Binary Search on Answer Technique in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the search boundaries for binary search on answer.

DSA Python

low, high = [1], max(arr)

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

A1

Blen(arr)

C-1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Starting low at 1 instead of 0

Using length of array as low boundary

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

2fill in blank

medium

Complete the code to calculate the middle value in binary search on answer.

DSA Python

mid = [1] + (high - low) // 2

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

Ahigh

Blow

C0

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using (low + high) // 2 directly

Using high - low without adding low

Mid is calculated from low plus half the distance to high to avoid overflow.

3fill in blank

hard

Fix the error in the condition to check feasibility in binary search on answer.

DSA Python

if can_achieve(mid) [1] True:

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

A!=

B=

C==

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '=' instead of '==' in if condition

Using '!=' which reverses logic

We must compare can_achieve(mid) to True using '==', not assign with '='.

4fill in blank

hard

Fill both blanks to update boundaries after feasibility check in binary search on answer.

DSA Python

if can_achieve(mid):
    low = [1]
else:
    high = [2]

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

Amid + 1

Bmid - 1

Cmid

Dmid * 2

Attempts:

3 left

💡 Hint

Common Mistakes

Setting low = mid instead of mid + 1

Setting high = mid instead of mid - 1

If mid is possible, we move low up to mid + 1 to search higher. Otherwise, high moves down to mid - 1.

5fill in blank

hard

Fill all three blanks to return the final answer after binary search on answer loop.

DSA Python

while low <= high:
    mid = low + (high - low) // 2
    if can_achieve(mid):
        result = [1]
        low = [2]
    else:
        high = [3]

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

Amid

Bmid + 1

Cmid - 1

Dlow

Attempts:

3 left

💡 Hint

Common Mistakes

Not updating result inside the loop

Incorrectly updating low or high boundaries

When mid is feasible, store mid as result and move low up to mid + 1. Otherwise, move high down to mid - 1.

Practice

1. What is the main idea behind the Binary Search on Answer technique?

easy

A. We apply binary search to split the input into two equal halves repeatedly.

B. We sort the input array and then apply binary search to find an element.

C. We guess an answer and check if it satisfies the problem, then adjust the guess using binary search.

D. We use binary search to find the middle element of a sorted list.

Solution

1. Step 1: Understand the technique's purpose

   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 using binary search to find the optimal answer.

3. Final Answer:

   We guess an answer and check if it satisfies the problem, then adjust the guess using binary search. -> Option C

4. Quick Check:

   Guess and check with binary search = We guess an answer and check if it satisfies the problem, then adjust the guess using binary search. [OK]

Hint: Think of guessing answers and narrowing down with binary search [OK]

Common Mistakes:

• Confusing with normal binary search on sorted arrays
• Thinking it only finds elements, not answers
• Ignoring the check step after guessing

2. Which of the following is the correct syntax to start a binary search on answer in Python for an integer answer between low and high?

easy

A. while low <= high:
mid = (low + high) // 2

B. for mid in range(low, high):

C. if low < high:
mid = (low + high) / 2

D. while low < high:
mid = (low + high) // 2 + 1

Solution

1. Step 1: Identify the binary search loop condition

   Binary search typically uses a while loop with condition low <= high to cover all possible values.

2. Step 2: Calculate mid correctly

   Mid is calculated as integer division (low + high) // 2 to get the middle integer index or value.

3. Final Answer:

   while low <= high: mid = (low + high) // 2 -> Option A

4. Quick Check:

   Correct loop and mid calculation = while low <= high: mid = (low + high) // 2 [OK]

Hint: Use while low <= high and integer division for mid [OK]

Common Mistakes:

• Using for loop instead of while
• Using float division instead of integer division
• Wrong loop condition like low < high

3. Consider the following code snippet using binary search on answer to find the minimum maximum load for splitting an array into m subarrays:

def can_split(nums, m, max_load):
    count, current_sum = 1, 0
    for num in nums:
        if current_sum + num > max_load:
            count += 1
            current_sum = num
            if count > m:
                return False
        else:
            current_sum += num
    return True

nums = [7, 2, 5, 10, 8]
m = 2
low, high = max(nums), sum(nums)
while low < high:
    mid = (low + high) // 2
    if can_split(nums, m, mid):
        high = mid
    else:
        low = mid + 1
print(low)

What is the output of this code?

medium

A. 15

B. 18

C. 20

D. 10

Solution

1. Step 1: Understand the problem and initial bounds

   We want to split nums into 2 subarrays minimizing the maximum sum. low = max(nums) = 10, high = sum(nums) = 32.

2. Step 2: Trace binary search iterations

   - mid=21: True ([7,2,5]=14, [10,8]=18 <=21), high=21
   - mid=15: False (requires 3 subarrays), low=16
   - mid=18: True, high=18
   - mid=17: False, low=18
   Final low=18.

3. Final Answer:

   18 -> Option B

4. Quick Check:

   Minimum max load for 2 splits = 18 [OK]

Hint: Check mid as max load and validate splits [OK]

Common Mistakes:

• Not setting low as max element
• Confusing count of subarrays
• Off-by-one errors in loop conditions

4. The following code tries to find the minimum maximum time to complete tasks by workers using binary search on answer. Identify the bug:

def can_finish(tasks, workers, max_time):
    count = 1
    current_time = 0
    for t in tasks:
        if t > max_time:
            return False
        if current_time + t <= max_time:
            current_time += t
        else:
            count += 1
            current_time = t
    return count <= workers

tasks = [3, 6, 7, 11]
workers = 2
low, high = 0, sum(tasks)
while low < high:
    mid = (low + high) // 2
    if can_finish(tasks, workers, mid):
        high = mid
    else:
        low = mid + 1
print(low)

medium

A. The current_time should reset to 0 instead of t after incrementing count

B. The function should return count < workers instead of <= workers

C. The loop condition should be while low <= high

D. Initial low should be max(tasks), not 0

Solution

1. Step 1: Check initial binary search bounds

   Low is set to 0, but minimum max_time must be at least max(tasks) because no task can be split.

2. Step 2: Understand effect of wrong low

   Starting low at 0 causes unnecessary checks and may cause infinite loop or wrong answer.

3. Final Answer:

   Initial low should be max(tasks), not 0 -> Option D

4. Quick Check:

   Low bound must be max task time [OK]

Hint: Set low to max task time to avoid invalid guesses [OK]

Common Mistakes:

• Setting low to 0 instead of max(tasks)
• Wrong loop condition causing infinite loops
• Resetting current_time incorrectly

5. You want to find the minimum largest sum when splitting an array nums into k subarrays. However, some elements can be zero or negative. How should you modify the binary search on answer approach to handle this correctly?

hard

A. Set low to min(nums) and high to sum of absolute values of nums, then apply binary search with adjusted checks.

B. Ignore negative values and treat them as zero, then apply normal binary search on answer.

C. Use normal binary search with low as max(nums) and high as sum(nums) without changes.

D. Sort the array first, then apply binary search on answer with low as 0 and high as sum(nums).

Solution

1. Step 1: Identify problem with negative and zero values

   Negative values can reduce sums, so max(nums) may be negative or zero, and sum(nums) may be less than max(nums).

2. Step 2: Adjust binary search bounds

   Set low to minimum element (which can be negative) and high to sum of absolute values to cover all possible sums.

3. Step 3: Modify check function

   Check function must handle negative sums correctly and validate splits accordingly.

4. Final Answer:

   Set low to min(nums) and high to sum of absolute values of nums, then apply binary search with adjusted checks. -> Option A

5. Quick Check:

   Adjust bounds for negatives and zeros [OK]

Hint: Adjust low/high bounds when negatives exist [OK]

Common Mistakes:

• Using max(nums) as low when negatives exist
• Ignoring negative values in sums
• Assuming sums are always positive