Maximum Product Subarray in DSA Go - 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.

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
}

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: Single loop traversing the array once.
• How many times: Exactly once for each element after the first.

How Execution Grows With Input

As the array size grows, the number of steps grows directly with it.

Input Size (n)	Approx. Operations
10	About 10 steps
100	About 100 steps
1000	About 1000 steps

Pattern observation: The steps increase in a straight line as input size increases.

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 products of subarrays, the time must be quadratic O(n²)."

[OK] Correct: The code cleverly keeps track of max and min products as it goes, so it only needs one pass through the array, not checking all subarrays separately.

Interview Connect

Understanding this linear time approach shows you can optimize problems that seem to require checking many combinations, a valuable skill in interviews and real projects.

Self-Check

"What if we used a nested loop to check every subarray product instead? 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 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