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 BufferedReader br;
private boolean eof;
RW(String inputFile, String outputFile) {
br = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(new OutputStreamWriter(System.out));
File f = new File(inputFile);
if (f.exists() && f.canRead()) {
try {
br = new BufferedReader(new FileReader(inputFile));
out = new PrintWriter(new FileWriter(outputFile));
} catch (IOException e) {
e.printStackTrace();
}
}
}
private String nextLine() {
String s = "";
try {
s = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return s;
}
private String next() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} 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)
View More Similar Problems
Inserting a Node Into a Sorted Doubly Linked List
Given a reference to the head of a doubly-linked list and an integer ,data , create a new DoublyLinkedListNode object having data value data and insert it at the proper location to maintain the sort. Example head refers to the list 1 <-> 2 <-> 4 - > NULL. data = 3 Return a reference to the new list: 1 <-> 2 <-> 4 - > NULL , Function Description Complete the sortedInsert function
View Solution →Reverse a doubly linked list
This challenge is part of a tutorial track by MyCodeSchool Given the pointer to the head node of a doubly linked list, reverse the order of the nodes in place. That is, change the next and prev pointers of the nodes so that the direction of the list is reversed. Return a reference to the head node of the reversed list. Note: The head node might be NULL to indicate that the list is empty.
View Solution →Tree: Preorder Traversal
Complete the preorder function in the editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's preorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the preOrder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the tree's
View Solution →Tree: Postorder Traversal
Complete the postorder function in the editor below. It received 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's postorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the postorder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the
View Solution →Tree: Inorder Traversal
In this challenge, you are required to implement inorder traversal of a tree. Complete the inorder function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values. Input Format Our hidden tester code passes the root node of a binary tree to your $inOrder* func
View Solution →Tree: Height of a Binary Tree
The height of a binary tree is the number of edges between the tree's root and its furthest leaf. For example, the following binary tree is of height : image Function Description Complete the getHeight or height function in the editor. It must return the height of a binary tree as an integer. getHeight or height has the following parameter(s): root: a reference to the root of a binary
View Solution →
