Maximum Product Subarray in DSA C

Mental Model

We want to find the largest product of a continuous part of the array. Because multiplying negatives can flip signs, we keep track of both the biggest and smallest products so far.

Analogy: Imagine walking on a path where stepping on a slippery stone (negative number) can flip your direction. To reach the farthest point (maximum product), you remember both your farthest forward and backward steps because a backward step might help you jump forward later.

Array: [2] -> [-3] -> [4] -> [-1] -> [0] -> [5]
Tracking max_prod and min_prod at each step

Dry Run Walkthrough

Input: array: [2, -3, 4, -1, 0, 5]

Goal: Find the maximum product of any continuous subarray

Step 1: Start with first element: max_prod = 2, min_prod = 2, result = 2

max_prod=2, min_prod=2, result=2

Why: Initialize tracking with first element to start comparisons

Step 2: Process -3: swap max_prod and min_prod because -3 is negative; max_prod = max(-3, 2 * -3) = -3; min_prod = min(-3, 2 * -3) = -6; result = max(2, -3) = 2

max_prod=-3, min_prod=-6, result=2

Why: Negative flips max and min, so swap to track correct values

Step 3: Process 4: max_prod = max(4, -3 * 4) = 4; min_prod = min(4, -6 * 4) = -24; result = max(2, 4) = 4

max_prod=4, min_prod=-24, result=4

Why: Positive number updates max and min normally

Step 4: Process -1: swap max_prod and min_prod; max_prod = max(-1, -24 * -1) = 24; min_prod = min(-1, 4 * -1) = -4; result = max(4, 24) = 24

max_prod=24, min_prod=-4, result=24

Why: Negative flips max and min again, potentially increasing max product

Step 5: Process 0: max_prod = max(0, 24 * 0) = 0; min_prod = min(0, -4 * 0) = 0; result = max(24, 0) = 24

max_prod=0, min_prod=0, result=24

Why: Zero resets product tracking

Step 6: Process 5: max_prod = max(5, 0 * 5) = 5; min_prod = min(5, 0 * 5) = 0; result = max(24, 5) = 24

max_prod=5, min_prod=0, result=24

Why: Positive number updates max product, but result remains max so far

Result:

Maximum product subarray = 24

Annotated Code

DSA C

#include <stdio.h>

int max(int a, int b) {
    return (a > b) ? a : b;
}

int min(int a, int b) {
    return (a < b) ? a : b;
}

int maxProduct(int* nums, int numsSize) {
    if (numsSize == 0) return 0;
    int max_prod = nums[0];
    int min_prod = nums[0];
    int result = nums[0];

    for (int i = 1; i < numsSize; i++) {
        if (nums[i] < 0) {
            int temp = max_prod;
            max_prod = min_prod;
            min_prod = temp;
        }
        max_prod = max(nums[i], max_prod * nums[i]);
        min_prod = min(nums[i], min_prod * nums[i]);
        if (max_prod > result) {
            result = max_prod;
        }
    }
    return result;
}

int main() {
    int nums[] = {2, -3, 4, -1, 0, 5};
    int size = sizeof(nums) / sizeof(nums[0]);
    int max_product = maxProduct(nums, size);
    printf("Maximum product subarray = %d\n", max_product);
    return 0;
}

if (nums[i] < 0) { int temp = max_prod; max_prod = min_prod; min_prod = temp; }

swap max and min product when current number is negative to handle sign flip

max_prod = max(nums[i], max_prod * nums[i]);

update max product ending at current index

min_prod = min(nums[i], min_prod * nums[i]);

update min product ending at current index

if (max_prod > result) { result = max_prod; }

update overall maximum product found so far

Output

Maximum product subarray = 24

Complexity Analysis

Time: O(n) because we scan the array once, updating max and min products at each step

Space: O(1) because we use only a few variables to track products, no extra arrays

vs Alternative: Compared to checking all subarrays (O(n^2)), this approach is much faster and uses constant space

Edge Cases

empty array

returns 0 as no subarray exists

DSA C

if (numsSize == 0) return 0;

array with one element

returns that element as max product

DSA C

int max_prod = nums[0]; int min_prod = nums[0]; int result = nums[0];

array with zeros

resets product tracking to zero, allowing new subarray start

DSA C

max_prod = max(nums[i], max_prod * nums[i]); min_prod = min(nums[i], min_prod * nums[i]);

When to Use This Pattern

When asked for the maximum product of a continuous subarray, track both max and min products because negatives can flip signs and affect the result.

Common Mistakes

Mistake: Not swapping max and min products when encountering a negative number
Fix: Swap max and min before updating them when current number is negative

Mistake: Only tracking max product and ignoring min product
Fix: Track both max and min products to handle negative numbers correctly

Summary

Finds the maximum product of any continuous subarray in an array.

Use when you need the largest product from a sequence of numbers that can include negatives and zeros.

Keep track of both maximum and minimum products at each step because negatives can flip the sign and change the maximum.

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