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Lecture 3 : Hard problems on array

Maximum Product Subarray in an Array

Nums = [1,2,3,4,5,0]
Output: 120
Explaination : 1×2×3×4×5

Nums = [1,2,-3,0,-4,-5]
Output : 20
Explaination : -5 x -4

Brute Force

• multiply all continuous combinations
• result = max(all)

def max_product(nums):
    n = len(nums)
    result = nums[0]

    for i in range(n):
        product =1
        for j in range(i+1,n):
            product *= nums[j] 
            result = max(result,product)

    return result

print(max_product([1,2,3,4,5,0]))
print(max_product([1,2,-3,0,-4,-5]))

120
20

Complexity

• O(N^2) : N = len(nums)
• O(1)

Two pass product

• Canculate running product in forward pass and backward pass
• Keep track of 0 ,
• max(prod)

def max_product(nums):
    n = len(nums)
    result = nums[0]

    # forward pass
    prod = 1
    for i in range(n):
        prod *= nums[i]
        result = max(result,prod)

        if nums[i] ==0:
            prod = 1
            result = max(result,0)

    # backward pass
    prod = 1
    for i in range(n-1,-1,-1):
        prod *= nums[i]
        result = max(result,prod)

        if nums[i] ==0:
            prod = 1
            result = max(result,0)

    return result

print(max_product([1,2,3,4,5,0]))
print(max_product([1,2,-3,0,-4,-5]))

120
20

Complexity

• O(N) : N = len(nums)

• O(1)

Approach 3 (Kadane's Algorithm) MOST CLEANEST

• Track min and max prod ending at each position because a negative can turn a minimum to maximum

def max_product(nums):
    result = nums[0]
    max_ending_here = nums[0]
    min_ending_here = nums[0]

    for num in nums[1:]:
        max_p,min_p = max_ending_here,min_ending_here
        max_ending_here = max(num,num*max_p,num* min_p)
        min_ending_here = min(num,num*max_p,num* min_p)
        result = max(result,max_ending_here)

    return result

print(max_product([1,2,3,4,5,0]))
print(max_product([1,2,-3,0,-4,-5]))
print(max_product([-4,-3,-2]))

120
20
12