Maximum Product Subarray in DSA Java - 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.

public int maxProduct(int[] nums) {
    int maxProd = nums[0];
    int minProd = nums[0];
    int result = nums[0];
    for (int i = 1; i < nums.length; i++) {
        int tempMax = maxProd;
        maxProd = Math.max(nums[i], Math.max(maxProd * nums[i], minProd * nums[i]));
        minProd = Math.min(nums[i], Math.min(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 each element of the array once.
• How many times: The loop runs exactly once for each element after the first, so (n - 1) times for an array of size n.

How Execution Grows With Input

As the array size grows, the number of steps grows roughly the same amount, because we check each element once.

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

Pattern observation: The steps increase linearly as the input size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the maximum product grows in direct proportion to the size of the input array.

Common Mistake

[X] Wrong: "Because we check products of subarrays, the time must be quadratic or worse."

[OK] Correct: The code cleverly keeps track of max and min products at each step, so it does not check all subarrays separately, avoiding extra loops.

Interview Connect

Understanding this linear time approach shows you can solve tricky problems efficiently by tracking important values as you go, a skill highly valued in interviews.

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 integer array?

easy

A. The contiguous subarray with the largest sum of its elements

B. The contiguous subarray with the largest product of its elements

C. The longest subarray with all positive numbers

D. The subarray with the smallest product of its elements

Solution

1. Step 1: Understand the problem statement

   The problem asks for the contiguous subarray within an array that has the maximum product of its elements.

2. Step 2: Differentiate from similar problems

   This is different from maximum sum subarray; here we multiply elements, not add.

3. Final Answer:

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

4. Quick Check:

   Maximum Product Subarray = Largest product subarray [OK]

Hint: Look for product, not sum, in contiguous subarrays [OK]

Common Mistakes:

• Confusing product with sum
• Ignoring negative numbers impact
• Assuming non-contiguous subarrays

2. Which of the following Java code snippets correctly initializes variables to track the maximum and minimum products for the Maximum Product Subarray problem?

easy

A. int maxProd = 1, minProd = 1; int result = 0;

B. int maxProd = 0, minProd = 0; int result = 0;

C. int maxProd = Integer.MIN_VALUE, minProd = Integer.MAX_VALUE; int result = 1;

D. int maxProd = 1, minProd = 1; int result = Integer.MIN_VALUE;

Solution

1. Step 1: Identify correct initial values for tracking products

   We start maxProd and minProd at 1 because multiplying by 1 doesn't change the product. Result starts at smallest integer to update correctly.

2. Step 2: Check each option's initialization

   int maxProd = 1, minProd = 1; int result = Integer.MIN_VALUE; correctly initializes maxProd and minProd to 1 and result to Integer.MIN_VALUE for proper comparison.

3. Final Answer:

   int maxProd = 1, minProd = 1; int result = Integer.MIN_VALUE; -> Option D

4. Quick Check:

   Initialize maxProd and minProd to 1, result to min int [OK]

Hint: Initialize max/min product to 1, result to smallest int [OK]

Common Mistakes:

• Initializing max/min product to 0 causes wrong multiplication
• Setting result to 0 misses negative max products
• Using maxProd = Integer.MIN_VALUE breaks multiplication logic

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

int maxProd = 1, minProd = 1, result = Integer.MIN_VALUE;
for (int num : nums) {
    if (num < 0) {
        int temp = maxProd;
        maxProd = minProd;
        minProd = temp;
    }
    maxProd = Math.max(num, maxProd * num);
    minProd = Math.min(num, minProd * num);
    result = Math.max(result, maxProd);
}
System.out.println(result);

medium

A. 4

B. 8

C. 6

D. -2

Solution

1. Step 1: Trace the loop for each element

   Start: maxProd=1, minProd=1, result=Integer.MIN_VALUE - num=2: maxProd=max(2,1*2)=2, minProd=min(2,1*2)=2, result=max(MIN_VALUE,2)=2 - num=3: maxProd=max(3,2*3)=6, minProd=min(3,2*3)=3, result=max(2,6)=6 - num=-2: num<0 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 - num=4: maxProd=max(4,-2*4)=max(4,-8)=4 minProd=min(4,-12*4)=min(4,-48)=-48 result=max(6,4)=6

2. Step 2: Final result after loop

   The maximum product found is 6.

3. Final Answer:

   6 -> Option C

4. Quick Check:

   Max product subarray = 2*3 = 6 [OK]

Hint: Swap max/min on negative, track max product each step [OK]

Common Mistakes:

• Not swapping max and min on negative number
• Confusing max and min updates
• Forgetting to update result each iteration

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

int maxProd = 1, minProd = 1, result = Integer.MIN_VALUE;
for (int num : nums) {
    if (num <= 0) {
        int temp = maxProd;
        maxProd = minProd;
        minProd = temp;
    }
    maxProd = Math.max(num, maxProd * num);
    minProd = Math.min(num, minProd * num);
    result = Math.max(result, maxProd);
}

medium

A. Change condition to num < 0 instead of num <= 0

B. Initialize maxProd and minProd to 0 instead of 1

C. Remove swapping maxProd and minProd

D. Update result only after the loop

Solution

1. Step 1: Analyze the condition for swapping maxProd and minProd

   The swap is needed only when the current number is negative because multiplying by a negative flips max and min.

2. Step 2: Identify why <= 0 causes error

   When num is zero, swapping maxProd and minProd is incorrect and breaks logic. So condition should be strictly < 0.

3. Final Answer:

   Change condition to num < 0 instead of num <= 0 -> Option A

4. Quick Check:

   Swap only on negative numbers [OK]

Hint: Swap max/min only when number is negative, not zero [OK]

Common Mistakes:

• Swapping on zero causes wrong product calculations
• Not swapping at all misses negative impact
• Updating result outside loop misses intermediate max

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

hard

A. 60

B. 1800

C. 0

D. -1

Solution

1. Step 1: Consider subarrays and their products

   Check subarrays: - [-1] = -1 - [-1, -3] = 3 - [-1, -3, -10] = -30 - [-3, -10] = 30 - [-3, -10, 0] = 0 - [0, 60] = 0*60=0 - [60] = 60 - [-1, -3, -10, 0, 60] = 0 - [-3, -10, 0, 60] = 0 - [-10, 0, 60] = 0 - [-1, -3, -10, 0] = 0

2. Step 2: Identify maximum product subarray

   Notice [-3, -10, 0, 60] includes zero, product zero. But [-1, -3, -10] product is -1 * -3 * -10 = -30 (negative). Check [-3, -10] = 30 (positive). Check [-1, -3] = 3. Check [-3, -10, 0, 60] zero product. Check [0, 60] = 0. Check [60] = 60. Check [-1, -3, -10, 0, 60] = 0. But what about [-1, -3, -10]? Actually, the product is -1 * -3 * -10 = -1 * 30 = -30 (negative). So max product is 60. But what about [-1, -3, -10, 0, 60]? Zero breaks subarray, so max product subarray is either before or after zero. So max product is 60.

3. Step 3: Reconsider negative products with even count

   Wait, the product of [-1, -3, -10] is -1 * -3 * -10 = -1 * 30 = -30 (negative). But what about [-1, -3, -10, 0, 60]? Zero breaks subarray. Check [-1, -3, -10] product is negative. Check [-3, -10] product is 30. Check [-1, -3] product is 3. Check [60] product is 60. So max product is 60. But what about [-1, -3, -10, 0, 60]? Zero breaks subarray. So max product is 60.

4. Step 4: Final check with code logic

   Using code logic: - Start maxProd=1, minProd=1, result=Integer.MIN_VALUE - num=-1: num<0 swap maxProd and minProd: maxProd=1, minProd=1 maxProd=max(-1,1*-1)=-1 minProd=min(-1,1*-1)=-1 result=max(MIN_VALUE,-1)=-1 - num=-3: num<0 swap maxProd and minProd: maxProd=-1, minProd=-1 maxProd=max(-3,-1*-3)=max(-3,3)=3 minProd=min(-3,-1*-3)=min(-3,3)=-3 result=max(-1,3)=3 - num=-10: num<0 swap maxProd and minProd: maxProd=-3, minProd=3 maxProd=max(-10,-3*-10)=max(-10,30)=30 minProd=min(-10,3*-10)=min(-10,-30)=-30 result=max(3,30)=30 - num=0: num not <0 no swap maxProd=max(0,30*0)=max(0,0)=0 minProd=min(0,-30*0)=min(0,0)=0 result=max(30,0)=30 - num=60: num not <0 no swap maxProd=max(60,0*60)=max(60,0)=60 minProd=min(60,0*60)=min(60,0)=0 result=max(30,60)=60 Final result is 60.

5. Final Answer:

   60 -> Option A

6. Quick Check:

   Maximum product subarray = 60 [OK]

Hint: Track max/min product carefully through negatives and zeros [OK]

Common Mistakes:

• Assuming product of all negatives is always max
• Ignoring zero breaks subarray
• Not updating max/min product correctly after zero