Arrays, Strings, and Pattern Techniques · Searching

Linear Search

Linear search checks items one at a time and is often the simplest correct approach when data is unsorted or small.

Enroll Now Text Us Contact Code Scholars Full DSA Course

Student Focus

Students learn when a simple scan is perfectly reasonable and when it becomes too slow.

Guided Lesson Notes

Understanding Linear Search

Linear Search focuses on contiguous sequences, index movement, and the state that can be reused while scanning. Linear search checks items one at a time and is often the simplest correct approach when data is unsorted or small.

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 Linear Search when the problem asks about a subarray, substring, range, pair, frequency, or a condition over consecutive values.

Visual Model

A small picture for Linear Search

i=0

i=1

i=2

i=3

i=4

i=5

i=6

i=7

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

Sequential scan
Early return conditions
Worst-case reasoning

Practice Prompts

Write linear search for values and objects.
Compare the number of checks for present and missing targets.

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 Linear Search

Trace a small input, focusing on sequential scan.

1Write a small input where Linear Search is clearly useful.
2Label the part of the input related to Sequential scan.
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
Indexed access	O(1)	Direct index lookup is fast when the position is known.
Full scan	O(n)	Most search, count, and validation tasks inspect each item once.
Middle insert/delete	O(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

Linear Search interview problem: Maximum Product Subarray

Interview medium

Problem

Given an integer array that may contain negative values and zeros, find the maximum product of a non-empty contiguous subarray.

Input

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

Output

Print the maximum product.

Sample Input

6
-2 3 -4 0 -1 -2

Sample Output

Why the sample works

The subarray -2, 3, -4 has product 24, which is the maximum.

Approach

1Track both the maximum and minimum product ending at the current index.
2A negative value can turn the minimum product into the new maximum product.
3Swap the two trackers when the current value is negative.
4Update the global best after processing each value.

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();
    long currentMax = sc.nextLong();
    long currentMin = currentMax;
    long answer = currentMax;

    for (int i = 1; i < n; i++) {
      long x = sc.nextLong();
      if (x < 0) {
        long temp = currentMax;
        currentMax = currentMin;
        currentMin = temp;
      }

      currentMax = Math.max(x, currentMax * x);
      currentMin = Math.min(x, currentMin * x);
      answer = Math.max(answer, currentMax);
    }

    System.out.println(answer);
  }
}

Python Solution

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

current_max = a[0]
current_min = a[0]
answer = a[0]

for x in a[1:]:
    if x < 0:
        current_max, current_min = current_min, current_max

    current_max = max(x, current_max * x)
    current_min = min(x, current_min * x)
    answer = max(answer, current_max)

print(answer)

Practice Challenge

Make the idea your own

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

Tutoring Connection

Need help with this topic?

Bring a question or code attempt to a tutoring session.

Reserve a Free Session Text Us Open Online Compiler