Maximum Product Subarray in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the maximum product variable.

DSA C

int maxProduct(int* nums, int numsSize) {
    int maxProd = [1];
    return maxProd;
}

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

A1

B0

C-1

Dnums[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing maxProd to 0 or 1 instead of nums[0].

The maximum product should start as the first element of the array to handle negative and positive values correctly.

2fill in blank

medium

Complete the code to update the maximum product during iteration.

DSA C

for (int i = 1; i < numsSize; i++) {
    maxProd = (maxProd > nums[i]) ? maxProd : [1];
}

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Anums[i]

BmaxProd * nums[i]

Cnums[i] * nums[i]

DmaxProd + nums[i]

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication or addition instead of direct comparison.

We compare the current maxProd with the current element nums[i] to update maxProd correctly.

3fill in blank

hard

Fix the error in the code to correctly track the minimum product for negative numbers.

DSA C

int temp = maxProd;
maxProd = (maxProd > 0) ? maxProd * nums[i] : [1];
minProd = (minProd < 0) ? minProd * nums[i] : nums[i];

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Atemp * nums[i]

BminProd * nums[i]

Cnums[i]

DmaxProd * nums[i]

Attempts:

3 left

💡 Hint

Common Mistakes

Using minProd or nums[i] instead of the stored maxProd value.

We use the previous maxProd (stored in temp) multiplied by nums[i] to update maxProd correctly when maxProd is not positive.

4fill in blank

hard

Fill both blanks to update maxProd and minProd correctly when nums[i] is negative.

DSA C

if (nums[i] < 0) {
    int temp = [1];
    [2] = maxProd;
    maxProd = minProd * nums[i];
    minProd = temp * nums[i];
}

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

AmaxProd

BminProd

Cnums[i]

Dtemp

Attempts:

3 left

💡 Hint

Common Mistakes

Not swapping maxProd and minProd correctly when nums[i] is negative.

We swap maxProd and minProd before updating them when nums[i] is negative to handle sign changes correctly.

5fill in blank

hard

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

DSA C

int maxProduct(int* nums, int numsSize) {
    int maxProd = nums[0], minProd = nums[0], result = nums[0];
    for (int i = 1; i < numsSize; i++) {
        if (nums[i] < 0) {
            int temp = [1];
            [2] = maxProd;
            maxProd = minProd * nums[i];
            minProd = temp * nums[i];
        } else {
            maxProd = (maxProd > 0) ? maxProd * nums[i] : nums[i];
            minProd = (minProd < 0) ? minProd * nums[i] : nums[i];
        }
        if (maxProd > result) {
            result = maxProd;
        }
    }
    return [3];
}

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

AmaxProd

BminProd

Cresult

Dnums[i]

Attempts:

3 left

💡 Hint

Common Mistakes

Returning maxProd or minProd instead of result.

Not swapping maxProd and minProd correctly.

The function swaps maxProd and minProd when nums[i] is negative, then updates result with the maximum product found.

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 and differentiate from similar problems

   The problem asks for a contiguous part of the array where the product of all elements is the highest possible. Unlike maximum sum subarray, this problem focuses on product, which can be affected by negative numbers and zeros.

2. Final Answer:

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

3. Quick Check:

   Maximum product subarray = largest product subarray [OK]

Hint: Focus on product, not sum, for this problem [OK]

Common Mistakes:

• Confusing product with sum
• Ignoring negative numbers impact
• Thinking subarray can be non-contiguous

2. Which of the following is the correct way to declare variables to track the maximum and minimum products at each step in C for the Maximum Product Subarray problem?

easy

A. int maxProd = 1, minProd = 1;

B. int maxProd, minProd = 0;

C. int maxProd = 0, minProd = 0;

D. int maxProd = -1, minProd = -1;

Solution

1. Step 1: Initialize maxProd and minProd properly and check alternatives

   Both maxProd and minProd should start at 1 because multiplying by 1 does not change the product. Other initializations use zero (nullifies products), negative values, or leave variables uninitialized leading to incorrect calculations or undefined behavior.

2. Final Answer:

   int maxProd = 1, minProd = 1; -> Option A

3. Quick Check:

   Initialize maxProd and minProd to 1 [OK]

Hint: Start max and min product as 1 to avoid zero multiplication [OK]

Common Mistakes:

• Initializing maxProd or minProd to 0
• Not initializing both variables
• Using negative initial values

3. What is the output of the following C code snippet for the array int arr[] = {2, 3, -2, 4};?

int maxProductSubarray(int* nums, int numsSize) {
    int maxProd = nums[0], minProd = nums[0], result = nums[0];
    for (int i = 1; i < numsSize; i++) {
        if (nums[i] < 0) {
            int temp = maxProd;
            maxProd = minProd;
            minProd = temp;
        }
        maxProd = (nums[i] > maxProd * nums[i]) ? nums[i] : maxProd * nums[i];
        minProd = (nums[i] < minProd * nums[i]) ? nums[i] : minProd * nums[i];
        if (maxProd > result) result = maxProd;
    }
    return result;
}

int main() {
    int arr[] = {2, 3, -2, 4};
    int size = sizeof(arr) / sizeof(arr[0]);
    printf("%d", maxProductSubarray(arr, size));
    return 0;
}

medium

A. 4

B. 6

C. -2

D. 24

Solution

1. Step 1: Trace the loop and determine final result

   Start with maxProd=2, minProd=2, result=2. - i=1 (3): maxProd=max(3,2*3)=6, minProd=min(3,2*3)=3, result=6 - i=2 (-2): swap to maxProd=3, minProd=6; maxProd=max(-2,3*(-2))=-2, minProd=min(-2,6*(-2))=-12, result=6 - i=3 (4): maxProd=max(4,-2*4)=4, minProd=min(4,-12*4)=-48, result=6 The maximum product found is 6 from subarray [2,3].

2. Final Answer:

   6 -> Option B

3. Quick Check:

   Max product subarray = 6 [OK]

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

Common Mistakes:

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

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

int maxProductSubarray(int* nums, int numsSize) {
    int maxProd = 1, minProd = 1, result = nums[0];
    for (int i = 0; i < numsSize; i++) {
        if (nums[i] < 0) {
            int temp = maxProd;
            maxProd = minProd;
            minProd = temp;
        }
        maxProd = (nums[i] > maxProd * nums[i]) ? nums[i] : maxProd * nums[i];
        minProd = (nums[i] < minProd * nums[i]) ? nums[i] : minProd * nums[i];
        if (maxProd > result) result = maxProd;
    }
    return result;
}

medium

A. Initialize maxProd and minProd with nums[0] and start loop from i=1

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

C. Initialize result to 1 instead of nums[0]

D. Change the loop to run from i=1 to numsSize-1 without other changes

Solution

1. Step 1: Check initialization and loop start; analyze other options

   maxProd and minProd should start as nums[0] to correctly handle the first element. Loop should start from i=1 to avoid reprocessing first element. Removing swap breaks handling of negative numbers. Initializing result to 1 is incorrect if nums[0] is negative. Changing loop start without fixing initialization causes errors.

2. Final Answer:

   Initialize maxProd and minProd with nums[0] and start loop from i=1 -> Option A

3. Quick Check:

   Proper initialization and loop start fix bug [OK]

Hint: Initialize with first element, loop from second element [OK]

Common Mistakes:

• Starting loop from 0 with maxProd=1
• Not swapping max and min on negative
• Wrong initialization of result

5. Given the array int arr[] = {-1, -3, -10, 0, 60};, what is the maximum product of any contiguous subarray?

hard

A. 1800

B. 0

C. 60

D. -1

Solution

1. Step 1: Analyze subarrays and their products

   Relevant products: [60]=60, [-3,-10]=30, [-1,-3]=3, [-1,-3,-10]=-30, [-10]=-10, subarrays crossing or including 0=0. No larger contiguous product exists (e.g. [-3,-10,60] skips 0, invalid).

2. Final Answer:

   60 -> Option C

3. Quick Check:

   Max product subarray = 60 [OK]

Hint: Check negative pairs and zeros carefully [OK]

Common Mistakes:

• Thinking non-contiguous subarrays like [-3,-10,60] are valid
• Choosing [-3,-10]=30 instead of 60
• Confusing zero as max product