Why Binary Search on Answer Technique in DSA Typescript?

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 fit all your items. You try sizes one by one, starting from the smallest. It takes forever and you get frustrated.

The Problem

Checking every possible size one by one is very slow and tiring. You might miss the best size or waste a lot of time trying sizes that are too small or too big.

The Solution

Binary Search on Answer Technique helps you guess the answer smartly. Instead of trying every size, you pick a middle size, check if it works, then decide to try bigger or smaller sizes. This way, you find the answer fast and without mistakes.

Before vs After

✗ Before

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

✓ After

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

What It Enables

This technique lets you quickly find the best answer in problems where the answer is a number and you can check if a guess is good or not.

Real Life Example

Finding the minimum time needed to complete all tasks when you can do multiple tasks at once, by guessing a time and checking if all tasks fit in it.

Key Takeaways

Manual checking is slow and error-prone.

Binary Search on Answer guesses answers smartly to save time.

It works when you can test if a guess is valid or not.

Practice

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

easy

A. We guess an answer and check if it satisfies the problem constraints, then adjust the guess.

B. We sort the input array before applying binary search.

C. We use binary search only on sorted arrays to find elements.

D. We divide the input into two halves and solve each recursively.

Solution

1. Step 1: Understand the technique's purpose

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

2. Step 2: Adjust guess based on check

   If the guess is valid, we try smaller or larger values accordingly to find the optimal answer.

3. Final Answer:

   We guess an answer and check if it satisfies the problem constraints, then adjust the guess. -> Option A

4. Quick Check:

   Guess and check approach = C [OK]

Hint: Think of guessing answers and verifying them stepwise [OK]

Common Mistakes:

• Confusing with normal binary search on arrays
• Assuming input must be sorted
• Thinking it divides input recursively

2. Which of the following is the correct TypeScript syntax to start a binary search on answer between low and high?

easy

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

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

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

D. if (low < high) { let mid = (low + high) >> 1; }

Solution

1. Step 1: Identify correct mid calculation

   Midpoint must be an integer index, so use Math.floor to avoid decimals.

2. Step 2: Confirm loop condition and syntax

   Use while loop with condition low <= high for binary search iteration.

3. Final Answer:

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

4. Quick Check:

   Integer mid with while loop = B [OK]

Hint: Use Math.floor for mid to avoid decimals in binary search [OK]

Common Mistakes:

• Using division without floor causing float mid
• Using for loop instead of while
• Using bitwise shift without understanding

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

function canShip(weights: number[], capacity: number, D: number): boolean {
  let days = 1, total = 0;
  for (const w of weights) {
    if (total + w > capacity) {
      days++;
      total = 0;
    }
    total += w;
  }
  return days <= D;
}

function shipWithinDays(weights: number[], D: number): number {
  let low = Math.max(...weights);
  let high = weights.reduce((a, b) => a + b, 0);
  while (low < high) {
    const 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?

medium

A. 15

B. 10

C. 20

D. 5

Solution

1. Step 1: Understand problem and code

   We find minimum ship capacity to ship weights in 5 days. Minimum capacity is max weight (10), max capacity is sum (55).

2. Step 2: Trace binary search and canShip checks

   Binary search narrows capacity to 15, which allows shipping in 5 days by grouping weights properly.

3. Final Answer:

   15 -> Option A

4. Quick Check:

   Minimum capacity = 15 [OK]

Hint: Check max weight as low and sum as high for binary search [OK]

Common Mistakes:

• Choosing max weight instead of sum as answer
• Confusing days count in canShip function
• Not using Math.floor for mid calculation

4. The following TypeScript code tries to find the maximum minimum distance between elements placed in positions array using binary search on answer. Identify the bug:

function canPlace(positions: number[], distance: number, m: number): boolean {
  let count = 1, lastPos = positions[0];
  for (let i = 1; i < positions.length; i++) {
    if (positions[i] - lastPos >= distance) {
      count++;
      lastPos = positions[i];
    }
  }
  return count >= m;
}

function maxMinDistance(positions: number[], m: number): number {
  positions.sort((a, b) => a - b);
  let low = 0, high = positions[positions.length - 1] - positions[0];
  while (low < high) {
    const mid = Math.floor((low + high + 1) / 2);
    if (canPlace(positions, mid, m)) low = mid;
    else high = mid - 1;
  }
  return low;
}

medium

A. Missing increment of count inside canPlace

B. Binary search mid calculation should not add 1

C. Binary search loop condition should be low <= high

D. No bug; code works correctly

Solution

1. Step 1: Analyze canPlace function

   It correctly counts placements with given minimum distance.

2. Step 2: Check binary search logic

   Mid uses upper mid to avoid infinite loop, loop condition low < high is valid, and updates low/high correctly.

3. Final Answer:

   No bug; code works correctly -> Option D

4. Quick Check:

   Code logic correct = A [OK]

Hint: Use upper mid to avoid infinite loop in binary search [OK]

Common Mistakes:

• Changing loop condition incorrectly
• Removing +1 in mid causing infinite loop
• Forgetting to sort positions array

5. You want to find the minimum maximum time to complete all tasks by k workers, where each worker does continuous tasks. Given tasks array of times, how would you apply binary search on answer to solve this?
Choose the best approach.

hard

A. Calculate average time and assign tasks evenly without binary search.

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

C. Use binary search on number of tasks per worker directly.

D. Sort tasks and assign largest tasks first to workers.

Solution

1. Step 1: Define search space for answer

   Minimum max time is max single task time; maximum is sum of all tasks.

2. Step 2: Check feasibility with mid value

   Check if tasks can be divided into ≤ k continuous groups with max sum ≤ mid.

3. Step 3: Adjust search space based on feasibility

   If feasible, try smaller max time; else increase it.

4. Final Answer:

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

5. Quick Check:

   Binary search on max time with group check = D [OK]

Hint: Binary search answer space, check feasibility with grouping [OK]

Common Mistakes:

• Sorting tasks instead of binary searching answer
• Trying to binary search on task count directly
• Ignoring continuous group constraint