Binary Search on Answer Technique in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the binary search boundaries.

DSA Javascript

let low = 0;
let high = [1];

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Aarr.length - 1

Barr.length

Carr.length + 1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Setting high to arr.length causes out-of-bound errors.

Setting high to 0 makes the search range empty.

We set high to arr.length - 1 because array indices go from 0 to length minus one.

2fill in blank

medium

Complete the code to calculate the middle index safely.

DSA Javascript

let mid = low + Math.floor([1] / 2);

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Ahigh / low

Bhigh + low

Chigh * low

Dhigh - low

Attempts:

3 left

💡 Hint

Common Mistakes

Using (high + low) / 2 can cause overflow in some languages.

Using multiplication or division incorrectly changes the midpoint.

We calculate mid as low + Math.floor((high - low) / 2) to avoid overflow and find the middle index.

3fill in blank

hard

Fix the error in the condition to move the low pointer.

DSA Javascript

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

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Amid + 1

Bmid - 1

Cmid

Dlow + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Setting low to mid causes infinite loops.

Setting low to mid - 1 moves backward incorrectly.

When canAchieve(mid) is true, we move low to mid + 1 to search higher values.

4fill in blank

hard

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

DSA Javascript

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

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Alow

Bhigh

Cmid

Dlow + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid in the condition causes wrong loop behavior.

Using low + 1 skips valid values.

The loop runs while low <= high to cover all possible values in the search range.

5fill in blank

hard

Fill all three blanks to return the correct final answer after binary search.

DSA Javascript

return [1] - 1;

// where [2] is the last successful mid
// and [3] is the variable updated when canAchieve(mid) is true

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

Alow

Bhigh

Dmid

Attempts:

3 left

💡 Hint

Common Mistakes

Returning high directly can be incorrect if low passed it.

Confusing which variable holds the last successful value.

After the loop, low is one more than the last successful mid, so we return low - 1.