# Game of Two Stacks

### Problem Statement :

```Alexa has two stacks of non-negative integers, stack A = [a0, a1, . . . , an-1 ] and stack B  = [b0, b1, . . . , 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  xgiven 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 g 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.

Output Format

For each of the g games, print an integer on a new line denoting the maximum possible score Nick can achieve without being disqualified.```

### Solution :

```                            ```Solution in C :

In C ++ :

// author: gary
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;

#define ALL(x) (x).begin(), x.end()

const int inf = 1e9;
const int maxn = 1e5 + 10;

ll a[maxn];
ll b[maxn];
int T, n, m, x;

int main() {
scanf("%d", &T);
while(T--) {
scanf("%d%d%d", &n, &m, &x);
for(int i = 1; i <= n; i++) scanf("%lld", a + i);
for(int i = 1; i <= m; i++) scanf("%lld", b + i);

for(int i = 1; i <= n; i++) a[i] += a[i-1];
for(int i = 1; i <= m; i++) b[i] += b[i-1];

int res = 0;
for(int i = 0, j = m; i <= n && a[i] <= x; i++) {
while(a[i] + b[j] > x && j >= 1)
j --;
res = max(res, i + j);
}
printf("%d\n", res);
}
return 0;
}

In Java :

import java.io.*;
import java.util.StringTokenizer;

public class Main {
private void solve() {
int t = rw.nextInt();
for (int t_id = 0; t_id < t; ++t_id) {
int n = rw.nextInt();
int m = rw.nextInt();
int x = rw.nextInt();
int[] a = new int[n];
int[] b = new int[m];
for (int i = 0; i < n; ++i)
a[i] = rw.nextInt();

for (int i = 0; i < m; ++i)
b[i] = rw.nextInt();

int sum = 0;
int c1 = 0;
for (; c1 < n; ) {
if (sum + a[c1] <= x) {
sum += a[c1];
++c1;
} else
break;
}
int c2 = 0;
for (; c2 < m; ) {
if (sum + b[c2] <= x) {
sum += b[c2];
++c2;
} else
break;
}
int ans = c1 + c2;
for (; c1 > 0; --c1) {
sum -= a[c1 - 1];
while (c2 < m) {
if (sum + b[c2] <= x) {
sum += b[c2];
++c2;
} else
break;
}
ans = Math.max(ans, c1 - 1 + c2);
}

rw.println(ans);
}

}

private RW rw;
private String FILE_NAME = "file";

public static void main(String[] args) {
new Main().run();
}

private void run() {
rw = new RW(FILE_NAME + ".in", FILE_NAME + ".out");
solve();
rw.close();
}

private class RW {
private StringTokenizer st;
private PrintWriter out;
private boolean eof;

RW(String inputFile, String outputFile) {
out = new PrintWriter(new OutputStreamWriter(System.out));

File f = new File(inputFile);
try {
out = new PrintWriter(new FileWriter(outputFile));
} catch (IOException e) {
e.printStackTrace();
}
}
}

private String nextLine() {
String s = "";
try {
} catch (IOException e) {
e.printStackTrace();
}
return s;
}

private String next() {
while (st == null || !st.hasMoreTokens()) {
try {
} catch (IOException e) {
eof = true;
return "-1";
}
}
return st.nextToken();
}

private long nextLong() {
return Long.parseLong(next());
}

private int nextInt() {
return Integer.parseInt(next());
}

private void println() {
out.println();
}

private void println(Object o) {
out.println(o);
}

private void print(Object o) {
out.print(o);
}

private void close() {
try {
br.close();
} catch (IOException e) {
e.printStackTrace();
}
out.close();
}
}
}

In C :

#include <stdio.h>

int main() {
long long int c, i, g, m, n, x, la, lb;
long long int a[100010], b[100010];
scanf("%lld", &g);
while (g--) {
la = lb = 0;
scanf("%lld%lld%lld", &n, &m, &x);
scanf("%lld", &a[1]);
for (i = 1; ++i <= n;) {
scanf("%lld", &a[i]);
a[i] += a[i - 1];
}
scanf("%lld", &b[1]);
for (i = 1; ++i <= m;) {
scanf("%lld", &b[i]);
b[i] += b[i - 1];
}
la = 1;
while (la <= n && a[la] <= x)
la++;
la--;
c = la;
lb = 1;
while (lb <= m && b[lb] <= x) {
if (la && b[lb] + a[la] > x)
la--;
else {
if (c < la + lb)
c = la + lb;
lb++;
}
}
printf("%lld\n", c);
}
return 0;
}

In Python3 :

#!/bin/python3

import sys

g = int(input().strip())
for a0 in range(g):
n,m,x = input().strip().split(' ')
n,m,x = [int(n),int(m),int(x)]
a = list(map(int, input().strip().split(' ')))
b = list(map(int, input().strip().split(' ')))

i = 0
while i < len(a) and x >= a[i]:
x -= a[i]
i += 1

ans = i
j = 0
for p in b:
j += 1
x -= p

while x < 0 and i > 0:
i -= 1
x += a[i]

if x >= 0: ans = max(ans, j + i)

print(ans)```
```

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