Maximum Product Subarray in DSA Javascript - 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 Javascript

let maxProduct = nums[1];

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

A1

B0

Clength

D[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Using nums0 instead of nums[0]

Initializing maxProduct to 0 or 1 which may be incorrect if array has negative numbers

We need to start with the first element of the array, which is nums[0].

2fill in blank

medium

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

DSA Javascript

for (let i = [1]; i < nums.length; i++) {

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

A1

B0

C2

Dnums[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Starting loop at 0 which repeats the first element

Using nums[0] as loop start which is invalid

We start looping from index 1 because index 0 is already used to initialize maxProduct.

3fill in blank

hard

Fix the error in updating the current maximum product when the current number is negative.

DSA Javascript

if (nums[i] < 0) {
  [maxEndingHere, minEndingHere] = [[1], [2]];
}

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

AminEndingHere

BmaxEndingHere

CmaxProduct

Attempts:

3 left

💡 Hint

Common Mistakes

Not swapping the variables

Swapping incorrectly by assigning in wrong order

When the current number is negative, we swap maxEndingHere and minEndingHere to handle sign change.

4fill in blank

hard

Fill both blanks to update maxEndingHere and minEndingHere correctly.

DSA Javascript

maxEndingHere = Math.max(nums[i], nums[i] [1] maxEndingHere, nums[i] [2] minEndingHere);

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

A*

B+

C-

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using + or - instead of *

Using / which is division and incorrect here

We multiply nums[i] with maxEndingHere and minEndingHere to find new max product candidates.

5fill in blank

hard

Fill all three blanks to update minEndingHere and maxProduct correctly.

DSA Javascript

minEndingHere = Math.min(nums[i], nums[i] [1] minEndingHere, nums[i] [2] maxEndingHere);
maxProduct = Math.max(maxProduct, [3]);

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

A*

B+

CmaxEndingHere

DminEndingHere

Attempts:

3 left

💡 Hint

Common Mistakes

Using + instead of *

Updating maxProduct with minEndingHere instead of maxEndingHere

We multiply nums[i] with minEndingHere and maxEndingHere to find new min product candidates, and update maxProduct with maxEndingHere.

Practice

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

easy

A. The subarray with the largest difference between elements

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

C. The subarray with the smallest product of its elements

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

Solution

1. Step 1: Understand the problem goal

   The problem asks for the contiguous subarray (elements next to each other) that has the maximum product when multiplied together.

2. Step 2: Differentiate from sum problems

   This is different from maximum sum subarray problems because it focuses on multiplication, which can be affected by negative numbers and zeros.

3. Final Answer:

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

4. Quick Check:

   Maximum Product Subarray = largest product subarray [OK]

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

Common Mistakes:

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

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

easy

A. let maxProd = Infinity, minProd = -Infinity, result = 0;

B. let maxProd = 0, minProd = 0, result = 0;

C. let maxProd = nums[0], minProd = nums[0], result = nums[0];

D. let maxProd = [], minProd = [], result = [];

Solution

1. Step 1: Understand variable roles

   maxProd and minProd track the max and min products ending at current index; result stores the overall max product.

2. Step 2: Correct initialization

   Initializing all three to the first element of the array ensures correct comparison and handling of negative numbers.

3. Final Answer:

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

4. Quick Check:

   Initialize with first element for correct tracking [OK]

Hint: Start tracking with first element, not zero or arrays [OK]

Common Mistakes:

• Initializing with zero causes wrong results if first element is negative
• Using arrays instead of numbers for tracking
• Using Infinity values incorrectly

3. What will be the output of the following JavaScript code?

const nums = [2, 3, -2, 4];
function maxProduct(nums) {
  let maxProd = nums[0], minProd = nums[0], result = nums[0];
  for (let i = 1; i < nums.length; i++) {
    if (nums[i] < 0) [maxProd, minProd] = [minProd, maxProd];
    maxProd = Math.max(nums[i], maxProd * nums[i]);
    minProd = Math.min(nums[i], minProd * nums[i]);
    result = Math.max(result, maxProd);
  }
  return result;
}
console.log(maxProduct(nums));

medium

A. 4

B. 6

C. -2

D. 8

Solution

1. Step 1: Trace the loop and variable updates

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

2. Step 2: Determine final result

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

3. Final Answer:

   6 -> Option B

4. Quick Check:

   Max product subarray = 6 [OK]

Hint: Swap max and min when negative to track products correctly [OK]

Common Mistakes:

• Not swapping max and min on negative numbers
• Confusing max product with sum
• Incorrectly updating result

4. Identify the bug in the following code for Maximum Product Subarray and select the fix:

function maxProduct(nums) {
  let maxProd = 0, minProd = 0, result = 0;
  for (let i = 0; i < nums.length; i++) {
    if (nums[i] < 0) [maxProd, minProd] = [minProd, maxProd];
    maxProd = Math.max(nums[i], maxProd * nums[i]);
    minProd = Math.min(nums[i], minProd * nums[i]);
    result = Math.max(result, maxProd);
  }
  return result;
}

medium

A. Initialize maxProd, minProd, and result with nums[0] instead of 0

B. Remove the swap of maxProd and minProd when nums[i] < 0

C. Start loop from i = 1 instead of 0

D. Use Math.min instead of Math.max for result update

Solution

1. Step 1: Check variable initialization

   Initializing maxProd, minProd, and result to 0 causes wrong results if nums[0] is negative or non-zero. They should start at nums[0].

2. Step 2: Confirm loop start and swap logic

   Loop starting at 0 is okay if initialization is correct. Swap on negative is needed to handle sign changes.

3. Final Answer:

   Initialize maxProd, minProd, and result with nums[0] instead of 0 -> Option A

4. Quick Check:

   Correct initialization fixes wrong product tracking [OK]

Hint: Always initialize tracking vars with first array element [OK]

Common Mistakes:

• Starting loop at 0 with zero initialization
• Removing swap logic on negative numbers
• Updating result with min instead of max

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: Trace subarrays and products

   Check subarrays:
   - [1] = 1
   - [1,-2] = -2
   - [1,-2,-3] = 6
   - [-2,-3] = 6
   - [7] = 7
   - [7,-8] = -56
   - [7,-8,-2] = 112 (max so far)
   Zero resets product.
   Other subarrays have smaller products.

2. Step 2: Identify max product

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

3. Final Answer:

   112 -> Option A

4. Quick Check:

   Max product subarray = 112 [OK]

Hint: Consider negative pairs and zeros to find max product [OK]

Common Mistakes:

• Ignoring zeros resetting product
• Not considering negative number pairs
• Only checking positive subarrays