Topic 4.2 — Traversing Arrays

Goal: use loops (especially for) to visit each element in an array, compute results, and apply common traversal patterns safely.

The big idea

Traversing an array means visiting elements one by one (often left to right) using an index.

On AP CSA, you’ll see a lot of “find the sum / count / max / average” style traversals. Most of them are just different versions of the same loop pattern.

Canonical traversal loop (memorize)

for (int i = 0; i < arr.length; i++) {
  // use arr[i]
}

Start at 0 (first index)
Stop at i < arr.length (last index is arr.length - 1)

Pattern 1: Sum (accumulator)

An accumulator variable stores a running total as you traverse.

int sum = 0;
for (int i = 0; i < nums.length; i++) {
  sum += nums[i];
}

Pattern 2: Count something

Count elements that match a condition.

int countEven = 0;
for (int i = 0; i < nums.length; i++) {
  if (nums[i] % 2 == 0) {
    countEven++;
  }
}

Pattern 3: Find max/min

Usually initialize with the first element, then compare.

int max = nums[0];
for (int i = 1; i < nums.length; i++) {
  if (nums[i] > max) {
    max = nums[i];
  }
}

Pattern 4: Compute average

Average = sum ÷ count. Watch out for integer division.

int sum = 0;
for (int i = 0; i < nums.length; i++) {
  sum += nums[i];
}
double avg = (double) sum / nums.length;

Classic bug: off-by-one

This tries to access arr[arr.length] (invalid).

for (int i = 0; i <= arr.length; i++) { // ❌
  System.out.println(arr[i]);
}

Fix: use i < arr.length.

Traversing while updating elements

You can modify an array during traversal by assigning to arr[i].

// Add 5 to every element
for (int i = 0; i < nums.length; i++) {
  nums[i] += 5;
}

This is a very common AP-style loop: “update each element based on a rule.”

Forward vs backward traversal

Sometimes you traverse from the end (especially when removing/shifting in other units).

for (int i = arr.length - 1; i >= 0; i--) {
  // use arr[i]
}

When you need the index

When updating: arr[i] = ...
When comparing neighbors: arr[i] vs arr[i+1]
When tracking the location of something (like the index of max)

int maxIndex = 0;
for (int i = 1; i < nums.length; i++) {
  if (nums[i] > nums[maxIndex]) {
    maxIndex = i;
  }
}

Quick self-check

Write a loop that prints every element in arr.
What’s wrong with i <= arr.length?
Write a loop that counts how many values are negative.
How do you compute an average without integer division problems?
How would you find the index of the smallest value?

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