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DSA Typescriptprogramming~10 mins

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

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

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

DSA Typescript
let 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 left pointer at 1 or negative index.
2fill in blank
medium

Complete the code to move the right pointer to the end of the array.

DSA Typescript
let right = [1];
Drag options to blanks, or click blank then click option'
A0
Barray.length
Carray.length - 1
D-1
Attempts:
3 left
💡 Hint
Common Mistakes
Setting right pointer to array.length causes out-of-bounds errors.
3fill in blank
hard

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

DSA Typescript
while (left [1] right) {
  // process pointers
}
Drag options to blanks, or click blank then click option'
A<
B<=
C>
D>=
Attempts:
3 left
💡 Hint
Common Mistakes
Using '<=' causes pointers to overlap and double count.
4fill in blank
hard

Fill both blanks to correctly move pointers based on sum comparison.

DSA Typescript
if (array[left] + array[right] [1] target) {
  left [2];
} else {
  right--;
}
Drag options to blanks, or click blank then click option'
A<
B>
C++
D--
Attempts:
3 left
💡 Hint
Common Mistakes
Using '>' instead of '<' in comparison.
Decrementing left pointer instead of incrementing.
5fill in blank
hard

Fill all three blanks to create a dictionary of numbers and their indices for brute force approach.

DSA Typescript
const numIndices = new Map<number, number>();
for (let i = 0; i [1] array.length; i[2]) {
  numIndices.set(array[i], [3]);
}
Drag options to blanks, or click blank then click option'
A<
Bi
C++
Darray[i]
Attempts:
3 left
💡 Hint
Common Mistakes
Using '>=' in loop condition.
Using 'array[i]' instead of index as map value.

Practice

(1/5)
1. Why does the two pointer technique often perform better than the brute force approach in problems involving sorted arrays?
easy
A. It reduces the number of comparisons by moving pointers inward based on conditions.
B. It uses recursion to explore all possibilities efficiently.
C. It sorts the array first to make searching faster.
D. It uses extra memory to store intermediate results.

Solution

  1. Step 1: Understand brute force approach

    Brute force checks all pairs, leading to many unnecessary comparisons.
  2. Step 2: Analyze two pointer technique advantage

    Two pointers move inward based on conditions, skipping unnecessary checks and reducing time.
  3. Final Answer:

    It reduces the number of comparisons by moving pointers inward based on conditions. -> Option A
  4. Quick Check:

    Two pointer skips pairs, faster than brute force [OK]
Hint: Two pointers move inward to skip unnecessary checks [OK]
Common Mistakes:
  • Thinking recursion is always faster
  • Confusing sorting with two pointer technique
  • Assuming extra memory is used in two pointer
2. Which of the following is the correct syntax to initialize two pointers in TypeScript for an array named arr?
easy
A. let left = arr[0], right = arr[arr.length];
B. let left = 0, right = arr.length - 1;
C. let left = 1, right = arr.length;
D. let left = arr.length, right = 0;

Solution

  1. Step 1: Identify valid pointer indices

    Array indices start at 0 and end at length - 1.
  2. Step 2: Check each option

    let left = 0, right = arr.length - 1; correctly sets left to 0 and right to last index (length - 1). Others use invalid indices.
  3. Final Answer:

    let left = 0, right = arr.length - 1; -> Option B
  4. Quick Check:

    Array indices start at 0 [OK]
Hint: Array indices go from 0 to length-1 [OK]
Common Mistakes:
  • Starting pointer at 1 instead of 0
  • Using arr.length as index (out of bounds)
  • Swapping left and right pointers
3. Consider the following TypeScript code using two pointers to find if a sorted array contains two numbers that sum to 10:
const arr = [1, 2, 3, 7, 8, 9];
let left = 0;
let right = arr.length - 1;
let found = false;
while (left < right) {
  const sum = arr[left] + arr[right];
  if (sum === 10) {
    found = true;
    break;
  } else if (sum < 10) {
    left++;
  } else {
    right--;
  }
}
console.log(found);
What will be printed?
medium
A. Error
B. false
C. undefined
D. true

Solution

  1. Step 1: Trace pointer values and sums

    Start: left=0(arr[0]=1), right=5(arr[5]=9), sum=10 -> found=true, loop breaks.
  2. Step 2: Confirm output

    Since found is set to true, console.log prints true.
  3. Final Answer:

    true -> Option D
  4. Quick Check:

    1 + 9 = 10 found [OK]
Hint: Check sum at pointers, move accordingly [OK]
Common Mistakes:
  • Not breaking loop after finding sum
  • Confusing pointer increments
  • Assuming array is unsorted
4. The following TypeScript code tries to find two numbers in a sorted array that sum to 15 using two pointers, but it has a bug:
const arr = [2, 4, 6, 8, 10];
let left = 0;
let right = arr.length;
let found = false;
while (left < right) {
  const sum = arr[left] + arr[right];
  if (sum === 15) {
    found = true;
    break;
  } else if (sum < 15) {
    left++;
  } else {
    right--;
  }
}
console.log(found);
What is the bug?
medium
A. right pointer is initialized out of array bounds
B. left pointer should start at 1
C. sum comparison should be with 10 instead of 15
D. The loop condition should be left <= right

Solution

  1. Step 1: Check pointer initialization

    right is set to arr.length which is out of bounds (max index is length - 1).
  2. Step 2: Understand impact of bug

    Accessing arr[right] causes runtime error or undefined behavior.
  3. Final Answer:

    right pointer is initialized out of array bounds -> Option A
  4. Quick Check:

    Array indices max is length-1 [OK]
Hint: Right pointer must be length - 1, not length [OK]
Common Mistakes:
  • Starting left pointer at 1
  • Changing sum target incorrectly
  • Using <= instead of < in loop
5. You have a sorted array arr = [1, 3, 4, 5, 7, 10, 12]. Using the two pointer technique, how many unique pairs sum to 13?
hard
A. 4
B. 3
C. 2
D. 1

Solution

  1. Step 1: Initialize pointers and check sums

    left=0(arr[0]=1), right=6(arr[6]=12), sum=13 -> count=1, move left++ and right--
  2. Step 2: Continue checking pairs

    left=1(arr[1]=3), right=5(arr[5]=10), sum=13 -> count=2, move left++ and right--
  3. Step 3: Check next pair

    left=2(arr[2]=4), right=4(arr[4]=7), sum=11 < 13 -> move left++
  4. Step 4: Check final pair

    left=3(arr[3]=5), right=4(arr[4]=7), sum=12 < 13 -> move left++
  5. Step 5: Pointers crossed, stop

    Total pairs found = 2
  6. Final Answer:

    2 -> Option C
  7. Quick Check:

    Pairs (1,12) and (3,10) sum to 13 [OK]
Hint: Move pointers inward, count pairs summing to target [OK]
Common Mistakes:
  • Counting pairs with same elements twice
  • Not moving pointers correctly after sum check
  • Assuming pairs must be adjacent
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10 questions~5 mineasy hard