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

Insert a Node at the Tail of a Linked List

You are given the pointer to the head node of a linked list and an integer to add to the list. Create a new node with the given integer. Insert this node at the tail of the linked list and return the head node of the linked list formed after inserting this new node. The given head pointer may be null, meaning that the initial list is empty. Input Format: You have to complete the SinglyLink

View Solution →

Insert a Node at the head of a Linked List

Given a pointer to the head of a linked list, insert a new node before the head. The next value in the new node should point to head and the data value should be replaced with a given value. Return a reference to the new head of the list. The head pointer given may be null meaning that the initial list is empty. Function Description: Complete the function insertNodeAtHead in the editor below

View Solution →

Insert a node at a specific position in a linked list

Given the pointer to the head node of a linked list and an integer to insert at a certain position, create a new node with the given integer as its data attribute, insert this node at the desired position and return the head node. A position of 0 indicates head, a position of 1 indicates one node away from the head and so on. The head pointer given may be null meaning that the initial list is e

View Solution →

Delete a Node

Delete the node at a given position in a linked list and return a reference to the head node. The head is at position 0. The list may be empty after you delete the node. In that case, return a null value. Example: list=0->1->2->3 position=2 After removing the node at position 2, list'= 0->1->-3. Function Description: Complete the deleteNode function in the editor below. deleteNo

View Solution →

Print in Reverse

Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing

View Solution →

Reverse a linked list

Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty. Example: head references the list 1->2->3->Null. Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null. Function Descriptio

View Solution →