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];

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

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);

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

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;
}

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

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;
  }
}

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

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.

Practice

1. What is the main idea behind the Binary Search on Answer technique?

easy

A. We guess an answer and use binary search to find the smallest or largest valid value.

B. We sort the array and then apply binary search to find an element.

C. We use binary search to find the middle element in a sorted list.

D. We apply binary search to split the array into two halves repeatedly.

Solution

1. Step 1: Understand the technique's goal

   Binary Search on Answer guesses a possible answer and checks if it meets the problem's conditions.

2. Step 2: Use binary search on the answer range

   We narrow down the answer by checking mid-values until we find the smallest or largest valid answer.

3. Final Answer:

   We guess an answer and use binary search to find the smallest or largest valid value. -> Option A

4. Quick Check:

   Binary Search on Answer = Guess and check with binary search [OK]

Hint: Think of binary searching the answer, not the array [OK]

Common Mistakes:

• Confusing it with normal binary search on sorted arrays
• Trying to binary search elements instead of answer values
• Not defining a valid check function for guesses

2. Which of the following is the correct syntax to start a binary search on answer in JavaScript for a range from low to high?

easy

A. for (let mid = low; mid <= high; mid++) { }

B. while (low <= high) { let mid = Math.floor((low + high) / 2); }

C. if (low < high) { mid = (low + high) / 2; }

D. do { mid = (low + high) / 2; } while (low < high);

Solution

1. Step 1: Identify the binary search loop structure

   Binary search uses a while loop checking low <= high to continue searching.

2. Step 2: Calculate mid correctly

   Mid is calculated as Math.floor((low + high) / 2) to get the middle integer index.

3. Final Answer:

   while (low <= high) { let mid = Math.floor((low + high) / 2); } -> Option B

4. Quick Check:

   Correct binary search loop syntax = while (low <= high) { let mid = Math.floor((low + high) / 2); } [OK]

Hint: Use while loop with low <= high and floor mid [OK]

Common Mistakes:

• Using for loop instead of while for binary search
• Not flooring mid value causing float indices
• Using incorrect loop conditions like low < high

3. Consider the following code snippet using binary search on answer to find the minimum capacity to ship packages within D days:

function canShip(weights, capacity, D) {
  let days = 1, currentLoad = 0;
  for (let w of weights) {
    if (currentLoad + w > capacity) {
      days++;
      currentLoad = 0;
    }
    currentLoad += w;
  }
  return days <= D;
}

function shipWithinDays(weights, D) {
  let low = Math.max(...weights);
  let high = weights.reduce((a, b) => a + b, 0);
  while (low < high) {
    let mid = Math.floor((low + high) / 2);
    if (canShip(weights, mid, D)) {
      high = mid;
    } else {
      low = mid + 1;
    }
  }
  return low;
}

console.log(shipWithinDays([1,2,3,4,5,6,7,8,9,10], 5));

What is the output of this code?

medium

A. 20

B. 10

C. 15

D. 5

Solution

1. Step 1: Understand the problem and code

   The code finds the minimum ship capacity to ship weights in 5 days. It uses binary search between max weight (10) and sum (55).

2. Step 2: Trace binary search narrowing

   Checking mid values, the minimum capacity that allows shipping in 5 days is 15.

3. Final Answer:

   15 -> Option C

4. Quick Check:

   Minimum capacity to ship in 5 days = 15 [OK]

Hint: Check max weight and sum, binary search between them [OK]

Common Mistakes:

• Choosing max weight (10) as answer instead of minimum capacity
• Choosing total sum (55) instead of minimum valid capacity
• Misunderstanding the days condition in canShip

4. The following code tries to find the maximum length of a rope to cut at least k pieces using binary search on answer. Find the bug:

function maxLength(ropes, k) {
  let low = 0;
  let high = Math.max(...ropes);
  while (low <= high) {
    let mid = Math.floor((low + high) / 2);
    let count = 0;
    for (let r of ropes) {
      count += Math.floor(r / mid);
    }
    if (count >= k) {
      low = mid + 1;
    } else {
      high = mid - 1;
    }
  }
  return high;
}

medium

A. Using while (low <= high) instead of while (low < high) is wrong.

B. Not sorting ropes before binary search causes wrong answer.

C. Returning high instead of low is incorrect.

D. Starting low from 0 causes division by zero error.

Solution

1. Step 1: Identify potential division by zero

   Starting low at 0 means mid can be 0, causing division by zero in Math.floor(r / mid).

2. Step 2: Fix low initialization

   Set low to 1 to avoid division by zero and ensure valid rope length guesses.

3. Final Answer:

   Starting low from 0 causes division by zero error. -> Option D

4. Quick Check:

   low = 0 causes divide by zero [OK]

Hint: Avoid zero as divisor in binary search on answer [OK]

Common Mistakes:

• Ignoring division by zero possibility
• Changing loop condition unnecessarily
• Returning wrong boundary without reason

5. You want to find the minimum maximum time to complete all tasks by k workers, where each worker can do consecutive tasks only. Given task times array tasks, how would you apply binary search on answer to solve this?

Choose the best approach:

hard

A. Binary search on time range; check if tasks can be split into ≤ k groups with max sum ≤ mid.

B. Sort tasks and binary search for the smallest task time.

C. Use binary search on number of tasks each worker gets, ignoring task times.

D. Apply binary search on total tasks count to assign tasks evenly.

Solution

1. Step 1: Define the answer range

   The minimum max time is at least max task time, max is sum of all tasks.

2. Step 2: Check feasibility with mid value

   Check if tasks can be split into ≤ k groups where each group's sum ≤ mid.

3. Step 3: Use binary search to narrow answer

   Adjust low and high based on feasibility until minimum max time is found.

4. Final Answer:

   Binary search on time range; check if tasks can be split into ≤ k groups with max sum ≤ mid. -> Option A

5. Quick Check:

   Binary search on max time with group feasibility check [OK]

Hint: Binary search answer range, check group splits feasibility [OK]

Common Mistakes:

• Binary searching on tasks count instead of time
• Ignoring consecutive task grouping constraint
• Sorting tasks unnecessarily