.

O(n log n) time and O(n) extra space.

Java Solution

import java.util.*;

public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    int[][] intervals = new int[n][2];
    for (int i = 0; i < n; i++) {
      intervals[i][0] = sc.nextInt();
      intervals[i][1] = sc.nextInt();
    }

    Arrays.sort(intervals, Comparator.comparingInt(a -> a[0]));
    List<int[]> merged = new ArrayList<>();

    for (int[] interval : intervals) {
      if (merged.isEmpty() || interval[0] > merged.get(merged.size() - 1)[1]) {
        merged.add(interval);
      } else {
        int[] last = merged.get(merged.size() - 1);
        last[1] = Math.max(last[1], interval[1]);
      }
    }

    System.out.println(merged.size());
    for (int[] interval : merged) {
      System.out.println(interval[0] + " " + interval[1]);
    }
  }
}

Python Solution

n = int(input())
intervals = [tuple(map(int, input().split())) for _ in range(n)]
intervals.sort()

merged = []
for start, end in intervals:
    if not merged or start > merged[-1][1]:
        merged.append([start, end])
    else:
        merged[-1][1] = max(merged[-1][1], end)

print(len(merged))
for start, end in merged:
    print(start, end)

Practice Challenge

Make the idea your own

Create a two-minute explanation of Quicksort: define it, trace one example, name one edge case, and give the runtime of the main operation.

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