Binary Search on Answer Technique in DSA C - Execution Trace

Concept Flow - Binary Search on Answer Technique

Set low and high bounds

↓

Calculate mid = (low + high) / 2

↓

Check if mid is a valid answer

↓

Update high

↓

Repeat until low > high

↓

Return best answer found

This flow shows how we guess an answer in a range, check if it works, then narrow the range until we find the best answer.

Execution Sample

DSA C

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

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

Execution Table

Step	Operation	low	high	mid	isValid(mid)	ans	Range Updated	Reason	Visual State
1	Initialize	1	10	-	-	-1	-	-	Range: [1..10], ans=-1
2	Calculate mid	1	10	5	No	-1	low = mid + 1 = 6	5 invalid, search higher	Range: [6..10], ans=-1
3	Calculate mid	6	10	8	Yes	8	high = mid - 1 = 7	8 valid, search lower	Range: [6..7], ans=8
4	Calculate mid	6	7	6	No	8	low = mid + 1 = 7	6 invalid, search higher	Range: [7..7], ans=8
5	Calculate mid	7	7	7	Yes	7	high = mid - 1 = 6	7 valid, search lower	Range: [7..6], ans=7
6	Check loop condition	7	6	-	-	7	-	low > high, stop	Final answer = 7

💡 Loop ends when low (7) becomes greater than high (6), final answer is 7.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	After Step 5	Final
low	1	6	6	7	7	7
high	10	10	7	7	6	6
mid	-	5	8	6	7	-
ans	-1	-1	8	8	7	7

Key Moments - 3 Insights

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

Because we want to find the smallest valid answer, so we try to see if there's a smaller valid number by moving 'high' below mid (see step 3 and 5 in execution_table).

Why does the loop stop when low > high?

The search space becomes invalid (empty), meaning we have checked all possible answers. This is shown at step 6 where low=7 and high=6.

What happens if isValid(mid) is false?

We move 'low' above mid to search higher values, because mid and below are invalid (see step 2 and 4).

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'ans' after step 3?

A 5

B -1

C 8

D 7

💡 Hint

Check the 'ans' column in row for step 3 in execution_table.

At which step does the loop condition become false?

A Step 6

B Step 5

C Step 4

D Step 3

💡 Hint

Look at the 'Reason' and 'Range Updated' columns in execution_table for when low > high.

If isValid(mid) was always false, what would happen to 'ans'?

A It would keep updating to mid

B It would remain -1

C It would become 0

D It would become high

💡 Hint

Refer to 'ans' updates in execution_table rows where isValid(mid) is true or false.

Concept Snapshot

Binary Search on Answer Technique:
- Set low and high bounds for possible answers.
- Calculate mid = (low + high) / 2.
- Check if mid is valid (problem-specific).
- If valid, update answer and move high to mid - 1 to find smaller valid.
- If invalid, move low to mid + 1 to find valid higher.
- Repeat until low > high, then return answer.

Full Transcript

Binary Search on Answer Technique is a method to find the best answer by guessing a value in a range and checking if it meets the problem's condition. We start with low and high bounds, calculate mid, and check if mid is valid. If valid, we save mid as a possible answer and try to find a smaller valid answer by moving high below mid. If invalid, we move low above mid to try larger values. We repeat this until low is greater than high, meaning no more values to check. The last saved answer is the best valid answer. This technique is useful when the answer is numeric and can be checked with a yes/no function. The execution table shows each step with variable values and decisions, helping visualize how the search narrows down the answer.

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