Binary Search on Answer Technique in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the low boundary for binary search.

DSA Typescript

let low = [1];

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

A0

B1

C-1

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Starting low at 1 or negative values without reason.

We start binary search from 0 as the lowest possible answer.

2fill in blank

medium

Complete the code to calculate the middle point in binary search.

DSA Typescript

let mid = (low [1] high) / 2;

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

A*

B+

C-

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction or multiplication instead of addition.

We add low and high, then divide by 2 to find the middle.

3fill in blank

hard

Fix the error in the condition to update the high boundary.

DSA Typescript

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

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

Amid + 1

Bmid - 1

Cmid

Dlow

Attempts:

3 left

💡 Hint

Common Mistakes

Setting high to mid + 1 or low incorrectly.

If mid is achievable, we move high to mid to search lower values.

4fill in blank

hard

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

DSA Typescript

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

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

A<=

B<

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= in loop condition or subtracting 1 from low incorrectly.

The loop runs while low is less than high, and low updates by adding 1.

5fill in blank

hard

Fill all three blanks to create a function that uses binary search on answer technique.

DSA Typescript

function binarySearchAnswer(low: number, high: number, [1]: (mid: number) => boolean): number {
  while (low [2] high) {
    let mid = Math.floor((low + high) / 2);
    if ([3](mid)) {
      high = mid;
    } else {
      low = mid + 1;
    }
  }
  return low;
}

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

AcanAchieve

B<=

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= in loop condition or wrong function name calls.

The function takes a check function 'canAchieve', loops while low < high, and calls canAchieve(mid) inside.

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