Binary Search on Answer Technique in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the low boundary for binary search.

DSA Typescript

let low = [1];

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

A0

B1

C-1

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Starting low at 1 or negative values without reason.

We start binary search from 0 as the lowest possible answer.

2fill in blank

medium

Complete the code to calculate the middle point in binary search.

DSA Typescript

let mid = (low [1] high) / 2;

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

A*

B+

C-

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction or multiplication instead of addition.

We add low and high, then divide by 2 to find the middle.

3fill in blank

hard

Fix the error in the condition to update the high boundary.

DSA Typescript

if (canAchieve(mid)) {
  high = [1];
} else {
  low = mid + 1;
}

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

Amid + 1

Bmid - 1

Cmid

Dlow

Attempts:

3 left

💡 Hint

Common Mistakes

Setting high to mid + 1 or low incorrectly.

If mid is achievable, we move high to mid to search lower values.

4fill in blank

hard

Fill both blanks to complete the binary search loop condition and update low.

DSA Typescript

while (low [1] high) {
  let mid = (low + high) / 2;
  if (canAchieve(mid)) {
    high = mid;
  } else {
    low = mid [2] 1;
  }
}

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

A<=

B<

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= in loop condition or subtracting 1 from low incorrectly.

The loop runs while low is less than high, and low updates by adding 1.

5fill in blank

hard

Fill all three blanks to create a function that uses binary search on answer technique.

DSA Typescript

function binarySearchAnswer(low: number, high: number, [1]: (mid: number) => boolean): number {
  while (low [2] high) {
    let mid = Math.floor((low + high) / 2);
    if ([3](mid)) {
      high = mid;
    } else {
      low = mid + 1;
    }
  }
  return low;
}

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

AcanAchieve

B<=

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= in loop condition or wrong function name calls.

The function takes a check function 'canAchieve', loops while low < high, and calls canAchieve(mid) inside.