Maximum Product Subarray in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the maximum product with the first element of the array.

DSA Python

def max_product_subarray(nums):
    max_prod = [1]
    return max_prod

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

Amax(nums)

B0

C1

Dnums[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing max_prod to 0 or 1 which may be incorrect if the first element is negative or less than 1.

We start by setting max_prod to the first element because the maximum product subarray must include at least one element.

2fill in blank

medium

Complete the code to iterate through the array starting from the second element.

DSA Python

for i in [1](1, len(nums)):
    pass

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

Arange

Blist

Cenumerate

Dfilter

Attempts:

3 left

💡 Hint

Common Mistakes

Using list or enumerate which do not directly generate index ranges.

Starting loop from 0 instead of 1.

We use range(1, len(nums)) to loop through the array starting from index 1.

3fill in blank

hard

Fix the error in updating the current maximum product by choosing the correct function to get the maximum.

DSA Python

curr_max = max(nums[i], curr_max * nums[i], [1] * nums[i])

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

Acurr_max

Bcurr_min

Cmax_prod

Dnums[i-1]

Attempts:

3 left

💡 Hint

Common Mistakes

Using curr_max instead of curr_min, missing cases where negative times negative is positive.

We use curr_min because multiplying a negative minimum product by a negative number can become a new maximum.

4fill in blank

hard

Fill both blanks to update current minimum product and maximum product so far.

DSA Python

curr_min = min(nums[i], [1] * nums[i], [2] * nums[i])
max_prod = max(max_prod, curr_max)

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

Acurr_min

Bcurr_max

Cmax_prod

Dnums[i]

Attempts:

3 left

💡 Hint

Common Mistakes

Using max_prod or nums[i] incorrectly in the min function.

Not updating max_prod after curr_max changes.

curr_min updates by considering nums[i], curr_min * nums[i], and curr_max * nums[i]. max_prod updates with the maximum found so far.

5fill in blank

hard

Fill all three blanks to swap current max and min when nums[i] is negative.

DSA Python

if nums[i] < 0:
    [1], [2] = [3], curr_max

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

Acurr_max

Bcurr_min

Attempts:

3 left

💡 Hint

Common Mistakes

Not swapping or swapping incorrectly causing wrong max/min tracking.

When nums[i] is negative, we swap curr_max and curr_min to correctly track max and min products.

Practice

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

easy

A. The subarray with the smallest product of its elements

B. The contiguous subarray with the largest sum of its elements

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

D. The subarray with the largest difference between elements

Solution

1. Step 1: Understand the problem statement

   The problem asks for a contiguous part of the array where the product of all elements is the highest possible.

2. Step 2: Compare options with problem goal

   Only The contiguous subarray with the largest product of its elements matches the goal of finding the maximum product of a contiguous subarray.

3. Final Answer:

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

4. Quick Check:

   Maximum product subarray = largest product contiguous segment [OK]

Hint: Look for largest product, not sum or difference [OK]

Common Mistakes:

• Confusing product with sum
• Ignoring the contiguous subarray condition
• Thinking about smallest product instead

2. Which of the following Python code snippets correctly initializes variables to track the maximum product subarray?

easy

A. max_prod = None; min_prod = None; result = None

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

C. max_prod = float('inf'); min_prod = float('-inf'); result = 0

D. max_prod = nums[0]; min_prod = nums[0]; result = nums[0]

Solution

1. Step 1: Understand initialization for max and min products

   We start with the first element as initial max, min, and result because the subarray must include at least one element.

2. Step 2: Check each option's correctness

   max_prod = nums[0]; min_prod = nums[0]; result = nums[0] correctly initializes all three variables to the first element of the array, which is standard practice.

3. Final Answer:

   max_prod = nums[0]; min_prod = nums[0]; result = nums[0] -> Option D

4. Quick Check:

   Initialize with first element to track products [OK]

Hint: Initialize with first element, not zero or None [OK]

Common Mistakes:

• Initializing with zero causes wrong results if array has negatives
• Using None causes errors in multiplication
• Using infinity values breaks logic

3. What is the output of the following code?

def maxProduct(nums):
    max_prod = min_prod = result = nums[0]
    for n in nums[1:]:
        candidates = (n, max_prod * n, min_prod * n)
        max_prod = max(candidates)
        min_prod = min(candidates)
        result = max(result, max_prod)
    return result

print(maxProduct([2,3,-2,4]))

medium

A. 24

B. 6

C. -2

D. 4

Solution

1. Step 1: Trace the loop with input [2,3,-2,4]

   Start: max_prod = min_prod = result = 2 Iteration 1 (n=3): candidates = (3, 2*3=6, 2*3=6) max_prod=6, min_prod=3, result=6 Iteration 2 (n=-2): candidates = (-2, 6*(-2)=-12, 3*(-2)=-6) max_prod=-2, min_prod=-12, result=6 Iteration 3 (n=4): candidates = (4, -2*4=-8, -12*4=-48) max_prod=4, min_prod=-48, result=6

2. Step 2: Determine final result

   The maximum product found during iterations is 6.

3. Final Answer:

   6 -> Option B

4. Quick Check:

   Maximum product subarray = 6 [OK]

Hint: Track max and min products each step [OK]

Common Mistakes:

• Ignoring min product when negative flips sign
• Only tracking max product causes wrong answer
• Not updating result each iteration

4. Find the error in this code snippet for maximum product subarray:

def maxProduct(nums):
    max_prod = min_prod = result = 0
    for n in nums:
        temp_max = max_prod
        max_prod = max(n, max_prod * n, min_prod * n)
        min_prod = min(n, temp_max * n, min_prod * n)
        result = max(result, max_prod)
    return result

medium

A. Initializing max_prod, min_prod, and result to 0 instead of nums[0]

B. Using temp_max instead of max_prod directly

C. Not iterating from second element

D. Returning result instead of max_prod

Solution

1. Step 1: Check variable initialization

   Initializing max_prod, min_prod, and result to 0 causes errors if all numbers are negative or first element is not zero.

2. Step 2: Compare with correct initialization

   Correct approach initializes these variables to nums[0] to handle negative and positive values properly.

3. Final Answer:

   Initializing max_prod, min_prod, and result to 0 instead of nums[0] -> Option A

4. Quick Check:

   Initialize with first element, not zero [OK]

Hint: Always initialize with first element, not zero [OK]

Common Mistakes:

• Starting max/min product at zero causes wrong results
• Not handling negative numbers properly
• Forgetting to update result each iteration

5. Given the array nums = [1, -2, -3, 0, 7, -8, -2], what is the maximum product of any contiguous subarray?

hard

A. 112

B. 48

C. 7

D. 0

Solution

1. Step 1: Identify subarrays with max product

   Check subarrays: - [1, -2, -3] product = 1 * -2 * -3 = 6 - [7] product = 7 - [7, -8, -2] product = 7 * -8 * -2 = 112 (since -8 * -2 = 16, 7 * 16 = 112) - Others are smaller or zero

2. Step 2: Confirm maximum product

   The largest product found is 112 from subarray [7, -8, -2].

3. Final Answer:

   112 -> Option A

4. Quick Check:

   Max product subarray = 112 [OK]

Hint: Check negative pairs that flip sign to positive [OK]

Common Mistakes:

• Ignoring negative number pairs that increase product
• Stopping at zero without checking after
• Only considering single elements