Coding Problem 4

-
           Product of elements of Array
                                                          asked by Uber

Ques. Given an array of integers, return a new array such that each element at index i of new array is the product of all the numbers in the original array except that one at i.
For example, if our input was [1, 2, 3, 4, 5] then the expected output would be [120, 60, 40, 30, 24]. If input is [3, 2, 1] then the expected output would be [2, 3, 6].

Solution.
We need to multiply all the elements of array except the one on the index at which we are currently. At index one we will multiply all the elements except the one at index 1. There can be many ways to solve this and different complexities. We can traverse the array two times using nested for loops. Lets see how-

inp = [1, 2, 3, 4, 5]        # input list
out = []                     # output list
for i in range(0, len(inp)):
product = 1
for j in range(0, len(inp)):
if j == i:
continue
product*=inp[j]
out.append(product)
print(out)  

Output: [120, 60, 40, 30, 24]

But the complexity is O(n^2) which is high. 
Well we have another efficient solution. We can take the product of all elements and then at the time of output we just divide the product with the element at current index and we can get desired result. Lets see this solution-

inp = [1, 2, 3, 4, 5]
out = []
product = 1

# taking product of all elements of input
for i in inp:
product*=i

# traverse the list and divide the product with current element.
for i in inp:
temp = product//i
out.append(temp)
print(out)      

Output: [120, 60, 40, 30, 24]

Now in this case we have two for loops which makes complexity of O(2n) which is equal to O(n).
If there is more efficient solution then let me know.

Comments

Post a Comment

Popular posts from this blog

Coding Problem 10

-
Image
                                Minimum Cost Required                                                        asked by Facebook Ques. A builder is looking to build a row of N houses that can be of K different colors. He has a goal of minimizing cost while ensuring that no two neighbouring houses are of same color. Given an N by K matrix where the nth row and kth column represents the cost to build the nth house with kth color return the minimum cost which achieves this goal. Solution.   Suppose we have cost matrix like this - [ [2, 1, 1],   [1, 10, 3],   [1, 2, 100]  ] Each row represents the house and each column represents the color so the values are the cost of painting the color for a house. Since we know that indexes starts from 0 so cost of pain...
Read more

Coding Problem 3

-
             Sum of Numbers in List                                                      asked by Google Ques. Given a list of numbers and a integer K return whether any two numbers from the list will add upto K or not. For example - Given list is [10, 15, 3, 7] and K = 17, then return True because 10 + 7 = 17. Solution.   So basically we have to find the two numbers that are present in the list and sum of both is equal to K. So we can simply run two loops on the list and check each pair of number if they are equal to K or not. This method is Brute-force and complexity is O(n^2). What we can do here is run the loop over the list and then subtract that number from the target and check whether the remaining number is present in the list or not. Lets see the code  def find_numbers(inp_list, K):    ...
Read more

Coding Problem 9

-
                                    Maximum in Subarray                                                          asked by Google Ques. Given an array of integers and a number k, where 1<= k <= length of array, compute the maximum values of each subarray of length k. For example - given array = [10, 5, 2, 7, 8, 7] and k = 3, we should get [10, 7, 8, 8], since  10 = max(10, 5, 2) 7 = max(5, 2, 7) 8 = max(2, 7, 8) 8 = max(7, 8, 7) Solution. This question has a simple solution. In array we will create a window which consist of number of elements equal to k and can find the maximum among them and append that in output list. But while traversing the array we have to keep something very important in mind. We will not traverse from index 0 to nth index...
Read more