Sorting Algorithms · Sorting

Insertion Sort

Insertion sort builds a sorted prefix by inserting each new value into its correct place.

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Student Focus

This sort is a good place to teach invariants and practical tradeoffs.

Guided Lesson Notes

Understanding Insertion Sort

Insertion Sort focuses on ordering data so later operations become simpler, faster, or easier to prove correct. Insertion sort builds a sorted prefix by inserting each new value into its correct place.

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

Visual Model

A small picture for Insertion Sort

Key Ideas

Sorted prefix invariant
Shifting values
Nearly sorted performance

Practice Prompts

Trace insertion sort on a nearly sorted list.
Explain why it can be efficient for small or mostly sorted data.

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 Insertion Sort

Trace a small input, focusing on sorted prefix invariant.

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

Complexity Check

Costs students should be able to explain

Operation	Typical Cost	Reason
Simple comparison sorts	O(n^2)	Selection, insertion, and bubble sort compare many pairs on larger inputs.
Divide-and-conquer sorts	O(n log n)	Merge sort and average quicksort reduce the problem by splitting.
Non-comparison sorts	O(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

Insertion Sort 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

Sample Output

Why the sample works

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

Approach

1Sort intervals by starting point.
2Keep the current merged interval.
3If the next interval starts before the current end, extend the end.
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 Insertion Sort: define it, trace one example, name one edge case, and give the runtime of the main operation.

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