Why Binary Search on Answer Technique in DSA Javascript?

The Big Idea

What if you could find the perfect answer without trying every possibility?

The Scenario

Imagine you want to find the smallest size of a box that can hold all your items, but you only know the total number of items and their sizes. You try different box sizes one by one, starting from very small to very large, to see which fits. This takes a lot of time and effort.

The Problem

Checking every possible box size manually is slow and tiring. It wastes time because you try many sizes that are obviously too small or too big. It's easy to make mistakes or lose track of which sizes you already tested.

The Solution

The Binary Search on Answer technique helps by smartly guessing the box size. Instead of checking every size, it picks a middle size, checks if it fits, and then decides to try bigger or smaller sizes. This way, it quickly narrows down to the best size without wasting time.

Before vs After

✗ Before

let size = 1;
while (!fits(size)) {
  size++;
}
return size;

✓ After

let low = 1, high = maxSize;
while (low < high) {
  let mid = Math.floor((low + high) / 2);
  if (fits(mid)) high = mid;
  else low = mid + 1;
}
return low;

What It Enables

This technique enables solving complex problems by efficiently guessing the answer, saving time and effort.

Real Life Example

Finding the minimum time needed to complete a set of tasks when you know the speed limits but not the exact time, by guessing and checking smartly.

Key Takeaways

Manual checking is slow and error-prone.

Binary Search on Answer guesses smartly to find the best answer fast.

It works by narrowing down the search space step-by-step.

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