Sharing answer codes of mine about HackerRank: Game of Two Stacks.

# HackerRank: Game of Two Stacks (in Data Structures)

## Problem Statement

Alexa has two stacks of non-negative integers, stack \(A=[a_0,a_1,...,a_{n-1}]\) and stack \(B=[b_0,b_1,...,b_{m-1}]\) where index 0 denotes the top of the stack. Alexa challenges Nick to play the following game:

- In each move, Nick can remove one integer from the top of either stack \(A\) or stack \(B\).
- Nick keeps a running sum of the integers he removes from the two stacks.
- Nick is disqualified from the game if, at any point, his running sum becomes greater than some integer \(x\) given at the beginning of the game.
- Nick’s final score is the total number of integers he has removed from the two stacks.

Given \(A, B,\) and \(x\) for \(y\) games, find the maximum possible score Nick can achieve (i.e., the maximum number of integers he can remove without being disqualified) during each game and print it on a new line.

## Answer Code (in Python3)

- Time complexity: \(O(n*m)\)

```
#!/bin/python3
import sys
g = int(input().strip())
for a0 in range(g):
n, m, x = map(int, input().strip().split(' '))
a = list(map(int, input().strip().split(' ')))
b = list(map(int, input().strip().split(' ')))
sum, count, max_count = 0, 0, 0
tempA = []
# Inverse 'a' and 'b' list to use as stack
a.reverse()
b.reverse()
# Pop from stack A and sum until it exceeds the limit
while len(a)!=0:
if sum + a[-1] <= x:
sum += a[-1]
count += 1
tempA.append(a.pop()) # Save pop-ed element from stack A
else:
break
max_count = count # Save current max_count
# Pop from stack B and plus it with 'sum'
while len(b)!=0:
sum += b.pop()
count += 1
# If 'sum' exceeds the limit, discard one from tempA
while sum > x and len(tempA)!=0:
sum -= tempA.pop()
count -= 1
if sum <= x and max_count < count:
max_count = count
elif sum > x:
break
print (max_count)
```