Binary Search on Answer Technique in DSA Typescript - Execution Trace

Concept Flow - Binary Search on Answer Technique

Set low and high bounds for answer

↓

Calculate mid = (low + high) / 2

↓

Check if mid is a valid answer

↓

Update high = mid

↓

Repeat until low >= high

↓

Return low as answer

This flow shows how we guess an answer between low and high, check if it works, then narrow the range until we find the best answer.

Execution Sample

DSA Typescript

function canPlace(mid: number): boolean {
  // Check if mid satisfies problem condition
  return true; // placeholder implementation
}

let low = 1, high = 10;
while (low < high) {
  let mid = Math.floor((low + high) / 2);
  if (canPlace(mid)) high = mid;
  else low = mid + 1;
}
return low;

This code tries to find the smallest number between 1 and 10 that satisfies canPlace.

Execution Table

Step	Operation	low	high	mid	canPlace(mid)	Action	Range After Action
1	Initialize low and high	1	10	-	-	-	[1, 10]
2	Calculate mid	1	10	5	false	low = mid + 1	[6, 10]
3	Calculate mid	6	10	8	true	high = mid	[6, 8]
4	Calculate mid	6	8	7	true	high = mid	[6, 7]
5	Calculate mid	6	7	6	false	low = mid + 1	[7, 7]
6	Check low < high	7	7	-	-	Stop loop	[7, 7]
7	Return answer	7	7	-	-	Answer = 7	[7]

💡 Loop stops when low equals high, meaning the smallest valid answer is found.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	Final
low	1	6	6	6	7	7	7
high	10	10	8	7	7	7	7
mid	-	5	8	7	6	-	-
canPlace(mid)	-	false	true	true	false	-	-

Key Moments - 3 Insights

Why do we update 'low' to mid + 1 when canPlace(mid) is false?

Because mid does not satisfy the condition, so the answer must be greater than mid. This is shown in step 2 and 5 where low moves past mid.

Why do we update 'high' to mid when canPlace(mid) is true?

Because mid satisfies the condition, so the answer could be mid or smaller. We narrow the search by setting high to mid, as seen in steps 3 and 4.

Why does the loop stop when low equals high?

Because the search space is reduced to one number, which must be the smallest valid answer. This is shown in step 6.

Visual Quiz - 3 Questions

Look at the execution table, what is the value of 'mid' at step 4?

A 7

B 6

C 8

D 5

💡 Hint

Check the 'mid' column in the execution table at step 4.

At which step does 'low' first become greater than 5?

A Step 3

B Step 5

C Step 2

D Step 4

💡 Hint

Look at the 'low' values in the variable tracker after each step.

If canPlace(mid) was always true, what would happen to 'high'?

A It would stay the same

B It would decrease

C It would increase

D It would become equal to low immediately

💡 Hint

Refer to the action column in the execution table when canPlace(mid) is true.

Concept Snapshot

Binary Search on Answer Technique:
- Set low and high bounds for possible answers.
- Calculate mid = (low + high) / 2.
- Check if mid satisfies the condition (canPlace).
- If yes, move high to mid to find smaller answer.
- If no, move low to mid + 1 to find larger answer.
- Repeat until low == high, which is the answer.

Full Transcript

Binary Search on Answer Technique is a way to find the smallest or largest value that meets a condition. We start with a low and high range. We pick the middle value and check if it works. If it does, we try smaller values by moving the high down. If it doesn't, we try bigger values by moving the low up. We repeat this until low and high meet. The final low is the answer. This method helps solve problems where direct search is slow but checking a guess is fast.

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