Maximum Product Subarray in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the maximum product variable.

DSA Rust

let mut max_product = [1];

Drag options to blanks, or click blank then click option'

Anums[0]

B0

C1

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing max_product to 0 or 1 instead of the first element.

Using a fixed number like -1 which may not be in the array.

We start with the first element of the array as the initial maximum product.

2fill in blank

medium

Complete the code to update the current maximum product at each step.

DSA Rust

max_product = max(max_product, [1]);

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Anums[i]

Bcurrent_max

Ccurrent_min

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using nums[i] directly instead of the current maximum product.

Using current_min which tracks minimum product, not maximum.

The current maximum product is compared with the global max_product to update it.

3fill in blank

hard

Fix the error in swapping current_max and current_min when the current number is negative.

DSA Rust

if nums[i] < 0 {
    let temp = [1];
    [2] = current_max;
    current_max = temp;
}

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Acurrent_min

Bcurrent_max

Cnums[i]

Dtemp

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping with the wrong variable.

Not using a temporary variable causing overwrite.

When the current number is negative, we swap current_max and current_min to handle sign changes.

4fill in blank

hard

Fill both blanks to update current_max and current_min with the current number.

DSA Rust

current_max = max(nums[i], current_max [1] nums[i]);
current_min = min(nums[i], current_min [2] nums[i]);

Drag options to blanks, or click blank then click option'

A*

B+

C-

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using addition or subtraction instead of multiplication.

Using division which is incorrect here.

We multiply current_max and current_min by nums[i] to update them correctly.

5fill in blank

hard

Fill all three blanks to complete the loop that calculates the maximum product subarray.

DSA Rust

for i in 1..nums.len() {
    if nums[i] < 0 {
        let temp = [1];
        [2] = current_max;
        current_max = temp;
    }
    current_max = max(nums[i], current_max [3] nums[i]);
    current_min = min(current_min * nums[i], nums[i]);
    max_product = max(max_product, current_max);
}

Drag options to blanks, or click blank then click option'

Acurrent_min

Bcurrent_max

C*

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Not swapping current_max and current_min correctly.

Using addition instead of multiplication in updates.

This loop updates current_max and current_min correctly, swapping when negative and multiplying to track products.

Practice

1. What does the "Maximum Product Subarray" problem ask you to find in an array of integers?

easy

A. The subarray with the smallest product of its elements

B. The subarray with the largest sum of its elements

C. The longest subarray with all positive numbers

D. The contiguous subarray with the largest product of its elements

Solution

1. Step 1: Understand the problem goal

   The problem asks for a contiguous part of the array where the product of all numbers is as large as possible.

2. Step 2: Differentiate from similar problems

   This is different from sum problems or longest subarray problems; it focuses on product, not sum or length.

3. Final Answer:

   The contiguous subarray with the largest product of its elements -> Option D

4. Quick Check:

   Maximum product subarray = largest product contiguous segment [OK]

Hint: Look for the subarray with the highest multiplication result [OK]

Common Mistakes:

• Confusing product with sum
• Ignoring negative numbers impact
• Thinking about longest subarray instead

2. Which Rust code snippet correctly initializes variables to track the maximum and minimum products for the Maximum Product Subarray problem?

easy

A. let (mut max_prod, mut min_prod, mut result) = (nums[0], nums[0], nums[0]);

B. let max_prod = 0; let min_prod = 0; let result = 0;

C. let max_prod = None; let min_prod = None; let result = None;

D. let (max_prod, min_prod, result) = (0, 0, 0);

Solution

1. Step 1: Identify correct initialization

   We must start with the first element of the array for max, min, and result to handle negative and positive values properly.

2. Step 2: Check Rust syntax for mutable variables

   Variables must be mutable (mut) to update during iteration, and tuple assignment is correct.

3. Final Answer:

   let (mut max_prod, mut min_prod, mut result) = (nums[0], nums[0], nums[0]); -> Option A

4. Quick Check:

   Initialize with first element and mut keyword [OK]

Hint: Initialize with first element and use mut for updates [OK]

Common Mistakes:

• Using zero instead of first element
• Not making variables mutable
• Using Option types unnecessarily

3. What is the output of the following Rust code for the Maximum Product Subarray problem?

let nums = vec![2, 3, -2, 4];
let (mut max_prod, mut min_prod, mut result) = (nums[0], nums[0], nums[0]);
for &num in &nums[1..] {
    if num < 0 {
        let temp = max_prod;
        max_prod = min_prod;
        min_prod = temp;
    }
    max_prod = max_prod.max(num).max(max_prod * num);
    min_prod = min_prod.min(num).min(min_prod * num);
    result = result.max(max_prod);
}
println!("{}", result);

medium

A. 6

B. 8

C. 4

D. -2

Solution

1. Step 1: Trace iteration and variable updates

   Start: max_prod=2, min_prod=2, result=2 - num=3: num>0, max_prod=max(2,3,2*3)=6, min_prod=min(2,3,2*3)=2, result=max(2,6)=6 - num=-2: num<0, swap max_prod and min_prod (max_prod=2, min_prod=6), max_prod=max(2,-2,2*-2)=2, min_prod=min(6,-2,6*-2)=-12, result=max(6,2)=6 - num=4: num>0, max_prod=max(2,4,2*4)=8, min_prod=min(-12,4,-12*4)=-48, result=max(6,8)=8

2. Step 2: Check final result printed

   The code prints the final result variable which is 8 after the last iteration.

3. Final Answer:

   8 -> Option B

4. Quick Check:

   Final max product = 8 [OK]

Hint: Swap max and min on negative, track max product [OK]

Common Mistakes:

• Not swapping max and min on negative number
• Confusing max and min updates
• Forgetting to update result

4. Identify the bug in this Rust code snippet for Maximum Product Subarray:

let nums = vec![2, -5, -2, -4, 3];
let (mut max_prod, mut min_prod, mut result) = (nums[0], nums[0], nums[0]);
for &num in &nums[1..] {
    if num > 0 {
        let temp = max_prod;
        max_prod = min_prod;
        min_prod = temp;
    }
    max_prod = max_prod.max(num).max(max_prod * num);
    min_prod = min_prod.min(num).min(min_prod * num);
    result = result.max(max_prod);
}
println!("{}", result);

medium

A. The loop should start from index 0, not 1

B. Variables max_prod and min_prod are not mutable

C. The swap condition should be for negative numbers, not positive

D. The max and min updates use wrong methods max() and min()

Solution

1. Step 1: Analyze the swap condition

   The code swaps max_prod and min_prod only if num > 0, but it should swap when num is negative to handle sign changes correctly.

2. Step 2: Confirm other parts are correct

   Variables are mutable, loop starts from index 1 correctly, and max/min methods are used properly.

3. Final Answer:

   The swap condition should be for negative numbers, not positive -> Option C

4. Quick Check:

   Swap on negative numbers only [OK]

Hint: Swap max and min only when current number is negative [OK]

Common Mistakes:

• Swapping on positive numbers
• Not swapping at all
• Incorrect loop start index

5. Given the array nums = [-1, -3, -10, 0, 60], what is the maximum product subarray and its product?

hard

A. Subarray: [60], Product: 60

B. Subarray: [-1, -3, -10], Product: -30

C. Subarray: [-3, -10, 0], Product: 0

D. Subarray: [-1, -3, -10, 0, 60], Product: 0

Solution

1. Step 1: Evaluate products of subarrays

   Check subarrays: - [-1, -3, -10]: product = -1 * -3 * -10 = -30 - [60]: product = 60 - [-3, -10, 0]: product = 0 - Entire array: product = 0 The highest product is 60 from subarray [60].

2. Step 2: Confirm no larger product exists

   Other subarrays with zeros or negatives produce smaller or zero products.

3. Final Answer:

   Subarray: [60], Product: 60 -> Option A

4. Quick Check:

   Max product is 60 from single element [OK]

Hint: Check single elements when zeros reset product [OK]

Common Mistakes:

• Ignoring single positive elements after zero
• Multiplying negatives without sign check
• Assuming longer subarray always better