Why Binary Search on Answer Technique in DSA C++?

The Big Idea

What if you could find the perfect answer without trying every possibility?

The Scenario

Imagine you want to find the smallest size of a box that can hold all your items, but you only have a list of item sizes and no direct formula. You try different box sizes one by one, starting from very small to very large, checking each time if all items fit.

The Problem

This trial-and-error method is slow and tiring because you might test many sizes unnecessarily. It's easy to get stuck testing sizes that are too small or too big without a clear way to narrow down your search quickly.

The Solution

The Binary Search on Answer technique helps by smartly guessing the box size in the middle of your current range. Then it checks if the guess works. If it does, it tries smaller sizes; if not, it tries bigger sizes. This way, it quickly zooms in on the best size without testing every possibility.

Before vs After

✗ Before

int size = 1;
while (!fits(size)) {
    size++;
}
return size;

✓ After

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

What It Enables

This technique enables you to solve complex optimization problems efficiently by turning them into a search for the best answer within a range.

Real Life Example

For example, when deciding the minimum time needed to complete all tasks with limited workers, you can use this technique to find the smallest time that still allows all tasks to finish.

Key Takeaways

Manual checking is slow and inefficient for large ranges.

Binary Search on Answer narrows down the answer quickly by testing midpoints.

It transforms optimization problems into search problems for easier solving.

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