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Merge transactions in bank sheets in the order of their occurrence such that their sum remains positive
  • Difficulty Level : Hard
  • Last Updated : 03 Mar, 2021

Given an array arr[] consisting of N integers 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:-

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);
    }
}
Output:
100 400 -300 2000 -500 -1000 -500

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

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