Maximum Product Subarray in DSA Go - 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 Go

maxProduct := nums[1]

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

A<0>

B(0)

C{0}

D[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses instead of square brackets for indexing.

Using curly braces which are invalid for indexing.

We use nums[0] to get the first element of the slice in Go.

2fill in blank

medium

Complete the code to update the maximum product if the current product is greater.

DSA Go

if currMax > [1] {
    maxProduct = currMax
}

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

AmaxProduct

BcurrMin

Cnums[i]

DprevMax

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing currMax with currMin instead of maxProduct.

Using an undefined variable like prevMax.

We compare currMax with maxProduct to update the maximum product found so far.

3fill in blank

hard

Fix the error in swapping currMax and currMin when nums[i] is negative.

DSA Go

if nums[i] < 0 {
    [1] = currMax
    currMax = currMin
    currMin = temp
}

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

AcurrMax

BcurrMin

Ctemp

Dnums[i]

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping without using a temporary variable causes incorrect values.

Assigning currMin to temp instead of currMax.

We need to store currMax in a temporary variable before swapping.

4fill in blank

hard

Fill both blanks to update currMax and currMin with the maximum and minimum of nums[i] and their products.

DSA Go

currMax = max(nums[i], currMax[1]nums[i])
currMin = min(nums[i], currMin[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 nums[i] with currMax and currMin to update the products.

5fill in blank

hard

Fill all three blanks to complete the function that returns the maximum product subarray.

DSA Go

func maxProduct(nums []int) int {
    maxProduct := nums[1]
    currMax, currMin := nums[0], nums[0]
    for i := 1; i < len(nums); i++ {
        if nums[i] < 0 {
            [2] = currMax
            currMax = currMin
            currMin = [3]
        }
        currMax = max(nums[i], currMax*nums[i])
        currMin = min(nums[i], currMin*nums[i])
        if currMax > maxProduct {
            maxProduct = currMax
        }
    }
    return maxProduct
}

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

A[0]

Btemp

D(0)

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses instead of square brackets for indexing.

Not using a temporary variable for swapping.

We initialize maxProduct with nums[0], use temp to hold currMax before swapping, and assign currMin to temp during swap.

Practice

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

easy

A. The longest subarray with positive numbers

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

C. The maximum single element in the array

D. The contiguous subarray with the largest product of its 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: Differentiate from similar problems

   Unlike maximum sum subarray, here we multiply elements, so negative numbers and zeros affect the product differently.

3. Final Answer:

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

4. Quick Check:

   Maximum product subarray = contiguous subarray with max product [OK]

Hint: Look for contiguous elements with max multiplication [OK]

Common Mistakes:

• Confusing product with sum
• Ignoring negative numbers impact
• Thinking about non-contiguous elements

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

easy

A. maxProd, minProd, result := 1, 1, 1

B. maxProd, minProd, result := 0, 0, 0

C. maxProd, minProd, result := nums[0], nums[0], nums[0]

D. maxProd, minProd, result := -1, -1, -1

Solution

1. Step 1: Understand initialization for max product subarray

   We start with the first element as initial max, min, and result to handle negative and positive values correctly.

2. Step 2: Check each option's initialization

   maxProd, minProd, result := nums[0], nums[0], nums[0] uses nums[0] which is correct; others use fixed values that may not match array's first element.

3. Final Answer:

   maxProd, minProd, result := nums[0], nums[0], nums[0] -> Option C

4. Quick Check:

   Initialize with first element for correct tracking [OK]

Hint: Initialize with first element, not zero or one [OK]

Common Mistakes:

• Initializing with zero causes wrong results if nums[0] is negative
• Using 1 ignores negative or zero first elements
• Not initializing minProd separately

3. What is the output of the following Go code for the input array nums := []int{2, 3, -2, 4}?

func maxProduct(nums []int) int {
    maxProd, minProd, result := nums[0], nums[0], nums[0]
    for i := 1; i < len(nums); i++ {
        if nums[i] < 0 {
            maxProd, minProd = minProd, maxProd
        }
        maxProd = max(nums[i], maxProd*nums[i])
        minProd = min(nums[i], minProd*nums[i])
        if maxProd > result {
            result = maxProd
        }
    }
    return result
}

func max(a, b int) int {
    if a > b {
        return a
    }
    return b
}

func min(a, b int) int {
    if a < b {
        return a
    }
    return b
}

medium

A. 6

B. 4

C. -2

D. 8

Solution

1. Step 1: Trace the loop and update maxProd, minProd, and result

   Start with maxProd=2, minProd=2, result=2 - i=1 (3): nums[i]=3 > 0, maxProd=max(3,2*3)=6, minProd=min(3,2*3)=3, result=6 - i=2 (-2): nums[i]<0, swap maxProd and minProd (maxProd=3, minProd=6), maxProd=max(-2,3*(-2))=-2, minProd=min(-2,6*(-2))=-12, result=6 - i=3 (4): nums[i]>0, maxProd=max(4,-2*4)=4, minProd=min(4,-12*4)=-48, result=6

2. Step 2: Determine final result

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

3. Final Answer:

   6 -> Option A

4. Quick Check:

   Max product subarray = 6 [OK]

Hint: Swap max and min when negative number found [OK]

Common Mistakes:

• Not swapping max and min on negative number
• Confusing max and min updates
• Ignoring update of result

4. Identify the bug in this Go code snippet for maximum product subarray:

func maxProduct(nums []int) int {
    maxProd, minProd, result := nums[0], nums[0], nums[0]
    for i := 1; i < len(nums); i++ {
        if nums[i] > 0 {
            maxProd = max(maxProd*nums[i], nums[i])
            minProd = min(minProd*nums[i], nums[i])
        } else {
            maxProd = max(minProd*nums[i], nums[i])
            minProd = min(maxProd*nums[i], nums[i])
        }
        if maxProd > result {
            result = maxProd
        }
    }
    return result
}

medium

A. Swapping maxProd and minProd is missing when nums[i] is negative

B. Initialization of maxProd, minProd, result is incorrect

C. Loop should start from i = 0 instead of 1

D. max and min functions are not defined

Solution

1. Step 1: Analyze handling of negative numbers

   When nums[i] is negative, maxProd and minProd should be swapped before updating because multiplying by negative flips max and min.

2. Step 2: Check the code's else block

   The code updates maxProd and minProd without swapping, causing incorrect results for negative numbers.

3. Final Answer:

   Swapping maxProd and minProd is missing when nums[i] is negative -> Option A

4. Quick Check:

   Swap max and min on negative input [OK]

Hint: Always swap max and min before multiplying negative number [OK]

Common Mistakes:

• Forgetting to swap maxProd and minProd
• Updating maxProd before minProd causing wrong minProd
• Starting loop at wrong index

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

hard

A. Subarray [-1, -3, -10] with product -30

B. Subarray [60] with product 60

C. Subarray [-3, -10, 0, 60] with product 0

D. Subarray [-10, 0, 60] with product 0

Solution

1. Step 1: Evaluate products of possible subarrays

   Check subarrays: - [-1, -3, -10]: product = -1 * -3 * -10 = -30 - [60]: product = 60 - [-3, -10, 0, 60]: product = 0 (due to zero) - [-10, 0, 60]: product = 0 (due to zero)

2. Step 2: Identify maximum product

   The highest product is 60 from subarray [60].

3. Final Answer:

   Subarray [60] with product 60 -> Option B

4. Quick Check:

   Max product = 60 from single element [OK]

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

Common Mistakes:

• Multiplying negative numbers without counting sign flips
• Ignoring zero resets product
• Assuming longer subarray always has max product