Binary Search on Answer Technique in DSA Rust - 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, then narrow the range based on validity until the answer is found.

Execution Sample

DSA Rust

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

This code finds the smallest valid answer between low and high using binary search on the answer.

Execution Table

Step	Operation	low	high	mid	is_valid(mid)?	Action	Range After Action
1	Initialize	1	10				[1, 10]
2	Calculate mid	1	10	5	No	low = mid + 1	[6, 10]
3	Calculate mid	6	10	8	Yes	high = mid	[6, 8]
4	Calculate mid	6	8	7	No	low = mid + 1	[8, 8]
5	Check low < high	8	8			Stop	[8, 8]
6	Return answer	8	8			Answer = 8	[8]

💡 Loop ends when low equals high, which is the smallest valid answer.

Variable Tracker

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

Key Moments - 3 Insights

Why do we update high = mid when is_valid(mid) is true?

Because mid is a valid answer, we try to find a smaller valid answer by moving high down to mid, as shown in step 3 of the execution_table.

Why do we set low = mid + 1 when is_valid(mid) is false?

Because mid is not valid, all answers less than or equal to mid are invalid, so we move low up to mid + 1 to search higher values, as seen in step 2 and 4.

Why does the loop stop when low equals high?

When low equals high, the search range has narrowed to one value, which is the smallest valid answer, as shown in step 5.

Visual Quiz - 3 Questions

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

A 8

B 5

C 7

D 6

💡 Hint

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

At which step does low become equal to high?

A Step 4

B Step 3

C Step 5

D Step 2

💡 Hint

Look at 'low' and 'high' columns in execution_table rows to find when they match.

If is_valid(mid) was true at step 2, what would be the new value of high?

A 6

B 5

C 10

D 1

💡 Hint

Refer to the action column in execution_table and how high changes when is_valid(mid) is true.

Concept Snapshot

Binary Search on Answer Technique:
- Start with low and high bounds of answer range.
- Calculate mid = (low + high) / 2.
- Check if mid is valid.
- If valid, set high = mid to find smaller valid answer.
- Else, set low = mid + 1 to search higher.
- Repeat until low == high; low is the answer.

Full Transcript

Binary Search on Answer Technique helps find the smallest or largest value that satisfies a condition. We start with a range low to high. We pick the middle value mid and check if it is valid. If yes, we move the high bound down to mid to try smaller values. If no, we move the low bound up to mid plus one to try larger values. We repeat this until low equals high. That value is the answer. The execution table shows each step with low, high, mid, and decisions. The variable tracker shows how low, high, and mid change. Key moments explain why we update bounds as we do. The quiz tests understanding of mid values and loop termination.

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