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Arrays, Strings, and Pattern Techniques · Pattern Techniques

Two Pointers

Two-pointer methods use two moving indices to scan sorted arrays, strings, or windows without unnecessary nested loops.

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Student Focus

We teach students to state exactly why each pointer moves, which prevents guess-and-check coding.

Guided Lesson Notes

Understanding Two Pointers

Two Pointers focuses on contiguous sequences, index movement, and the state that can be reused while scanning. Two-pointer methods use two moving indices to scan sorted arrays, strings, or windows without unnecessary nested loops.

Model: picture the input as boxes with numbered positions; every algorithm decision should say which positions are being read, updated, skipped, or remembered.

Invariant: the variables beside the array must summarize exactly the part of the array that has already been processed.

Implementation: careful loops, boundary checks, and small helper variables for sums, counts, positions, or best answers.

Use Two Pointers when the problem asks about a subarray, substring, range, pair, frequency, or a condition over consecutive values.

Visual Model

A small picture for Two Pointers

i=0

4

i=1

1

i=2

7

i=3

2

i=4

5

i=5

9

i=6

3

i=7

6

Array and string techniques usually become clear when each index has a job: scan, compare, count, enter a range, leave a range, or store a best answer.

Key Ideas

  • Left and right pointer movement
  • Sorted-data assumptions
  • Invariant-based reasoning

Practice Prompts

  • Solve pair-sum on a sorted list.
  • Remove duplicates from a sorted array in place.

Vocabulary

Terms students should be able to say clearly

Index

The numeric position used to access an item directly.

Window

A contiguous section of the array or string currently being considered.

Prefix

Information accumulated from the start of the sequence up to a position.

Boundary

The first or last valid position included in the current scan.

State

The running information kept while the loop moves.

Invariant

The rule that makes the running state trustworthy after each iteration.

Worked Example

Worked example: tracing Two Pointers

Trace a small input, focusing on left and right pointer movement.

  1. 1Write a small input where Two Pointers is clearly useful.
  2. 2Label the part of the input related to Left and right pointer movement.
  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
Indexed accessO(1)Direct index lookup is fast when the position is known.
Full scanO(n)Most search, count, and validation tasks inspect each item once.
Middle insert/deleteO(n)Items often need to shift to keep order.

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

Two Pointers interview problem: Container With Most Water

Interview medium

Problem

Given n vertical line heights, choose two lines that hold the largest possible water container.

Input

The first line contains n. The second line contains n integers representing heights.

Output

Print the maximum container area.

Sample Input

9
1 8 6 2 5 4 8 3 7

Sample Output

49

Why the sample works

The best pair is height 8 at index 1 and height 7 at index 8, giving width 7 and area 49.

Approach

  1. 1Start one pointer at each end.
  2. 2Compute the area from the shorter boundary.
  3. 3Move the pointer at the shorter boundary inward because the taller boundary cannot improve while width shrinks.
  4. 4Keep the best area seen.

O(n) time and O(1) 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[] h = new int[n];
    for (int i = 0; i < n; i++) h[i] = sc.nextInt();

    int left = 0;
    int right = n - 1;
    long best = 0;

    while (left < right) {
      long area = (long) Math.min(h[left], h[right]) * (right - left);
      best = Math.max(best, area);

      if (h[left] < h[right]) {
        left++;
      } else {
        right--;
      }
    }

    System.out.println(best);
  }
}

Python Solution

n = int(input())
h = list(map(int, input().split()))

left, right = 0, n - 1
best = 0

while left < right:
    best = max(best, min(h[left], h[right]) * (right - left))
    if h[left] < h[right]:
        left += 1
    else:
        right -= 1

print(best)

Practice Challenge

Make the idea your own

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

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