Binary Search on Answer Technique in DSA Java - 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

↓

Move high = mid

↓

Repeat until low >= high

↓

Return low as the answer

Start with a range for possible answers, check the middle value, adjust range based on validity, repeat until range narrows to the answer.

Execution Sample

DSA Java

int low = 1, high = 10;
while (low < high) {
  int mid = (low + high) / 2;
  if (isValid(mid)) high = mid;
  else low = mid + 1;
}
return low;

This code finds the smallest valid number between 1 and 10 using binary search on answer.

Execution Table

Step	Operation	low	high	mid	isValid(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			Exit loop	[7, 7]
7	Return answer	7	7			Answer = 7	[7]

💡 Loop ends when low equals high, which is the smallest valid answer.

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
isValid(mid)		false	true	true	false

Key Moments - 3 Insights

Why do we set high = mid when isValid(mid) is true instead of high = mid - 1?

Because mid itself can be the valid answer, we keep it in the range by setting high = mid. See step 3 and 4 in execution_table where high moves to mid.

Why do we set low = mid + 1 when isValid(mid) is false?

Because mid is not valid, the answer must be greater than mid, so we move low to mid + 1. See step 2 and 5 in execution_table where low moves up.

When does the loop stop and why?

The loop stops when low equals high, meaning the range has narrowed to one number, which is the smallest valid answer. See step 6 in execution_table.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of mid at step 4?

A 8

B 7

C 6

D 5

💡 Hint

Check the 'mid' column in execution_table row with Step 4.

At which step does low first change from its initial value?

A Step 5

B Step 3

C Step 2

D Step 4

💡 Hint

Look at the 'low' column in execution_table and variable_tracker after Step 2.

If isValid(mid) was true at step 2, what would be the new value of high?

A 5

B 10

C 6

D 1

💡 Hint

When isValid(mid) is true, high is set to mid. Check step 3 action.

Concept Snapshot

Binary Search on Answer Technique:
- Define low and high as answer bounds.
- Calculate mid = (low + high) / 2.
- Check if mid is valid.
- If valid, set high = mid (keep mid).
- Else, set low = mid + 1 (exclude mid).
- Repeat until low == high.
- Return low as smallest valid answer.

Full Transcript

Binary Search on Answer Technique starts by setting a range low to high for possible answers. We calculate mid as the middle of this range. We check if mid is a valid answer using a problem-specific condition. If mid is valid, we move the high bound down to mid to keep mid in the range. If mid is not valid, we move the low bound up to mid plus one to exclude mid. We repeat this process until low equals high, meaning the range has narrowed to one number. This number is the smallest valid answer. The execution table shows each step with values of low, high, mid, and the decision made. Key moments clarify why we keep mid when valid and exclude it when invalid. The visual quiz tests understanding of mid values and bound changes. This technique efficiently finds answers in problems where answers lie in a range and can be checked for validity.

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 array and then apply binary search to find an element.

C. We use binary search to find the maximum element in an unsorted array.

D. We divide the array into two halves and solve each half separately.

Solution

1. Step 1: Understand the technique's approach

   Binary Search on Answer guesses a possible answer 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 guesses accordingly to find the best 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 = We guess an answer and check if it satisfies the problem constraints, then adjust the guess. [OK]

Hint: Guess answer, check condition, adjust guess accordingly [OK]

Common Mistakes:

• Confusing with normal binary search on sorted arrays
• Thinking it only finds elements, not answers
• Assuming it divides data instead of answer range

2. Which of the following is the correct way to initialize the search range for binary search on answer when finding minimum maximum load among workers?

easy

A. low = 0; high = maximum single load

B. low = minimum load; high = maximum load

C. low = 0; high = sum of all loads

D. low = average load; high = sum of all loads

Solution

1. Step 1: Define search range for answer

   Minimum maximum load can be between 0 and sum of all loads because no worker can have load more than total.

2. Step 2: Set low and high accordingly

   Set low = 0 and high = sum of all loads to cover all possible answers.

3. Final Answer:

   low = 0; high = sum of all loads -> Option C

4. Quick Check:

   Search range covers all loads = low = 0; high = sum of all loads [OK]

Hint: Set low to minimum possible, high to maximum possible answer [OK]

Common Mistakes:

• Setting low to minimum load instead of 0
• Setting high to max single load instead of sum
• Using average load as boundary

3. Consider the following Java code snippet using binary search on answer to find minimum maximum pages assigned to students:

int[] pages = {10, 20, 30, 40};
int low = 40, high = 100;
while (low < high) {
  int mid = (low + high) / 2;
  if (canAllocate(pages, mid, 2)) {
    high = mid;
  } else {
    low = mid + 1;
  }
}
System.out.println(low);

// canAllocate checks if pages can be divided among 2 students with max pages mid

What will be the output?

medium

A. 70

B. 60

C. 90

D. 100

Solution

1. Step 1: Understand the problem and initial range

   Pages array is {10,20,30,40}, low=40 (max single), high=100 (sum). We want minimum max pages for 2 students.

2. Step 2: Trace binary search iterations

   Mid = 70: canAllocate true (10+20+30=60 <=70 for 1st student, 40 for 2nd). So high=70.
   Mid = 55: canAllocate false (10+20+30=60 >55). So low=56.
   Mid = 63: canAllocate true (10+20+30=60 <=63, 40 for 2nd). high=63.
   Mid = 59: canAllocate false (60 >59). low=60.
   Mid = 61: canAllocate true. high=61.
   Mid = 60: canAllocate true. high=60.
   Mid = 60 again, low=60, high=60 ends loop.

3. Final Answer:

   60 -> Option B

4. Quick Check:

   Minimum max pages = 60 [OK]

Hint: Check mid feasibility, adjust low/high until low meets high [OK]

Common Mistakes:

• Not updating low/high correctly after canAllocate
• Misunderstanding canAllocate logic
• Incorrect mid calculation or loop condition

4. The following code attempts to find the minimum largest sum of subarrays using binary search on answer. Identify the bug:

int[] nums = {7, 2, 5, 10, 8};
int m = 2;
int low = 0, high = 32;
while (low <= high) {
  int mid = (low + high) / 2;
  if (canSplit(nums, mid, m)) {
    high = mid - 1;
  } else {
    low = mid + 1;
  }
}
System.out.println(low);

// canSplit checks if nums can be split into m subarrays with max sum <= mid

medium

A. The initial low should be max element, not 0

B. The loop condition should be low < high, not low <= high

C. The mid calculation should use (low + high + 1) / 2

D. The canSplit function should return false when sum equals mid

Solution

1. Step 1: Check initial low value

   Low is set to 0, but minimum largest sum can't be less than max element (10).

2. Step 2: Correct low initialization

   Set low = max element (10) to ensure valid search range and avoid infinite loop or wrong answer.

3. Final Answer:

   The initial low should be max element, not 0 -> Option A

4. Quick Check:

   Low must start at max element [OK]

Hint: Start low at max element to avoid invalid guesses [OK]

Common Mistakes:

• Starting low at 0 causing invalid mid values
• Using wrong loop condition causing infinite loop
• Misunderstanding canSplit return values

5. You want to find the minimum time required to complete all jobs by k workers, where each job takes a certain time. Using binary search on answer, which approach correctly finds the answer? int[] jobs = {3, 6, 7, 11}; int k = 2; Options:

hard

A. Sort jobs ascending; assign jobs alternately to workers; return max load.

B. Set low = 0, high = max job time; assign jobs greedily without checking mid; return high.

C. Set low = sum of all jobs, high = max job time; use binary search to find minimum sum.

D. Set low = max job time, high = sum of all jobs; check if jobs can be assigned within mid time using k workers; adjust low/high accordingly.

Solution

1. Step 1: Define search boundaries

   Low should be max job time (11), high sum of all jobs (3+6+7+11=27) to cover all possible times.

2. Step 2: Use binary search with feasibility check

   Check if jobs can be assigned to k=2 workers within mid time. Adjust low/high based on feasibility.

3. Final Answer:

   Set low = max job time, high = sum of all jobs; check if jobs can be assigned within mid time using k workers; adjust low/high accordingly. -> Option D

4. Quick Check:

   Binary search on answer with feasibility check = Set low = max job time, high = sum of all jobs; check if jobs can be assigned within mid time using k workers; adjust low/high accordingly. [OK]

Hint: Set low=max job, high=sum; check feasibility; adjust range [OK]

Common Mistakes:

• Setting wrong low/high boundaries
• Not checking feasibility properly
• Assigning jobs without binary search