Why Two Pointer Technique Beats Brute Force in DSA Java - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the left pointer at the start of the array.

DSA Java

int left = [1];

Drag options to blanks, or click blank then click option'

A1

B0

C-1

Darray.length

Attempts:

3 left

💡 Hint

Common Mistakes

Starting left pointer at 1 instead of 0.

The left pointer starts at index 0, the beginning of the array.

2fill in blank

medium

Complete the code to initialize the right pointer at the end of the array.

DSA Java

int right = [1] - 1;

Drag options to blanks, or click blank then click option'

Aarray.length()

B0

Carray.length

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using array.length() which is invalid in Java.

The right pointer starts at the last index, which is array.length - 1.

3fill in blank

hard

Fix the error in the while loop condition to continue while left is less than right.

DSA Java

while ([1] < right) {
    // process
}

Drag options to blanks, or click blank then click option'

Aright - 1

Bright

Cleft + 1

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Using right < right which is always false.

The loop should run while left pointer is less than right pointer.

4fill in blank

hard

Fill both blanks to move pointers correctly: increment left and decrement right.

DSA Java

if (sum < target) {
    [1]++;
} else {
    [2]--;
}

Drag options to blanks, or click blank then click option'

Aleft

Bright

Csum

Dtarget

Attempts:

3 left

💡 Hint

Common Mistakes

Incrementing right pointer or decrementing left pointer incorrectly.

When sum is less than target, move left pointer right to increase sum. Otherwise, move right pointer left to decrease sum.

5fill in blank

hard

Fill all three blanks to create a two-pointer loop that finds pairs summing to target.

DSA Java

while ([1] < [2]) {
    int sum = array[[1]] + array[[2]];
    if (sum == target) {
        return true;
    } else if (sum < target) {
        [1]++;
    } else {
        [2]--;
    }
}
return false;

Drag options to blanks, or click blank then click option'

Aleft

Bright

Csum

Dtarget

Attempts:

3 left

💡 Hint

Common Mistakes

Using sum or target as loop variables.

The loop runs while left is less than right. sum is calculated using left and right indices. If sum is less than target, increment left; else decrement right.

Practice

1. What is the main advantage of the two pointer technique over brute force in array problems?

easy

A. It uses more memory to store intermediate results

B. It reduces time complexity by avoiding nested loops

C. It always sorts the array first

D. It requires recursion to work

Solution

1. Step 1: Understand brute force approach

   Brute force often uses nested loops, leading to O(n²) time complexity.

2. Step 2: Analyze two pointer technique

   Two pointers move through the array in a single pass, reducing time complexity to O(n).

3. Final Answer:

   It reduces time complexity by avoiding nested loops -> Option B

4. Quick Check:

   Two pointer = faster than brute force [OK]

Hint: Two pointers scan array once, avoiding double loops [OK]

Common Mistakes:

• Thinking two pointer always sorts array
• Assuming it uses more memory
• Believing recursion is required

2. Which of the following is the correct syntax to initialize two pointers in Java for an array arr?

easy

A. int left = arr.length, right = 0;

B. int left = 1, right = arr.length;

C. int left = 0, right = arr.length - 1;

D. int left = -1, right = arr.length + 1;

Solution

1. Step 1: Identify valid pointer positions

   Array indices start at 0 and end at length-1.

2. Step 2: Check initialization correctness

   Left pointer at 0 and right pointer at arr.length - 1 are valid.

3. Final Answer:

   int left = 0, right = arr.length - 1; -> Option C

4. Quick Check:

   Pointers start at array ends [OK]

Hint: Pointers start at 0 and last index (length-1) [OK]

Common Mistakes:

• Starting right pointer at arr.length (out of bounds)
• Starting left pointer at 1 (skips first element)
• Using negative or invalid indices

3. What will be the output of the following Java code snippet?

int[] arr = {1, 2, 3, 4, 5};
int left = 0, right = arr.length - 1;
int sum = 0;
while (left < right) {
  sum += arr[left] + arr[right];
  left++;
  right--;
}
System.out.println(sum);

medium

A. 10

B. 25

C. 15

D. 12

Solution

1. Step 1: Trace pointer movements and sums

   Pairs added: (1+5)=6, (2+4)=6; loop stops when left=right=2.

2. Step 2: Calculate total sum

   Sum = 6 + 6 = 12; loop ends before adding middle element 3.

3. Step 3: Check if middle element added

   Middle element 3 is not added because left == right stops loop.

4. Final Answer:

   12 -> Option D

5. Quick Check:

   Sum pairs from ends = 12 [OK]

Hint: Sum pairs while left < right, middle element excluded [OK]

Common Mistakes:

• Adding middle element when left == right
• Summing all elements instead of pairs
• Miscounting loop iterations

4. Identify the bug in the following two pointer code that tries to find if a pair sums to 10:

int[] arr = {1, 3, 5, 7, 9};
int left = 0, right = arr.length - 1;
boolean found = false;
while (left <= right) {
  int sum = arr[left] + arr[right];
  if (sum == 10) {
    found = true;
    break;
  } else if (sum < 10) {
    left++;
  } else {
    right++;
  }
}
System.out.println(found);

medium

A. Right pointer should decrement, not increment, when sum > 10

B. Left pointer should decrement when sum < 10

C. Loop condition should be left < right, not left <= right

D. Sum calculation is incorrect

Solution

1. Step 1: Understand pointer movement logic

   If sum is less than target, move left pointer right to increase sum; if sum is greater, move right pointer left to decrease sum.

2. Step 2: Identify incorrect pointer update

   Code incorrectly increments right pointer when sum > 10; it should decrement right pointer.

3. Final Answer:

   Right pointer should decrement, not increment, when sum > 10 -> Option A

4. Quick Check:

   Decrease right pointer when sum too big [OK]

Hint: Decrease right pointer if sum too large, increase left if too small [OK]

Common Mistakes:

• Incrementing right pointer instead of decrementing
• Using wrong loop condition
• Miscalculating sum

5. Given a sorted array arr of integers, how can the two pointer technique be used to find all unique pairs that sum to a target value k efficiently?

hard

A. Use two pointers at start and end, move inward adjusting pointers based on sum compared to k

B. Use nested loops to check every pair for sum equal to k

C. Sort the array first, then use binary search for each element to find complement

D. Use recursion to generate all pairs and check sums

Solution

1. Step 1: Initialize two pointers

   Set one pointer at start (left) and one at end (right) of the sorted array.

2. Step 2: Move pointers based on sum comparison

   Calculate sum of elements at pointers; if sum equals k, record pair and move both pointers; if sum less than k, move left pointer right; if sum greater than k, move right pointer left.

3. Step 3: Avoid duplicates by skipping equal elements

   After finding a pair, skip over duplicate elements to ensure unique pairs.

4. Final Answer:

   Use two pointers at start and end, move inward adjusting pointers based on sum compared to k -> Option A

5. Quick Check:

   Two pointer moves inward to find pairs efficiently [OK]

Hint: Move pointers inward based on sum vs target to find pairs fast [OK]

Common Mistakes:

• Using nested loops causing O(n²) time
• Not skipping duplicates leading to repeated pairs
• Using binary search unnecessarily increasing complexity