Given a set of N nuts of different sizes and N bolts of different sizes. There is a one-one mapping between nuts and bolts. Match nuts and bolts efficiently.
Comparison of a nut to another nut or a bolt to another bolt is not allowed. It means nut can only be compared with bolt and bolt can only be compared with nut to see which one is bigger/smaller.
Input:
The first line contains 'T' denoting the number of testcases. Then follows description of T testcases:
Each case begins with a single positive integer N denoting the number of nuts/bolts. Then follows the array of nuts, each element separated by a space. And finally the bolts array, again, each element is separated by a space here. Array of Nuts/Bolts can only consist of the following elements:{'@', '#', '$', '%', '^', '&', '~', '*', '!'}. And no element can be repeated.
Output:
For each test case, output the matched array of nuts and bolts in separate lines, where each element in the array is separated by a space. Print the elements in the following order ! # $ % & * @ ^ ~
Constraints:
1 <= T <= 70
1 <= N <= 9
Example:
Input:
2
5
@ % $ # ^
% @ # $ ^
9
^ & % @ # * $ ~ !
~ # @ % & * $ ^ !
Output:
# $ % @ ^
# $ % @ ^
! # $ % & * @ ^ ~
! # $ % & * @ ^ ~
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lst=['!' ,'#' ,'$' ,'%' ,'&' ,'*', '@', '^' ,'~']
t = int(input())
for _ in range(t):
n = int(input())
nuts = input().split()
bolts = input().split()
nutrst=[]
boltrst=[]
for i in lst:
for j in nuts:
if(j ==i):
nutrst.append(j)
for j in bolts:
if(j ==i):
boltrst.append(j)
print(*nutrst)
print(*boltrst
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