Binary Search on Answer Technique in DSA Rust - Build from Scratch

Binary Search on Answer Technique

📖 Scenario: You are helping a factory manager decide the minimum time needed to produce a certain number of items using multiple machines working in parallel. Each machine produces items at its own speed.This is like finding the smallest time where the total items produced by all machines meet or exceed the target.

🎯 Goal: Build a Rust program that uses the binary search on answer technique to find the minimum time required to produce at least the target number of items.

📋 What You'll Learn

Create a vector called machines with the exact values [1, 2, 3, 4, 5] representing the time each machine takes to produce one item.

Create a variable called target and set it to 10 representing the total items to produce.

Implement a function called can_produce that takes time and returns true if machines can produce at least target items in that time.

Use binary search on the answer space to find the minimum time to produce at least target items.

Print the minimum time found.

💡 Why This Matters

🌍 Real World

Factories and production lines often need to find the minimum time to meet production goals using machines with different speeds.

💼 Career

This technique is useful for software engineers working on optimization problems, scheduling, and resource allocation.

1

Create the machines vector

Create a vector called machines with the exact values [1, 2, 3, 4, 5] representing the time each machine takes to produce one item.

DSA Rust

# Create the machines vector with values [1, 2, 3, 4, 5]
// Your code here

Hint

Use let machines = vec![1, 2, 3, 4, 5]; to create the vector.

2

Set the target number of items

Create a variable called target and set it to 10 representing the total items to produce.

DSA Rust

fn main() {
    let machines = vec![1, 2, 3, 4, 5];
    // Create a variable called target and set it to 10
    // Your code here
}

Hint

Use let target = 10; to set the target.

3

Implement the can_produce function

Implement a function called can_produce that takes time: u64, the machines: &Vec, and target: u64, and returns true if the total items produced by all machines in time is at least target. Use integer division to calculate items per machine.

DSA Rust

fn main() {
    let machines = vec![1, 2, 3, 4, 5];
    let target = 10;
}

// Implement can_produce function here
// Your code here

Hint

Loop over each machine's time and sum time / machine_time. Return true if sum ≥ target, else false.

4

Find and print the minimum time using binary search

Use binary search on the answer space between low = 0 and high = 1_000_000_000_000 to find the minimum time where can_produce(time, &machines, target) is true. Print the minimum time found using println!("{}", min_time);.

DSA Rust

fn main() {
    let machines = vec![1, 2, 3, 4, 5];
    let target = 10;

    // Use binary search to find minimum time
    // Your code here
}

fn can_produce(time: u64, machines: &Vec, target: u64) -> bool {
    let mut total = 0u64;
    for &machine_time in machines.iter() {
        total += time / machine_time;
        if total >= target {
            return true;
        }
    }
    false
}

Hint

Use a while loop with low and high. Update min_time when can_produce is true. Print min_time at the end.

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