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Binary Search on Answer Technique in DSA Java - Cheat Sheet & Quick Revision

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Recall & Review
beginner
What is the main idea behind the Binary Search on Answer Technique?
It is a method to find an unknown value by guessing an answer range and using binary search to narrow down the correct answer based on a condition check.
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beginner
In Binary Search on Answer, what do you need to define before starting the search?
You need to define the search space (minimum and maximum possible answers) and a condition function that tells if a guess is valid or not.
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intermediate
Why is Binary Search on Answer useful for optimization problems?
Because it helps find the best possible answer efficiently by checking feasibility of guesses instead of trying all possibilities.
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intermediate
What is the role of the condition function in Binary Search on Answer?
It checks if the current guess satisfies the problem constraints, guiding whether to search higher or lower in the answer range.
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beginner
Give a simple example where Binary Search on Answer can be applied.
Finding the minimum maximum load to divide tasks among workers so that no worker has to do more than that load.
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What does Binary Search on Answer technique primarily search over?
AA tree structure
BA range of possible answers
CA linked list
DAn array of sorted numbers
What must the condition function return to move the search to the lower half?
AAn exception
BFalse, meaning the guess is invalid
CNull
DTrue, meaning the guess is valid
Which of these problems is a good candidate for Binary Search on Answer?
AFinding the square root of a number
BSorting an array
CFinding the shortest path in a graph
DChecking if a number is prime
What is the time complexity of Binary Search on Answer if the search space size is N?
AO(log N)
BO(N)
CO(N log N)
DO(1)
What is the first step in applying Binary Search on Answer?
ABuild a tree
BSort the input data
CDefine the minimum and maximum possible answers
DInitialize a queue
Explain how Binary Search on Answer works and why it is useful.
Think about guessing answers and checking if they fit the problem.
You got /4 concepts.
    Describe the steps to implement Binary Search on Answer for a problem.
    Focus on setting up the search and checking guesses.
    You got /5 concepts.

      Practice

      (1/5)
      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
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