Maximum Product Subarray in DSA Rust - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Maximum Product Subarray Calculation

What is the output of the following Rust code that finds the maximum product of a contiguous subarray?

DSA Rust

use std::vec::Vec;

fn max_product(nums: Vec<i32>) -> i32 {
    let mut max_prod = nums[0];
    let mut min_prod = nums[0];
    let mut result = nums[0];

    for &num in nums.iter().skip(1) {
        let cand1 = max_prod * num;
        let cand2 = min_prod * num;
        max_prod = num.max(cand1).max(cand2);
        min_prod = num.min(cand1).min(cand2);
        result = result.max(max_prod);
    }
    result
}

fn main() {
    let nums = vec![2, 3, -2, 4];
    println!("{}", max_product(nums));
}

A 6

B 4

C -2

D 8

💡 Hint

Think about the subarray that gives the maximum product, considering negative numbers can flip signs.

The maximum product subarray is [2, 3] with product 6.

The code tracks max and min products to handle negative numbers.

❓ Predict Output

intermediate

Result of Maximum Product Subarray with Negative Numbers

What will be printed by this Rust program that calculates the maximum product subarray?

DSA Rust

use std::vec::Vec;

fn max_product(nums: Vec<i32>) -> i32 {
    let mut max_prod = nums[0];
    let mut min_prod = nums[0];
    let mut result = nums[0];

    for &num in nums.iter().skip(1) {
        let cand1 = max_prod * num;
        let cand2 = min_prod * num;
        max_prod = num.max(cand1).max(cand2);
        min_prod = num.min(cand1).min(cand2);
        result = result.max(max_prod);
    }
    result
}

fn main() {
    let nums = vec![-2, 0, -1];
    println!("{}", max_product(nums));
}

A 0

B -2

C -1

D 1

💡 Hint

Remember zero resets the product chain.

The maximum product subarray is [0] with product 0.

Negative numbers and zero reset the product tracking.

🔧 Debug

advanced

Identify the Logical Error in Maximum Product Subarray Code

This Rust code aims to find the maximum product subarray but produces wrong results. What is the logical error?

DSA Rust

use std::vec::Vec;

fn max_product(nums: Vec<i32>) -> i32 {
    let mut max_prod = 1;
    let mut min_prod = 1;
    let mut result = i32::MIN;

    for &num in nums.iter() {
        if num == 0 {
            max_prod = 1;
            min_prod = 1;
            continue;
        }
        let temp = max_prod * num;
        max_prod = max_prod.max(min_prod) * num;
        min_prod = min_prod.min(temp);
        result = result.max(max_prod);
    }
    result
}

A The result variable is initialized incorrectly

B The initial values of max_prod and min_prod should be nums[0], not 1

C The code does not handle zero values properly

D max_prod and min_prod are not updated correctly before multiplication

💡 Hint

Check how max_prod and min_prod are updated inside the loop.

The code updates max_prod and min_prod incorrectly by multiplying before choosing max/min.

This causes wrong tracking of products.

❓ Predict Output

advanced

Output of Maximum Product Subarray with All Negative Numbers

What is the output of this Rust program that calculates the maximum product subarray for all negative numbers?

DSA Rust

use std::vec::Vec;

fn max_product(nums: Vec<i32>) -> i32 {
    let mut max_prod = nums[0];
    let mut min_prod = nums[0];
    let mut result = nums[0];

    for &num in nums.iter().skip(1) {
        let cand1 = max_prod * num;
        let cand2 = min_prod * num;
        max_prod = num.max(cand1).max(cand2);
        min_prod = num.min(cand1).min(cand2);
        result = result.max(max_prod);
    }
    result
}

fn main() {
    let nums = vec![-2, -3, -7];
    println!("{}", max_product(nums));
}

A -7

B -2

C 21

D 42

💡 Hint

Multiplying two negatives can produce a positive product.

The maximum product subarray is [-3, -7] with product 21.

The code tracks max and min products to handle negative numbers.

🧠 Conceptual

expert

Why Track Both Maximum and Minimum Products in Maximum Product Subarray?

In the maximum product subarray problem, why do we track both the maximum and minimum products ending at the current position?

A Because the minimum product helps to find the smallest subarray sum

B Because the minimum product can become maximum if multiplied by a negative number

C Because tracking both helps to find the longest subarray with positive product

D Because the minimum product is used to reset the maximum product when zero is encountered

💡 Hint

Think about how negative numbers affect multiplication.

Multiplying a negative minimum product by a negative number can produce a large positive maximum product.

Tracking both helps handle negative numbers correctly.

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