Array Traversal Patterns in DSA Java - Time & Space Complexity

Time Complexity: Array Traversal Patterns

O(n)

Understanding Time Complexity

When we go through an array, we want to know how long it takes as the array gets bigger.

We ask: How does the work grow when the array size grows?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code goes through each item in the array and prints it once.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Accessing and printing each element in the array.
• How many times: Exactly once for each element, so as many times as the array length.

How Execution Grows With Input

As the array gets bigger, the work grows in a straight line with the number of elements.

Input Size (n)	Approx. Operations
10	10 prints
100	100 prints
1000	1000 prints

Pattern observation: Doubling the array size doubles the work.

Final Time Complexity

Time Complexity: O(n)

This means the time to finish grows directly with the number of items in the array.

Common Mistake

[X] Wrong: "Since the loop is simple, it must be constant time O(1)."

[OK] Correct: The loop runs once for each element, so time grows with array size, not fixed.

Interview Connect

Understanding how simple loops grow with input size helps you explain your code clearly and shows you know how to reason about efficiency.

Self-Check

"What if we nested another loop inside to compare each element with every other? How would the time complexity change?"

Practice

1. What does the following Java code snippet do?
for (int i = 0; i < arr.length; i++) { System.out.print(arr[i] + " "); }

easy

A. Prints all elements of the array from end to start

B. Prints only the first element of the array

C. Prints all elements of the array from start to end

D. Prints only the last element of the array

Solution

1. Step 1: Understand the loop range

   The loop starts at index 0 and runs while i is less than the array length, so it covers all valid indices from start to end.

2. Step 2: Analyze the print statement inside the loop

   Each element at index i is printed followed by a space, so all elements are printed in order.

3. Final Answer:

   Prints all elements of the array from start to end -> Option C

4. Quick Check:

   Loop from 0 to length-1 prints all elements [OK]

Hint: Loop from 0 to length-1 prints all elements in order [OK]

Common Mistakes:

• Confusing loop direction and thinking it prints backwards
• Assuming it prints only one element
• Ignoring the loop condition

2. Which of the following is the correct syntax to traverse an array arr backwards in Java?

easy

A. for (int i = arr.length; i > 0; i--)

B. for (int i = 0; i < arr.length; i++)

C. for (int i = 1; i <= arr.length; i++)

D. for (int i = arr.length - 1; i >= 0; i--)

Solution

1. Step 1: Identify correct start and end indices for backward traversal

   Backward traversal starts at last index (arr.length - 1) and ends at 0 inclusive.

2. Step 2: Check loop condition and decrement

   The loop continues while i is greater than or equal to 0, decrementing i each time, correctly covering all indices backwards.

3. Final Answer:

   for (int i = arr.length - 1; i >= 0; i--) -> Option D

4. Quick Check:

   Start at last index, go down to 0 [OK]

Hint: Start at length-1, loop down to 0 with i-- [OK]

Common Mistakes:

• Starting loop at arr.length (out of bounds)
• Using i > 0 instead of i >= 0 (misses index 0)
• Starting loop at 1 instead of length-1

3. What is the output of the following code?

int[] arr = {2, 4, 6, 8};
for (int i = 1; i < arr.length; i += 2) {
    System.out.print(arr[i] + " ");
}

medium

A. 4 8

B. 2 6

C. 4 6 8

D. 2 4 6 8

Solution

1. Step 1: Identify loop indices

   The loop starts at i=1 and increments by 2, so i takes values 1 and 3 (since arr.length=4).

2. Step 2: Print elements at indices 1 and 3

   arr[1] = 4 and arr[3] = 8, so output is "4 8 "

3. Final Answer:

   4 8 -> Option A

4. Quick Check:

   Indices 1 and 3 printed: 4 and 8 [OK]

Hint: Loop i=1; i<length; i+=2 prints odd indices [OK]

Common Mistakes:

• Assuming loop prints all elements
• Confusing index increments
• Printing elements at even indices instead

4. Find the error in this code snippet that tries to print array elements backwards:

int[] arr = {1, 3, 5, 7};
for (int i = arr.length; i >= 0; i--) {
    System.out.print(arr[i] + " ");
}

medium

A. Loop starts at arr.length which is out of bounds

B. Loop condition should be i <= arr.length

C. Array index should start at 0

D. No error, code works fine

Solution

1. Step 1: Check starting index of loop

   arr.length is 4, but valid indices are 0 to 3. Starting at 4 causes ArrayIndexOutOfBoundsException.

2. Step 2: Correct starting index for backward traversal

   Loop should start at arr.length - 1 (3) to access last element safely.

3. Final Answer:

   Loop starts at arr.length which is out of bounds -> Option A

4. Quick Check:

   Array indices go from 0 to length-1 [OK]

Hint: Array indices max is length-1, not length [OK]

Common Mistakes:

• Starting loop at arr.length instead of arr.length-1
• Ignoring index out of bounds error
• Confusing loop condition signs

5. Given an integer array arr, which code snippet correctly prints all elements at even indices in reverse order?

hard

A. for (int i = arr.length - 1; i >= 0; i -= 2) { System.out.print(arr[i] + " "); }

B. for (int i = (arr.length % 2 == 0 ? arr.length - 2 : arr.length - 1); i >= 0; i -= 2) { System.out.print(arr[i] + " "); }

C. for (int i = 0; i < arr.length; i += 2) { System.out.print(arr[i] + " "); }

D. for (int i = 1; i < arr.length; i += 2) { System.out.print(arr[i] + " "); }

Solution

1. Step 1: Determine the highest even index to start from

   If array length is even, highest even index is length-2; if odd, it's length-1.

2. Step 2: Loop backwards by 2 steps to cover even indices

   Start from the highest even index and decrement by 2 to print all even indices in reverse order.

3. Final Answer:

   for (int i = (arr.length % 2 == 0 ? arr.length - 2 : arr.length - 1); i >= 0; i -= 2) { System.out.print(arr[i] + " "); } -> Option B

4. Quick Check:

   Start at highest even index, step down by 2 [OK]

Hint: Start at highest even index, decrement by 2 [OK]

Common Mistakes:

• Starting at length-1 without checking even index
• Using i -= 1 instead of i -= 2
• Printing odd indices instead of even