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def myf(r,c):  
    return r+c

f = np.vectorize(myf)

r_arr = np.meshgrid(range(10000),range(10000))
c_arr = np.meshgrid(range(10000),range(10000))

start = time.time()

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import time
import numpy as np

r_arr = np.meshgrid(range(10000),range(10000))
c_arr = np.meshgrid(range(10000),range(10000))

start = time.time()

z_arr = r_arr + c_arr

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tic
for i = 1:10000
    for j = 1:10000
        z = i +  j;
    end
end
toc

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import time
import numpy as np

start = time.time()

for r in range(10000):
  for c in range(10000):
      z = r + c

print(time.time() - start)

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def mergeLists(head1, head2):
    llist3 = SinglyLinkedList()
    while head1 or head2:
        if head1 == None:
            llist3.insert_node(head2.data)
            head2 = head2.next
        elif head2 == None:
            llist3.insert_node(head1.data)
            head1 = head1.next
        elif head1.data >= head2.data:

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def flippingBits(n):
    result = 0
    for i in range(31,-1,-1):
        if n//(2**i) == 1:
            n = n - 2**i
            x = 0
        else:
            x = 1
        result += 2**i * x

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def minTime(machines, goal): 
    # make a modest guess of what the days may be, and use it as a starting point
    efficiency = [1.0/x for x in machines]
    lower_bound = int(goal / sum(efficiency)) - 1
    upper_bound = lower_bound + max(machines) + 1
    
    while lower_bound < upper_bound -1:
        days = (lower_bound + upper_bound)//2
        produce = sum([days//x for x in machines])
        if produce >= goal:

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def pairs(k, arr):
    result = 0
    arr = sorted(arr)
    j = 1
    for i in range(len(arr)-1):
        while j<len(arr):
            if arr[j] - arr[i] == k:
                result += 1
                j += 1
            elif arr[j] - arr[i] > k:

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# Complete the pairs function below.
def pairs(k, arr):
    my_dict = {}
    result = 0
    for ele in arr:
        my_dict[ele] = 1
        if ele + k in my_dict:
            result += 1
        if ele -k in my_dict:
            result += 1

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def triplets(a, b, c):
    # use set to get unique values, and sort the list
    a = sorted(list(set(a)))
    b = sorted(list(set(b)))
    c = sorted(list(set(c)))
    #initiate the number of options in a and c
    ia, ic = 0, 0
    result = 0
    for ib in range(len(b)):
        while ia < len(a):