Maximum Product Subarray in DSA Go - Build from Scratch

Maximum Product Subarray

📖 Scenario: Imagine you are analyzing daily sales growth rates of a product over a week. Each day's growth rate can be positive (growth), negative (decline), or zero (no change). You want to find the period (contiguous days) where the product of growth rates is the highest, indicating the best continuous growth streak.

🎯 Goal: Build a Go program that finds the maximum product of any contiguous subarray within a given list of integers representing daily growth rates.

📋 What You'll Learn

Create a slice of integers named growthRates with the exact values: 2, 3, -2, 4

Create two integer variables maxProduct and minProduct initialized to the first element of growthRates

Use a for loop with index i starting from 1 to iterate over growthRates

Inside the loop, update maxProduct and minProduct to track the maximum and minimum products ending at current index

Create an integer variable result initialized to the first element of growthRates to store the maximum product found

Print the value of result at the end

💡 Why This Matters

🌍 Real World

Finding the maximum product subarray helps in analyzing periods of best continuous growth or decline in financial data, sales, or stock prices.

💼 Career

This algorithm is useful for software engineers working in data analysis, finance, or any field requiring time series data processing and optimization.

1

Create the growth rates slice

Create a slice of integers called growthRates with these exact values: 2, 3, -2, 4

DSA Go

# Create the growthRates slice with the values 2, 3, -2, 4
# Your code here

Hint

Use growthRates := []int{2, 3, -2, 4} to create the slice.

2

Initialize maxProduct, minProduct, and result

Create three integer variables maxProduct, minProduct, and result all initialized to the first element of growthRates

DSA Go

package main

func main() {
    growthRates := []int{2, 3, -2, 4}
    // Initialize maxProduct, minProduct, and result to growthRates[0]
    // Your code here
}

Hint

Use maxProduct := growthRates[0], minProduct := growthRates[0], and result := growthRates[0].

3

Calculate maximum product subarray

Use a for loop with index i starting from 1 to iterate over growthRates. Inside the loop, create a temporary variable tempMax to store the current maxProduct. Update maxProduct to the maximum of growthRates[i], maxProduct * growthRates[i], and minProduct * growthRates[i]. Update minProduct to the minimum of growthRates[i], tempMax * growthRates[i], and minProduct * growthRates[i]. Update result to the maximum of result and maxProduct.

DSA Go

package main

func main() {
    growthRates := []int{2, 3, -2, 4}
    maxProduct := growthRates[0]
    minProduct := growthRates[0]
    result := growthRates[0]
    // Use a for loop starting from i=1 to update maxProduct, minProduct, and result
    // Your code here
}

Hint

Remember to create helper functions max and min to compare two integers.

4

Print the maximum product

Print the value of result using fmt.Println(result)

DSA Go

package main

import "fmt"

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
}

func main() {
    growthRates := []int{2, 3, -2, 4}
    maxProduct := growthRates[0]
    minProduct := growthRates[0]
    result := growthRates[0]

    for i := 1; i < len(growthRates); i++ {
        tempMax := maxProduct
        maxProduct = max(growthRates[i], max(maxProduct*growthRates[i], minProduct*growthRates[i]))
        minProduct = min(growthRates[i], min(tempMax*growthRates[i], minProduct*growthRates[i]))
        result = max(result, maxProduct)
    }
    // Print the result
    // Your code here
}

Hint

Use fmt.Println(result) to print the maximum product.

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