Two Pointer Technique on Arrays in DSA C - Time & Space Complexity

Time Complexity: Two Pointer Technique on Arrays

O(n)

Understanding Time Complexity

We want to understand how fast the two pointer technique works on arrays.

Specifically, how the number of steps changes as the array gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int left = 0, right = n - 1;
while (left < right) {
    if (arr[left] + arr[right] == target) {
        // found pair
        break;
    } else if (arr[left] + arr[right] < target) {
        left++;
    } else {
        right--;
    }
}
    

This code uses two pointers moving from both ends of a sorted array to find a pair that sums to a target.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop moving pointers left and right.
• How many times: Each pointer moves at most n times combined, so up to n steps.

How Execution Grows With Input

As the array size grows, the total steps grow roughly in a straight line.

Input Size (n)	Approx. Operations
10	About 10 steps
100	About 100 steps
1000	About 1000 steps

Pattern observation: The steps increase directly with input size, not faster or slower.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the pair grows in a straight line with the array size.

Common Mistake

[X] Wrong: "Since there are two pointers, the time is O(n²)."

[OK] Correct: The pointers move towards each other and each element is visited at most once, so total steps are linear, not squared.

Interview Connect

Understanding this technique shows you can solve problems efficiently by smartly scanning arrays.

Self-Check

"What if the array was not sorted? How would the time complexity change?"

Practice

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

easy

A. Use one pointer to traverse and another to store the sum.

B. Use two arrays to store elements separately.

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

D. Use two pointers to swap elements randomly.

Solution

1. Step 1: Understand the technique

   The two pointer technique involves using two indices or pointers to scan the array, often from the start and end, to solve problems efficiently.

2. Step 2: Identify the correct description

   Use two indices to traverse the array from different ends or positions simultaneously. correctly describes using two pointers to traverse the array from different ends or positions simultaneously.

3. Final Answer:

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

4. Quick Check:

   Two pointers move from different positions = A [OK]

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

Common Mistakes:

• Thinking two pointers mean two separate arrays
• Confusing pointers with storing sums
• Assuming pointers swap elements randomly

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

easy

A. int left = 0, right = n - 1;

B. int left = 1, right = n;

C. int left = n, right = 0;

D. int left = n - 1, right = 0;

Solution

1. Step 1: Understand pointer initialization

   For two pointers starting at the beginning and end of an array, left should be 0 and right should be n - 1 because array indices start at 0.

2. Step 2: Check options

   int left = 0, right = n - 1; correctly initializes left = 0 and right = n - 1. Other options have incorrect indices.

3. Final Answer:

   int left = 0, right = n - 1; -> Option A

4. Quick Check:

   Start at 0 and end at n-1 = C [OK]

Hint: Array indices start at 0, so right = n - 1 [OK]

Common Mistakes:

• Starting left at 1 instead of 0
• Setting right to n (out of bounds)
• Swapping left and right initialization

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

int arr[] = {1, 3, 5, 7, 9};
int left = 0, right = 4;
int sum = 0;
while (left < right) {
    sum += arr[left] + arr[right];
    left++;
    right--;
}
printf("%d", sum);

medium

A. 25

B. 15

C. 30

D. 20

Solution

1. Step 1: Trace the loop iterations

   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 ends because left is not less than right.

2. Step 2: Calculate final sum

   Sum after loop = 20. Note that the middle element (index 2) is not added because left == right stops the loop.

3. Final Answer:

   20 -> Option D

4. Quick Check:

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

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

Common Mistakes:

• Including middle element when left == right
• Adding elements twice
• Incorrect loop condition

4. Identify the error in the following code snippet that uses the two pointer technique to find if a sorted array contains two numbers that sum to a target value.

int arr[] = {2, 4, 6, 8, 10};
int left = 0, right = 4;
int target = 14;
while (left <= right) {
    int sum = arr[left] + arr[right];
    if (sum == target) {
        printf("Found");
        break;
    } else if (sum < target) {
        left++;
    } else {
        right--;
    }
}

medium

A. The pointers left and right are initialized incorrectly.

B. The loop condition should be left < right, not left <= right.

C. The sum comparison logic is reversed.

D. The printf statement should be outside the loop.

Solution

1. Step 1: Analyze loop condition

   The loop uses left <= right, which allows pointers to meet or cross. This can cause incorrect checks or double counting.

2. Step 2: Correct loop condition

   Using left < right ensures pointers do not cross and only valid pairs are checked.

3. Final Answer:

   The loop condition should be left < right, not left <= right. -> Option B

4. Quick Check:

   Pointers must not cross: left < right [OK]

Hint: Use left < right to avoid pointer crossing [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Swapping pointer increment/decrement logic
• Incorrect pointer initialization

5. Given a sorted array arr of size n, how can you use the two pointer technique to find all unique pairs that sum to a target value t? Choose the best approach.

hard

A. Initialize two pointers at start and end; move pointers inward based on sum comparison; skip duplicates to find unique pairs.

B. Use nested loops to check all pairs and store sums in a hash set.

C. Sort the array and use binary search for each element to find complement.

D. Initialize both pointers at start and move both forward to find pairs.

Solution

1. Step 1: Use two pointers at array ends

   Start with left = 0 and right = n - 1. Calculate sum of elements at these pointers.

2. Step 2: Move pointers and skip duplicates

   If sum equals target, record pair and move both pointers skipping duplicates. If sum < target, move left pointer right. If sum > target, move right pointer left.

3. Final Answer:

   Initialize two pointers at start and end; move pointers inward based on sum comparison; skip duplicates to find unique pairs. -> Option A

4. Quick Check:

   Two pointers from ends + skip duplicates = D [OK]

Hint: Move pointers inward and skip duplicates for unique pairs [OK]

Common Mistakes:

• Using nested loops instead of two pointers
• Not skipping duplicates causing repeated pairs
• Moving both pointers forward instead of inward