Why Binary Search on Answer Technique in DSA Rust?

The Big Idea

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

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 the smallest, checking if all items fit.

The Problem

This trial-and-error method is slow and frustrating because you might test many sizes unnecessarily. It's easy to make mistakes or waste time checking sizes that are obviously too small or too big.

The Solution

Binary Search on Answer Technique smartly guesses the answer by checking the middle size, then narrowing down the search range based on whether the guess works or not. This way, it quickly zooms in on the correct size without testing every possibility.

Before vs After

✗ Before

let mut size = 1;
while !can_fit(size) {
    size += 1;
}
println!("Minimum size: {}", size);

✓ After

let mut low = 1;
let mut high = max_possible_size;
while low < high {
    let mid = (low + high) / 2;
    if can_fit(mid) {
        high = mid;
    } else {
        low = mid + 1;
    }
}
println!("Minimum size: {}", low);

What It Enables

This technique enables you to solve complex optimization problems efficiently by turning them into yes/no questions and searching answers quickly.

Real Life Example

Suppose you want to split a set of files into DVDs with limited space. Binary Search on Answer helps find the smallest DVD size needed to fit all files without wasting space.

Key Takeaways

Manual guessing is slow and error-prone.

Binary Search on Answer narrows down the solution fast.

Transforms problems into yes/no checks for efficient searching.

Practice

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

easy

A. We guess an answer and use binary search to find the smallest or largest valid value.

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

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

D. We apply binary search to find the maximum element in an unsorted array.

Solution

1. Step 1: Understand the technique purpose

   Binary Search on Answer guesses a possible answer and checks if it satisfies the problem's conditions.

2. Step 2: Use binary search to narrow answer range

   By checking mid-values and adjusting search space, we find the smallest or largest valid answer efficiently.

3. Final Answer:

   We guess an answer and use binary search to find the smallest or largest valid value. -> Option A

4. Quick Check:

   Binary Search on Answer = Guess + Check + Narrow [OK]

Hint: Think of guessing answers and checking validity with binary search [OK]

Common Mistakes:

• Confusing it with normal binary search on sorted arrays
• Thinking it finds elements, not answers
• Assuming it works only on arrays

2. Which of the following Rust code snippets correctly initializes the binary search boundaries for searching an answer between 1 and 100?

easy

A. let mut low = 0; let mut high = 101;

B. let mut low = 0; let mut high = 100;

C. let mut low = 1; let mut high = 101;

D. let mut low = 1; let mut high = 100;

Solution

1. Step 1: Identify correct search range

   The problem states searching between 1 and 100 inclusive, so low should be 1 and high 100.

2. Step 2: Check boundary initialization

   let mut low = 1; let mut high = 100; sets low = 1 and high = 100, which matches the problem range exactly.

3. Final Answer:

   let mut low = 1; let mut high = 100; -> Option D

4. Quick Check:

   Boundaries match problem range [OK]

Hint: Set low and high exactly to the problem's answer range [OK]

Common Mistakes:

• Starting low at 0 when range starts at 1
• Setting high to 101 which is outside the range
• Confusing inclusive and exclusive boundaries

3. Consider this Rust function using binary search on answer to find the minimum x such that x*x >= 10:

fn binary_search_answer() -> i32 {
    let mut low = 1;
    let mut high = 10;
    while low < high {
        let mid = (low + high) / 2;
        if mid * mid >= 10 {
            high = mid;
        } else {
            low = mid + 1;
        }
    }
    low
}

fn main() {
    println!("{}", binary_search_answer());
}

What is the output when this program runs?

medium

A. 4

B. 5

C. 10

D. 3

Solution

1. Step 1: Understand the search condition

   We want the smallest x where x*x >= 10. Squares: 3*3=9 (no), 4*4=16 (yes).

2. Step 2: Trace binary search steps

   Initial low=1, high=10 mid=5, 5*5=25 >=10, high=5 mid=3, 3*3=9 <10, low=4 mid=4, 4*4=16 >=10, high=4 Now low=4, high=4, loop ends, returns low=4.

3. Final Answer:

   4 -> Option A

4. Quick Check:

   Minimum x with x*x >= 10 is 4 [OK]

Hint: Check mid squared against target, adjust low/high accordingly [OK]

Common Mistakes:

• Returning mid instead of low after loop
• Off-by-one errors in updating low or high
• Confusing condition direction (>= vs >)

4. The following Rust code tries to find the maximum x such that x*x <= 20 using binary search on answer. What is the bug?

fn max_x() -> i32 {
    let mut low = 0;
    let mut high = 20;
    while low <= high {
        let mid = (low + high) / 2;
        if mid * mid <= 20 {
            low = mid;
        } else {
            high = mid - 1;
        }
    }
    high
}

medium

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

B. The update 'low = mid' can cause infinite loop when low == mid.

C. The function should return 'low' instead of 'high'.

D. The initial 'high' should be 21, not 20.

Solution

1. Step 1: Analyze loop update statements

   When mid equals low, setting low = mid does not change low, causing infinite loop.

2. Step 2: Identify fix for infinite loop

   To avoid infinite loop, update low to mid + 1 when condition holds.

3. Final Answer:

   The update 'low = mid' can cause infinite loop when low == mid. -> Option B

4. Quick Check:

   Update low to mid + 1 to progress loop [OK]

Hint: Always move low or high past mid to avoid infinite loops [OK]

Common Mistakes:

• Not updating low or high correctly causing infinite loops
• Returning wrong boundary after loop ends
• Confusing loop condition operators

5. You want to find the minimum time required to complete all tasks given an array of task durations and a maximum allowed time. Using binary search on answer, which approach correctly finds the minimum maximum time per worker if tasks must be assigned in order and each worker can do consecutive tasks only? Choose the correct high-level approach.

hard

A. Use binary search to find the worker count, then assign tasks randomly.

B. Sort tasks and binary search for the task with maximum duration to assign first.

C. Binary search the answer range of time, and for each mid, greedily assign tasks to workers checking if all tasks fit within mid time.

D. Apply binary search on the number of tasks, splitting tasks evenly among workers.

Solution

1. Step 1: Define answer range and check function

   We binary search over possible maximum time per worker. For each mid, we check if tasks can be assigned in order without exceeding mid time per worker.

2. Step 2: Greedy assignment check

   Assign tasks to workers greedily, moving to next worker when current sum exceeds mid. If all tasks fit within allowed workers, mid is valid.

3. Final Answer:

   Binary search the answer range of time, and for each mid, greedily assign tasks to workers checking if all tasks fit within mid time. -> Option C

4. Quick Check:

   Binary Search on Answer + Greedy check = Correct approach [OK]

Hint: Binary search answer, check feasibility with greedy assignment [OK]

Common Mistakes:

• Sorting tasks when order matters
• Binary searching worker count instead of time
• Assigning tasks randomly ignoring order