add output for max_product algorithm

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Vitor Oliveira 2021-09-03 12:35:10 -07:00 committed by GitHub
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#Given an integer array nums, find a contiguous non-empty subarray within the array that has the largest product, and return the product.
#It is guaranteed that the answer will fit in a 32-bit integer.
#A subarray is a contiguous subsequence of the array.
#Example 1:
#Input: nums = [2,3,-2,4]
#Output: 6
#Explanation: [2,3] has the largest product 6.
# Given an integer array nums, find a contiguous non-empty subarray within the array that has the largest product, and return the product.
# It is guaranteed that the answer will fit in a 32-bit integer.
# A subarray is a contiguous subsequence of the array.
#Example 2:
#Input: nums = [-2,0,-1]
#Output: 0
#Explanation: The result cannot be 2, because [-2,-1] is not a subarray.
# Example 1:
# Input: nums = [2,3,-2,4]
# Output: 6
# Explanation: [2,3] has the largest product 6.
#Constraints:
#1 <= nums.length <= 2 * 104
# Example 2:
# Input: nums = [-2,0,-1]
# Output: 0
# Explanation: The result cannot be 2, because [-2,-1] is not a subarray.
# Constraints:
# 1 <= nums.length <= 2 * 104
#-10 <= nums[i] <= 10
#The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.
#Dynamic Programming Approach (Kadane's Algorithm) - O(n) Time / O(1) Space
#Track both current minimum and current maximum (Due to possibility of multiple negative numbers)
#Answer is the highest value of current maximum
# The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.
# Dynamic Programming Approach (Kadane's Algorithm) - O(n) Time / O(1) Space
# Track both current minimum and current maximum (Due to possibility of multiple negative numbers)
# Answer is the highest value of current maximum
# @param {Integer[]} nums
# @return {Integer}
def max_product(nums)
return nums[0] if nums.length == 1
cur_min, cur_max, max = 1, 1, -11
nums.each do |val|
tmp_cur_max = cur_max
cur_max = [val, val*cur_max, val*cur_min].max
cur_min = [val, val*tmp_cur_max, val*cur_min].min
max = [max, cur_max].max
end
max
end
return nums[0] if nums.length == 1
cur_min = 1
cur_max = 1
max = -11
nums.each do |val|
tmp_cur_max = cur_max
cur_max = [val, val * cur_max, val * cur_min].max
cur_min = [val, val * tmp_cur_max, val * cur_min].min
max = [max, cur_max].max
end
max
end
nums = [2, 3, -2, 4]
puts max_product(nums)
# Output: 6
nums = [-2, 0, -1]
puts max_product(nums)
# Output: 0