Maximum Product Subarray in DSA Javascript - Time & Space Complexity

Time Complexity: Maximum Product Subarray

O(n)

Understanding Time Complexity

We want to understand how the time needed to find the maximum product of a subarray changes as the input array grows.

How does the number of steps grow when the array gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function maxProduct(nums) {
  let maxProd = nums[0];
  let minProd = nums[0];
  let result = nums[0];

  for (let i = 1; i < nums.length; i++) {
    let tempMax = maxProd;
    maxProd = Math.max(nums[i], maxProd * nums[i], minProd * nums[i]);
    minProd = Math.min(nums[i], tempMax * nums[i], minProd * nums[i]);
    result = Math.max(result, maxProd);
  }

  return result;
}
    

This code finds the maximum product of any contiguous subarray in the input array.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: A single loop that goes through the array once.
• How many times: The loop runs once for each element after the first, so (n - 1) times if n is the array length.

How Execution Grows With Input

As the array size grows, the number of steps grows roughly the same amount.

Input Size (n)	Approx. Operations
10	About 9 loop steps
100	About 99 loop steps
1000	About 999 loop steps

Pattern observation: The steps increase directly with the input size, so doubling the input roughly doubles the work.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the maximum product grows linearly with the size of the input array.

Common Mistake

[X] Wrong: "Because we check multiple products inside the loop, the time complexity is more than linear, maybe quadratic."

[OK] Correct: Inside the loop, the number of operations is fixed and does not depend on the input size, so the overall time still grows linearly.

Interview Connect

Understanding this linear time solution shows you can handle problems that need careful tracking of multiple values while keeping efficiency.

Self-Check

"What if we used nested loops to check every possible subarray product? How would the time complexity change?"

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