# Find the minimum time after which one can exchange notes in Python

Suppose there are n number of cashiers exchanging the money, at the moment, the i-th cashier had ki number of people in front of him/her. Now, the j-th person in the line to i-th cashier had m[i,j] notes. We have to find how early can one exchange his/her notes. We have to keep in mind that the cashier spent 5 seconds to scan a single note.After completing scanning of every note for the customer, he/she took 15 seconds to exchange the notes.

So, if the input is like Input : n = 6, k = [12, 12, 12, 12, 12, 12]

 7 8 9 7 9 6 10 9 9 6 7 8 10 7 10 9 8 9 9 9 9 6 5 6 9 8 8 9 8 6 7 9 10 6 6 7 7 6 9 6 6 9 8 9 6 6 8 9 9 8 7 6 5 10 8 10 7 6 6 8 8 7 6 5 7 9 7 9 6 5 5 7

then the output will be 585, as the cashier needs 5 secs for scanning every note of each customer, so add 5*m[I,j]. Now each cashier takes 15 seconds for every customer, so add 15*k[] to the answer. The minimum time taken after computing the time taken by each cashier will be the answer. So cashier m[5] takes the minimum time 585.

To solve this, we will follow these steps −

• n := size of k

• minimum := 99999

• for i in range 0 to n, do

• temp := k[i] * 15

• for j in range 0 to k[i], do

• temp := temp + m[i, j] * 5

• if temp < minimum, then

• minimum := temp

• return minimum

## Example

Let us see the following implementation to get better understanding −

Live Demo

def minTimeToExchange(k, m):
n = len(k)
minimum = 99999
for i in range(n):
temp = k[i] * 15
for j in range(k[i]):
temp += m[i][j] * 5
if temp < minimum:
minimum = temp
return minimum

k = [12, 12, 12, 12, 12, 12]
m = [
[7,8,9,7,9,6,10,9,9,6,7,8],
[10,7,10,9,8,9,9,9,9,6,5,6],
[9,8,8,9,8,6,7,9,10,6,6,7],
[7,6,9,6,6,9,8,9,6,6,8,9],
[9,8,7,6,5,10,8,10,7,6,6,8],
[8,7,6,5,7,9,7,9,6,5,5,7]]
print(minTimeToExchange(k, m))

## Input

[12, 12, 12, 12, 12, 12],
[[7,8,9,7,9,6,10,9,9,6,7,8],
[10,7,10,9,8,9,9,9,9,6,5,6],
[9,8,8,9,8,6,7,9,10,6,6,7],
[7,6,9,6,6,9,8,9,6,6,8,9],
[9,8,7,6,5,10,8,10,7,6,6,8],
[8,7,6,5,7,9,7,9,6,5,5,7]]

## Output

585