Skip to content
Related Articles
Open in App
Not now

Related Articles

Merge transactions in bank sheets in the order of their occurrence such that their sum remains positive

Improve Article
Save Article
  • Difficulty Level : Hard
  • Last Updated : 01 Feb, 2023
Improve Article
Save Article

Given an array arr[][] consisting of N lists representing N transactions, the task is to merge the given lists of transactions in the order of their occurrences, such that at any point of time, the sum of already performed transactions is non-negative. If found to negative, then print “-1”. Otherwise, print the merged list of transactions.

Examples:

Input: arr[][] = {{100 → 400 → -1000 → -500}, {-300 → 2000 → -500}}
Output: 100 → 400 → -300 → 2000 → -500 → -1000 → -500
Explanation: The sum at every instant of the above list of transactions is given by {100, 500, 200, 2200, 1700, 700, 200}, which has no negative values.

Input: arr[][] = [[100 → 400]]
Output: 100 400

Approach: The given problem can be visualized as a variation of merge K-sorted linked lists with criteria that the sum of the merged list of transactions at any instant should be non-negative. 
Follow the steps below to solve the problem:

Below is the implementation of the above approach:-

C++




// C++ code for the above approach
#include <iostream>
#include <queue>
#include <vector>
using namespace std;
 
// Structure of a Node in the linked list
struct Node {
  int val;
  Node* next;
  Node(int val)
    : val(val)
      , next(NULL)
    {
    }
};
 
struct Compare {
  bool operator()(Node* a, Node* b)
  {
    return a->val < b->val;
  }
};
 
// Function to merge the Bank sheets
void mergeSheets(vector<Node*>& lists)
{
  // Initialize Max_Heap
  priority_queue<Node*, vector<Node*>, Compare> pq;
 
  // Insert the first element of each list
  for (int i = 0; i < lists.size(); i++)
  {
 
    // If list is not NULL
    if (lists[i])
    {
 
      // Insert element in the priority queue
      pq.push(lists[i]);
    }
  }
 
  // Stores the output list
  Node* head = new Node(0);
  Node* curr = head;
 
  // Iterate until PQ is non-empty
  while (!pq.empty()) {
    auto t = pq.top();
    pq.pop();
    curr->next = t;
    curr = curr->next;
    if (curr->next) {
      pq.push(curr->next);
    }
  }
  curr = head->next;
 
  // Print the output list
  while (curr->next) {
    cout << curr->val << " ";
    curr = curr->next;
  }
  cout << curr->val << endl;
}
 
int main()
{
 
  int N = 2;
  vector<Node*> lists(N);
  lists[0] = new Node(100);
  lists[0]->next = new Node(400);
  lists[0]->next->next = new Node(-1000);
  lists[0]->next->next->next = new Node(-500);
 
  lists[1] = new Node(-300);
  lists[1]->next = new Node(2000);
  lists[1]->next->next = new Node(-500);
 
  // Function call
  mergeSheets(lists);
  return 0;
}

Java




// Java program for the above approach
 
import java.util.*;
 
// Structure of a Node
// in the Linked List
class Node {
 
    int val;
    Node next;
 
    // Constructor
    Node(int val)
    {
        this.val = val;
        this.next = null;
    }
}
 
class GFG {
 
    // Function to merge the Bank sheets
    public static void mergeSheets(
        Node lists[])
    {
        // Initialize Max_Heap
        PriorityQueue<Node> pq
 
            = new PriorityQueue<>(
                new Comparator<Node>() {
 
                    // Comparator Function
                    // to make it maxHeap
                    public int compare(Node a, Node b)
                    {
                        return b.val - a.val;
                    }
 
                });
 
        // Stores the output list
        Node p, head = new Node(0);
        p = head;
 
        // Insert the first element
        // of each list
        for (int i = 0;
             i < lists.length; i++) {
 
            // If the list is not NULL
            if (lists[i] != null) {
 
                // Insert element in
                // the priority queue
                pq.add(lists[i]);
            }
        }
 
        // Iterate until PQ is non-empty
        while (!pq.isEmpty()) {
            p.next = pq.poll();
            p = p.next;
 
            if (p.next != null)
                pq.add(p.next);
        }
 
        p = head.next;
 
        // Print the output list
        while (p.next != null) {
            System.out.print(p.val + " ");
            p = p.next;
        }
 
        System.out.print(p.val);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int N = 2;
 
        Node arr[] = new Node[N];
 
        arr[0] = new Node(100);
        arr[0].next = new Node(400);
        arr[0].next.next = new Node(-1000);
        arr[0].next.next.next = new Node(-500);
 
        arr[1] = new Node(-300);
        arr[1].next = new Node(2000);
        arr[1].next.next = new Node(-500);
 
        // Function Call
        mergeSheets(arr);
    }
}

Python3




# Python code for the above approach
import heapq
 
# Structure of a Node in the Linked List
class Node:
    def __init__(self, val):
        self.val = val
        self.next = None
# Function to merge the Bank sheets
 
 
def mergeSheets(lists):
   
    # Initialize Max_Heap
    pq = []
     
    # Insert the first element  of each list
    for i in range(len(lists)):
       
      # If list is not NULL
        if lists[i] != None:
           
          # Insert element in the priority queue
            heapq.heappush(pq, (-lists[i].val, lists[i]))
 
    # Stores the output list
    p = Node(0)
    head = p
     
# Iterate until PQ is non-empty
    while len(pq) > 0:
        p.next = heapq.heappop(pq)[1]
        p = p.next
        if p.next != None:
            heapq.heappush(pq, (-p.next.val, p.next))
 
    p = head.next
 
    # Print the output list
    while p.next != None:
        print(p.val, end=" ")
        p = p.next
    print(p.val)
 
 
# Driver code
N = 2
lists = [None]*N
lists[0] = Node(100)
lists[0].next = Node(400)
lists[0].next.next = Node(-1000)
lists[0].next.next.next = Node(-500)
 
lists[1] = Node(-300)
lists[1].next = Node(2000)
lists[1].next.next = Node(-500)
 
# Function Call
mergeSheets(lists)
 
# This code is contributed by lokeshpotta20.

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class Node
{
    public int val;
    public Node next;
 
    public Node(int val)
    {
        this.val = val;
        this.next = null;
    }
}
 
class Program
{
    public static Node MergeSheets(List<Node> lists)
    {
        // Initialize Max_Heap
        var pq = new List<Node>();
 
        // Insert the first element of each list
        for (int i = 0; i < lists.Count; i++)
        {
            // If list is not NULL
            if (lists[i] != null)
            {
                // Insert element in the priority queue
                pq.Add(lists[i]);
            }
        }
 
        // Sort the pq
        pq.Sort((a, b) => b.val.CompareTo(a.val));
 
        // Stores the output list
        Node head = new Node(0);
        Node p = head;
 
        // Iterate until PQ is non-empty
        while (pq.Count > 0)
        {
            p.next = pq[0];
            pq.RemoveAt(0);
            p = p.next;
            if (p.next != null)
            {
                pq.Add(p.next);
                pq.Sort((a, b) => b.val.CompareTo(a.val));
            }
        }
 
        return head.next;
    }
 
    public static void Main()
    {
        int N = 2;
        var lists = new List<Node>();
        lists.Add(null);
        lists.Add(null);
 
        lists[0] = new Node(100);
        lists[0].next = new Node(400);
        lists[0].next.next = new Node(-1000);
        lists[0].next.next.next = new Node(-500);
 
        lists[1] = new Node(-300);
        lists[1].next = new Node(2000);
        lists[1].next.next = new Node(-500);
 
        // Function Call
        Node p = MergeSheets(lists);
 
        // Print the output list
        while (p.next != null)
        {
            Console.Write(p.val + " ");
            p = p.next;
        }
        Console.Write(p.val);
    }
}
 
//This code is contributed by shivamsharma215

Output: 

100 400 -300 2000 -500 -1000 -500

 

Time Complexity: O(N * log K)
Auxiliary Space: O(K)


My Personal Notes arrow_drop_up
Related Articles

Start Your Coding Journey Now!