Two Pointer Technique on Arrays in DSA Java - Interactive Practice

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'

A0

B1

C-1

Darray.length

Attempts:

3 left

💡 Hint

Common Mistakes

Starting the left pointer at 1 instead of 0.

Using array.length which is out of bounds.

The left pointer should start at index 0, which is the first element 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.size

B0

Carray.length

Darray.length()

Attempts:

3 left

💡 Hint

Common Mistakes

Using array.length() which is a method for strings, not arrays.

Using array.size which is not valid for arrays.

The right pointer should start 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 (left [1] right) {
    // logic here
}

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

A>

B<=

C>=

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using >= or <= which causes the loop to run too long or miss cases.

Using > which never runs if left starts less than right.

The loop should run while left is less than right to avoid overlapping pointers.

4fill in blank

hard

Fill both blanks to return the correct pointer indices inside the loop.

DSA Java

if (array[left] + array[right] == target) {
    return new int[] { [1], [2] };
} else if (array[left] + array[right] < target) {
    left++;
} else {
    right--;
}

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

Aleft

Bright

Ctarget

Darray

Attempts:

3 left

💡 Hint

Common Mistakes

Returning the target or array instead of indices.

Swapping left and right in the return array.

Return the indices of the left and right pointers when their sum equals the target.

5fill in blank

hard

Fill all three blanks to create a map of numbers to their indices for a one-pass hashmap approach.

DSA Java

Map<Integer, Integer> map = new HashMap<>();
for (int [1] = 0; [1] < array.length; [1]++) {
    int complement = [2] - array[[1]];
    if (map.containsKey(complement)) {
        return new int[] { map.get(complement), [1] };
    }
    map.put(array[[1]], [1]);
}

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

Ai

Btarget

Cindex

Dsum

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'index' or 'sum' instead of 'i' for the loop variable.

Using 'sum' instead of 'target' for the complement calculation.

Use 'i' as the loop variable and index, and 'target' as the value to find complements.

Practice

1. What is the main idea behind the Two Pointer Technique when used on arrays?

easy

A. Use a single pointer to iterate through the array

B. Use two indices to traverse the array from different ends or positions simultaneously

C. Use recursion to divide the array into halves

D. Sort the array before processing

Solution

1. Step 1: Understand the technique's approach

   The two pointer technique uses two indices or pointers to scan the array from different positions, often from start and end.

2. Step 2: Compare with other options

   Sorting or recursion are different techniques; single pointer is not two pointer technique.

3. Final Answer:

   Use two indices to traverse the array from different ends or positions simultaneously -> Option B

4. Quick Check:

   Two pointers = two indices moving [OK]

Hint: Two pointers move from different sides or positions [OK]

Common Mistakes:

• Confusing with sorting or recursion
• Thinking it uses only one pointer
• Assuming it always requires sorted arrays

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

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 initialization

   In Java, array indices start at 0 and end at length-1, so left = 0 and right = arr.length - 1 is correct.

2. Step 2: Check other options for invalid indices

   Options B, C, and D use invalid indices outside array bounds.

3. Final Answer:

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

4. Quick Check:

   Start at 0 and end at length-1 [OK]

Hint: Array indices start at 0, end at length-1 [OK]

Common Mistakes:

• Using 1-based indexing
• Setting pointers outside array bounds
• Confusing length with last index

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

int[] arr = {1, 3, 5, 7, 9};
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. 15

B. 30

C. 25

D. 20

Solution

1. Step 1: Trace the loop iterations and sum calculation

   Initial pointers: left=0, right=4 - Iteration 1: sum=0+(1+9)=10, left=1, right=3 - Iteration 2: sum=10+(3+7)=20, left=2, right=2 Loop stops as left is not less than right.

2. Step 2: Final sum value

   Sum after loop is 20.

3. Final Answer:

   20 -> Option D

4. Quick Check:

   Sum pairs (1+9) + (3+7) = 20 [OK]

Hint: Add pairs from ends until pointers meet [OK]

Common Mistakes:

• Including middle element when left == right
• Adding elements multiple times
• Incorrect pointer updates

4. Identify the error in the following Java code using two pointers:

int[] arr = {2, 4, 6, 8};
int left = 0, right = arr.length;
while (left < right) {
    System.out.print(arr[left] + ",");
    left++;
    right--;
}

medium

A. The 'right' pointer is initialized out of array bounds

B. The loop condition should be 'left <= right'

C. The array should be sorted before using two pointers

D. The print statement should be outside the loop

Solution

1. Step 1: Check pointer initialization

   Array indices go from 0 to length-1, but right is set to arr.length, which is out of bounds.

2. Step 2: Validate other options

   Loop condition and print placement are valid; sorting is not required here.

3. Final Answer:

   The 'right' pointer is initialized out of array bounds -> Option A

4. Quick Check:

   Right pointer must be arr.length - 1 [OK]

Hint: Right pointer must be last valid index (length-1) [OK]

Common Mistakes:

• Setting right = arr.length instead of arr.length - 1
• Confusing loop conditions
• Assuming sorting is always needed

5. Given a sorted array arr and a target sum target, which two-pointer approach correctly finds if there exists a pair of numbers that add up to target?

int left = 0, right = arr.length - 1;
while (left < right) {
    int sum = arr[left] + arr[right];
    if (sum == target) {
        return true;
    } else if (sum < target) {
        left++;
    } else {
        right--;
    }
}
return false;

hard

A. Move left pointer right if sum is less than target, else move right pointer left

B. Always move left pointer right regardless of sum

C. Always move right pointer left regardless of sum

D. Move pointers randomly until sum matches

Solution

1. Step 1: Understand pointer movement logic

   If sum is less than target, increase sum by moving left pointer right; if sum is more, decrease sum by moving right pointer left.

2. Step 2: Confirm correctness of approach

   This approach works because array is sorted, so moving pointers adjusts sum towards target.

3. Final Answer:

   Move left pointer right if sum is less than target, else move right pointer left -> Option A

4. Quick Check:

   Adjust pointers based on sum vs target [OK]

Hint: Move pointers based on sum comparison with target [OK]

Common Mistakes:

• Moving pointers without checking sum
• Ignoring array sorted property
• Moving both pointers at once