Binary Search on Answer Technique in DSA Javascript - Build from Scratch

Binary Search on Answer Technique

📖 Scenario: Imagine you are organizing a party and want to divide a list of guests into groups so that no group has more than a certain number of people. You want to find the smallest possible maximum group size that allows you to split all guests into a given number of groups.

🎯 Goal: Build a program that uses the Binary Search on Answer Technique to find the smallest maximum group size to split guests into a fixed number of groups.

📋 What You'll Learn

Create an array called guests with exact values: [10, 20, 30, 40, 50]

Create a variable called maxGroups and set it to 3

Write a function called canSplit that takes maxSize and returns true if guests can be split into at most maxGroups groups with each group sum ≤ maxSize, else false

Use binary search between low = max guest size and high = sum of all guests to find the smallest maxSize where canSplit(maxSize) is true

Print the final smallest maximum group size

💡 Why This Matters

🌍 Real World

This technique helps in tasks like dividing workloads, scheduling, or packing where you want to find an optimal limit by guessing and checking efficiently.

💼 Career

Binary Search on Answer is a common pattern in coding interviews and real-world optimization problems in software engineering and data analysis.

1

Create the guests array

Create an array called guests with these exact values: [10, 20, 30, 40, 50]

DSA Javascript

# Create the guests array with exact values
// Your code here

Hint

Use const guests = [10, 20, 30, 40, 50]; to create the array.

2

Set the maximum number of groups

Create a variable called maxGroups and set it to 3

DSA Javascript

const guests = [10, 20, 30, 40, 50];
// Create maxGroups variable and set it to 3
// Your code here

Hint

Use const maxGroups = 3; to set the number of groups.

3

Write the canSplit function

Write a function called canSplit that takes a parameter maxSize and returns true if the guests array can be split into at most maxGroups groups where the sum of each group is ≤ maxSize, otherwise returns false

DSA Javascript

const guests = [10, 20, 30, 40, 50];
const maxGroups = 3;
// Write the canSplit function below
// Your code here

Hint

Use a loop to add guests to current group sum. If sum exceeds maxSize, start a new group. Return false if groups exceed maxGroups.

4

Use binary search to find smallest max group size and print it

Use binary search with variables low set to the maximum value in guests and high set to the sum of all guests. Find the smallest maxSize where canSplit(maxSize) returns true. Print the final maxSize using console.log

DSA Javascript

const guests = [10, 20, 30, 40, 50];
const maxGroups = 3;

function canSplit(maxSize) {
    let groups = 1;
    let currentSum = 0;
    for (let i = 0; i  maxSize) {
            return false;
        }
        if (currentSum + guests[i]  maxGroups) {
                return false;
            }
        }
    }
    return true;
}

// Use binary search to find smallest maxSize and print it
// Your code here

Hint

Use binary search between max guest size and total sum. Update result when canSplit(mid) is true. Print result at the end.

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