Binary Search on Answer Technique in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the binary search boundaries.

DSA C

int low = 0, high = [1];

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

A100

B1e9

C10

D50

Attempts:

3 left

💡 Hint

Common Mistakes

Setting high too low and missing the correct answer.

Using a negative number for high.

We set high to a large number like 1e9 to cover the maximum possible answer range.

2fill in blank

medium

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

DSA C

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

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

A+

B-

C*

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction instead of addition.

Forgetting parentheses causing wrong order of operations.

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

3fill in blank

hard

Fix the error in the condition to update the binary search range.

DSA C

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

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

A!=

B=

C<

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using single equals = instead of double equals ==.

Using less than < which is not a boolean check.

The condition must check equality with true using ==.

4fill in blank

hard

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

DSA C

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

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

A<=

B<

C==

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using assignment = instead of comparison.

Using wrong loop condition causing infinite loop.

The loop runs while low <= high or low < high. Here, low < high is used. The condition checks if canAchieve(mid) == true.

5fill in blank

hard

Fill all three blanks to complete the binary search function returning the minimum valid answer.

DSA C

int binarySearch() {
    int low = 0, high = [1];
    int result = -1;
    while (low [2] high) {
        int mid = (low + high) / 2;
        if (canAchieve(mid) [3] true) {
            result = mid;
            high = mid - 1;
        } else {
            low = mid + 1;
        }
    }
    return result;
}

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

A1000000000

B<=

C==

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using < instead of <= in loop condition.

Using assignment instead of comparison in if condition.

Setting high too low and missing answers.

Set high to a large number like 1,000,000,000. The loop runs while low <= high. The condition checks if canAchieve(mid) == true.

Practice

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

easy

A. Searching for the answer by repeatedly dividing the search space based on feasibility

B. Sorting the array before applying binary search

C. Using binary search only on sorted arrays

D. Applying binary search to find the maximum element in an array

Solution

1. Step 1: Understand the problem context

   Binary Search on Answer is used when the answer lies within a range and we can check feasibility for a candidate answer.

2. Step 2: Explain the technique

   We repeatedly pick a middle value in the answer range and check if it satisfies the problem's conditions, then narrow the range accordingly.

3. Final Answer:

   Searching for the answer by repeatedly dividing the search space based on feasibility -> Option A

4. Quick Check:

   Binary Search on Answer = Divide and check feasibility [OK]

Hint: Think of answer range, not array elements, for binary search [OK]

Common Mistakes:

• Confusing it with normal binary search on arrays
• Trying to apply without a feasibility check
• Assuming answer must be in the input array

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

easy

A. for (int mid = low; mid <= high; mid++) { /* check feasibility */ }

B. do { mid = (low + high) / 2; /* check feasibility */ } while (low < high);

C. if (low < high) { mid = (low + high) / 2; /* check feasibility */ }

D. while (low <= high) { int mid = (low + high) / 2; /* check feasibility */ }

Solution

1. Step 1: Identify the binary search loop structure

   Binary search uses a while loop with condition low <= high to narrow the search space.

2. Step 2: Confirm calculation of mid and feasibility check

   Mid is calculated as (low + high) / 2 inside the loop, then feasibility is checked to adjust low or high.

3. Final Answer:

   while (low <= high) { int mid = (low + high) / 2; /* check feasibility */ } -> Option D

4. Quick Check:

   Binary search loop = while(low <= high) [OK]

Hint: Use while loop with low <= high for binary search [OK]

Common Mistakes:

• Using for loop instead of while
• Incorrect loop condition like low < high
• Not updating low or high inside loop

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

int canSplit(int* arr, int n, int m, int maxLoad) {
    int count = 1, currentSum = 0;
    for (int i = 0; i < n; i++) {
        if (arr[i] > maxLoad) return 0;
        if (currentSum + arr[i] > maxLoad) {
            count++;
            currentSum = arr[i];
        } else {
            currentSum += arr[i];
        }
    }
    return count <= m;
}

int binarySearchAnswer(int* arr, int n, int m) {
    int low = 0, high = 0;
    for (int i = 0; i < n; i++) high += arr[i];
    int ans = high;
    while (low <= high) {
        int mid = (low + high) / 2;
        if (canSplit(arr, n, m, mid)) {
            ans = mid;
            high = mid - 1;
        } else {
            low = mid + 1;
        }
    }
    return ans;
}

int main() {
    int arr[] = {7, 2, 5, 10, 8};
    int n = 5, m = 2;
    printf("%d", binarySearchAnswer(arr, n, m));
    return 0;
}

What is the output of this program?

medium

A. 18

B. 15

C. 20

D. 10

Solution

1. Step 1: Understand the problem and code logic

   The code finds the minimum maximum load to split the array into 2 subarrays so that the largest sum among subarrays is minimized.

2. Step 2: Dry run with given array

   Possible splits: - [7,2,5] and [10,8] max sums 14 and 18 -> max 18 - [7,2] and [5,10,8] max sums 9 and 23 -> max 23 - [7] and [2,5,10,8] max sums 7 and 25 -> max 25 Minimum max load is 18.

3. Final Answer:

   18 -> Option A

4. Quick Check:

   Minimum max load = 18 [OK]

Hint: Check feasibility by simulating splits, pick minimal max sum [OK]

Common Mistakes:

• Choosing sum of all elements as answer
• Not considering feasibility function logic
• Confusing minimum max load with total sum

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

int canComplete(int* tasks, int n, int m, int maxTime) {
    int count = 1, currentTime = 0;
    for (int i = 0; i < n; i++) {
        if (tasks[i] > maxTime) return 0;
        if (currentTime + tasks[i] >= maxTime) {
            count++;
            currentTime = tasks[i];
        } else {
            currentTime += tasks[i];
        }
    }
    return count <= m;
}

medium

A. The count variable should start from 0, not 1

B. The condition should be currentTime + tasks[i] > maxTime, not >= maxTime

C. The function should return count >= m instead of count <= m

D. The loop should run from i = 1 instead of i = 0

Solution

1. Step 1: Analyze the feasibility condition

   The condition currentTime + tasks[i] >= maxTime causes a new segment even when sum equals maxTime, which is incorrect.

2. Step 2: Correct the condition

   It should be currentTime + tasks[i] > maxTime to start a new segment only when sum exceeds maxTime.

3. Final Answer:

   The condition should be currentTime + tasks[i] > maxTime, not >= maxTime -> Option B

4. Quick Check:

   Use > for splitting condition, not >= [OK]

Hint: Use strict greater than (>) to split segments [OK]

Common Mistakes:

• Using >= causes extra segments
• Starting count from 0 instead of 1
• Returning wrong comparison for count and m

5. You want to find the minimum maximum height to cut trees so that the total wood collected is at least m units. Given an array trees[] of heights, which approach correctly applies binary search on answer to solve this problem?

hard

A. Set low = min height, high = max height; check if sum of all trees is >= m; adjust low or high accordingly

B. Sort trees, then binary search for the tree with height closest to m

C. Set low = 0, high = max height in trees; check if cutting at mid yields enough wood; adjust low or high accordingly

D. Use binary search on the array indices to find the tree that yields maximum wood

Solution

1. Step 1: Define the search space for cutting height

   The cutting height can vary from 0 to the tallest tree height.

2. Step 2: Check feasibility by summing wood collected at mid height

   For a candidate mid, sum wood = sum of (tree height - mid) if positive; check if sum >= m.

3. Step 3: Adjust search space based on feasibility

   If enough wood is collected, try higher cut (low = mid + 1) to minimize cut height; else lower cut (high = mid - 1).

4. Final Answer:

   Set low = 0, high = max height in trees; check if cutting at mid yields enough wood; adjust low or high accordingly -> Option C

5. Quick Check:

   Binary search on cut height range with feasibility check [OK]

Hint: Binary search on cutting height, not on tree indices [OK]

Common Mistakes:

• Binary searching on sorted array instead of answer range
• Using sum of all trees without cutting logic
• Confusing indices with answer values