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The Maximum Product of Three Numbers of the Array
We have an array of integers including negative numbers. We need to find the greatest product of three numbers of the array.
For example, we have the array int_arr, containing numbers -10, -10, 1, 3, 2. The function that handles the array should return 300 as -10 * -10 * 3 = 300. We need to perform the task as efficiently as possible not forgetting the negative numbers.
There are many methods to solve this problem, but it is not so easy to achieve O(n) running time and O(1) storage costs. To solve the problem effectively we will create and monitor the status of the following variables:
When we go through the array to the end, high_product_3 will contain the answer, and other variables will be used as temporary buffers. High_product_2 and low_product_2 will contain the greatest product of the two and the smallest product of the two numbers respectively, and passing through the array, we will check the product of the current number and these variables (negative current with low_product_2 and the positive current with high_product_2). We need the high and the low to remember the minimum and maximum numbers in the array.
Here is the solution implemented in Python:
def high_product_three(int_arr): # check the amount of the elements if len(int_arr) < 3: raise Exception('There are less than 3 elements in the list') # start from the third element # 2 first elements will be assigned to # high_product_2 and low_product_2. high = max(int_arr, int_arr) low = min(int_arr, int_arr) high_product_2 = int_arr * int_arr low_product_2 = int_arr * int_arr # calculate the high_product_3 using the first three elements high_product_3 = int_arr * int_arr * int_arr # pass through the array from the 2nd element for current in int_arr[2:]: # check the ability to enlarge high_product_3 high_product_3 = max( high_product_3, current * high_product_2, current * low_product_2) # check the same about high_product_2 high_product_2 = max( high_product_2, current * high, current * low) # and low_product_2 low_product_2 = min( low_product_2, current * high, current * low) # is there a new maximum? high = max(high, current) # is there a new minimum? low = min(low, current) return high_product_3
The complexity of the algorithm is O(n) execution time and O(1) memory.
Thanks for your attention!