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 = mid

↓

Repeat until low >= high

↓

Return low as answer

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

Execution Sample

DSA C++

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 the answer.

Execution Table

Step	low	high	mid	isValid(mid)	Action	Range After Action
1	1	10	5	No	low = mid + 1	[6, 10]
2	6	10	8	Yes	high = mid	[6, 8]
3	6	8	7	No	low = mid + 1	[8, 8]
Exit	8	8	-	-	low >= high, stop	Answer = 8

💡 low equals high, so the smallest valid answer is found.

Variable Tracker

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

Key Moments - 3 Insights

Why do we update 'low' to 'mid + 1' when isValid(mid) is false?

Because mid is not a valid answer, so the answer must be greater than mid. See Step 1 and Step 3 in execution_table where low moves past mid.

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

Because mid might be the smallest valid answer, so we keep it by setting high = mid. See Step 2 where high moves down to mid.

Why does the loop stop when low equals high?

Because the search range has narrowed to a single number, which is the smallest valid answer. See Exit row in execution_table.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'mid' at Step 3?

A 8

B 7

C 5

D 6

💡 Hint

Check the 'mid' column in the row labeled Step 3 in execution_table.

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

A Step 1

B Step 3

C Step 2

D Step 4

💡 Hint

Look at the 'low' column in execution_table rows to see when it changes from 1.

If isValid(mid) returned true at Step 1, what would be the new value of 'high' after Step 1?

A 10

B 6

C 5

D 1

💡 Hint

When isValid(mid) is true, high is set to mid. Check Step 2 for example.

Concept Snapshot

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

Full Transcript

Binary Search on Answer Technique starts by setting a low and high range for possible answers. We calculate the middle value mid and check if it is valid. If mid is valid, we move the high bound down to mid to find a smaller valid answer. If mid is not valid, we move the low bound up to mid plus one to search higher values. This repeats until low equals high, which is the smallest valid answer. The example code shows this process with low starting at 1 and high at 10. The execution table traces each step, showing how low, high, and mid change, and what action is taken based on validity. Key moments explain why we update low or high in each case and why the loop stops when low equals high. The visual quiz tests understanding of mid values, variable changes, and hypothetical changes in validity. This technique helps solve problems where the answer lies within a range and can be checked for validity efficiently.

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 to find an element in a sorted array

D. Applying binary search only on arrays with unique elements

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: Apply binary search on the answer range

   We repeatedly pick a middle value and check if it satisfies the problem's conditions, narrowing the search space 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 = feasibility check [OK]

Hint: Think of binary search on answer as guessing and checking [OK]

Common Mistakes:

• Confusing it with normal binary search on arrays
• Not checking feasibility properly
• Trying to apply it on unsorted data without a valid range

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

easy

A. while (low <= high) { int mid = (low + high) / 2; }

B. for (int mid = low; mid <= high; mid++) { }

C. while (low < high) { int mid = low + (high - low) / 2; }

D. if (low < high) { mid = (low + high) / 2; }

Solution

1. Step 1: Identify the correct loop condition

   Binary search typically uses a loop with condition low < high to avoid infinite loops and handle edge cases.

2. Step 2: Calculate mid safely

   Using mid = low + (high - low) / 2 prevents integer overflow compared to (low + high) / 2.

3. Final Answer:

   while (low < high) { int mid = low + (high - low) / 2; } -> Option C

4. Quick Check:

   Safe mid calculation with proper loop [OK]

Hint: Use low + (high - low)/2 to avoid overflow [OK]

Common Mistakes:

• Using <= in loop causing infinite loops
• Calculating mid as (low + high)/2 risking overflow
• Using for loop instead of while for binary search

3. Consider the following code snippet that uses binary search on answer to find the minimum maximum load for dividing an array into k subarrays:

bool canDivide(const vector<int>& arr, int k, int maxLoad) {
    int currentSum = 0, count = 1;
    for (int num : arr) {
        if (num > maxLoad) return false;
        if (currentSum + num > maxLoad) {
            count++;
            currentSum = num;
        } else {
            currentSum += num;
        }
    }
    return count <= k;
}

int binarySearchAnswer(const vector<int>& arr, int k) {
    int low = 0, high = 0;
    for (int num : arr) high += num;
    while (low < high) {
        int mid = low + (high - low) / 2;
        if (canDivide(arr, k, mid)) high = mid;
        else low = mid + 1;
    }
    return low;
}

int main() {
    vector<int> arr = {7, 2, 5, 10, 8};
    int k = 2;
    cout << binarySearchAnswer(arr, k) << endl;
}

What is the output of this program?

medium

A. 15

B. 18

C. 20

D. 22

Solution

1. Step 1: Understand the problem and code

   The code finds the minimum maximum sum when dividing the array into 2 subarrays. The total sum is 7+2+5+10+8=32.

2. Step 2: Trace binary search on answer

   Check mid values and feasibility:
   - mid=16: canDivide returns true (possible, e.g., subarrays sums 14 and 18)
   - mid=15: canDivide returns true (possible, e.g., subarrays sums 14 and 15)
   - mid=14: canDivide returns false (cannot split into 2 with max sum 14)
   Binary search narrows down to 15 as minimum max load.

3. Final Answer:

   15 -> Option A

4. Quick Check:

   Minimum max load = 15 [OK]

Hint: Sum array, binary search between 0 and sum [OK]

Common Mistakes:

• Not checking if single element > mid
• Incorrectly updating low and high
• Misunderstanding feasibility function

4. The following code attempts to find the smallest maximum weight capacity to ship packages within D days using binary search on answer. Identify the bug:

bool canShip(const vector<int>& weights, int D, int capacity) {
    int days = 1, currentLoad = 0;
    for (int w : weights) {
        if (w > capacity) return false;
        if (currentLoad + w > capacity) {
            days++;
            currentLoad = w;
        } else {
            currentLoad += w;
        }
    }
    return days <= D;
}

int shipWithinDays(const vector<int>& weights, int D) {
    int low = 0, high = accumulate(weights.begin(), weights.end(), 0);
    while (low <= high) {
        int mid = low + (high - low) / 2;
        if (canShip(weights, D, mid)) high = mid - 1;
        else low = mid + 1;
    }
    return low;
}

medium

A. The loop condition should be low < high instead of low <= high

B. The mid calculation can cause integer overflow

C. The canShip function does not handle empty weights array

D. The initial value of low should be the maximum weight in the array, not 0

Solution

1. Step 1: Analyze initial search range

   Low should start at the maximum single package weight because capacity cannot be less than the heaviest package.

2. Step 2: Identify impact of wrong low value

   Starting low at 0 causes unnecessary checks and may cause incorrect results or longer runtime.

3. Final Answer:

   The initial value of low should be the maximum weight in the array, not 0 -> Option D

4. Quick Check:

   Low = max(weights) [OK]

Hint: Low must be max element, high sum of all [OK]

Common Mistakes:

• Starting low at 0 instead of max element
• Using <= in loop causing off-by-one errors
• Ignoring edge cases with large weights

5. You want to find the minimum time required to complete n tasks using m machines, where each machine can complete one task in a fixed time given by array times. You can assign multiple tasks to a machine but tasks are done sequentially. Using binary search on answer, which approach correctly finds the minimum time to finish all tasks?

Choose the correct approach:

hard

A. Binary search on number of machines from 1 to m, checking if tasks can be done in fixed time

B. Binary search on time range from 0 to max(times) * n, and check if total tasks done by all machines in mid time is at least n

C. Sort times and pick the machine with minimum time multiplied by n

D. Use binary search on tasks from 1 to n, checking if tasks fit in given time

Solution

1. Step 1: Define search space for binary search

   The minimum time is 0 and maximum time is max machine time * n tasks (worst case all tasks on slowest machine).

2. Step 2: Check feasibility at mid time

   Calculate total tasks all machines can complete in mid time by summing floor(mid / times[i]). If total tasks >= n, mid time is feasible.

3. Step 3: Narrow search space

   If feasible, try smaller time by setting high = mid; else low = mid + 1.

4. Final Answer:

   Binary search on time range from 0 to max(times) * n, and check if total tasks done by all machines in mid time is at least n -> Option B

5. Quick Check:

   Binary search on time with feasibility check [OK]

Hint: Search time range, count tasks done at mid [OK]

Common Mistakes:

• Binary searching on machines instead of time
• Not summing tasks correctly for feasibility
• Ignoring sequential task processing per machine