Min operations to empty an array by erasing any increasing subsequence

Given array A[] of size N, the task is to find the number of operations to empty the array by performing the following operation one or more times. In one operation choose any strictly increasing subsequence and delete it from A[].

Examples:

Input: A[] = {2, 1, 4, 5, 3}
Output: 2
Explanation: Following operations are performed to empty the array:

• Choosing increasing subsequence {2, 4, 5} and removing it from A[], A[] becomes {1, 3}
• Choosing increasing subsequence {1, 3} and removing it from A[], A[] becomes empty.

Input: A[] = {0, 0, 0, 0}
Output:  4

Approach: The idea is to use a Priority Queue data structure to solve this problem.

Iterate over the array and keep inserting new element to priority queue, if given element is inserted at non-starting position than delete element just previous to that position.

Below are the steps for the above approach:

• Create priority queue pq[] using a multiset container to store the array elements in sorted order.
• Iterate from 0 to N – 1.
  • For each iteration, insert the current element in pq[].
  • For each iteration, search the current element in pq[].
• If the current element is at the initial position in the priority queue, move to the next iteration, else erase the previous element in the priority queue.
• Return the size of pq[] which will be the answer. 

Below is the code for the above approach:

C++

// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;

// Function to count number of increasing
// subsequences that can be deleted
// to empty array
int findMinOp(int A[], int N)
{

    // Declaring priority queue
    multiset<int> pq;

    // Iterating from 0 to N
    for (int i = 0; i < N; i++) {

        // Insert current element
        // in priority queue
        pq.insert(A[i]);

        // Binary search same element
        // in priority queue
        auto it = pq.find(A[i]);

        // If it is first element in
        // sorted pq skip the current
        // iteration
        if (it == pq.begin())
            continue;

        it--;

        // Delete that previous
        // element from pq
        pq.erase(it);
    }

    // Return final answer which is
    // size of priority queue
    return pq.size();
}

// Driver Code
int main()
{
    int A[] = { 2, 1, 4, 5, 3 };
    int N = sizeof(A) / sizeof(A[0]);

    // Function Call
    cout << findMinOp(A, N) << endl;

    return 0;
}

Java

import java.util.*;

public class Main
{

  // Function to count the number of increasing
  // subsequences that can be deleted to empty the array
  static int findMinOp(int[] A, int N)
  {

    // Declaring priority queue
    PriorityQueue<Integer> pq = new PriorityQueue<>();

    // Iterating from 0 to N
    for (int i = 0; i < N; i++)
    {

      // Insert current element in priority queue
      pq.offer(A[i]);

      // Get an iterator pointing to the same element in
      // the priority queue
      Iterator<Integer> it = pq.iterator();
      int prev = Integer.MIN_VALUE;

      // Find the previous element in the priority queue
      while (it.hasNext()) {
        int curr = it.next();
        if (curr == A[i]) {
          break;
        }
        prev = curr;
      }

      // If it is the first element in the sorted
      // priority queue, skip the current iteration
      if (A[i] == prev) {
        continue;
      }

      // Delete that previous element from the priority queue
      pq.remove(prev);
    }

    // Return the final answer which is the size of the priority queue
    return pq.size();
  }

  // Driver code
  public static void main(String[] args) {
    int[] A = { 2, 1, 4, 5, 3 };
    int N = A.length;

    // Function call
    System.out.println(findMinOp(A, N));
  }
}

// This code is contributed by Prajwal Kandekar

Python3

# Python code to implement the approach
import heapq

# Function to count number of increasing
# subsequences that can be deleted
# to empty array
def findMinOp(A, N):

    # Declaring priority queue
    pq = []

    # Iterating from 0 to N
    for i in range(N):

        # Insert current element
        # in priority queue
        heapq.heappush(pq, A[i])

        # If it is first element in
        # sorted pq skip the current
        # iteration
        if pq.index(A[i]) == 0:
            continue

        # Delete that previous
        # element from pq
        pq.remove(A[i-1])

    # Return final answer which is
    # size of priority queue
    return len(pq)

# Driver Code
if __name__ == '__main__':
    A = [2, 1, 4, 5, 3]
    N = len(A)

    # Function Call
    print(findMinOp(A, N))

C#

// C# code to implement the approach

using System;
using System.Collections.Generic;

class GFG
{
    static int FindMinOp(int[] A, int N)
    {
        // Create a stack to track increasing subsequences
        Stack<int> stack = new Stack<int>();

        for (int i = 0; i < N; i++)
        {
            // If the stack is empty or the current element is greater than the top of the stack,
            // push the current element onto the stack
            if (stack.Count == 0 || A[i] > stack.Peek())
            {
                stack.Push(A[i]);
            }
            else
            {
                // If the current element is smaller than or equal
               // to the top of the stack,
              // pop elements from the stack until a
             // suitable position is found
                while (stack.Count > 0 && A[i] < stack.Peek())
                {
                    stack.Pop();
                }

                // Push the current element onto the stack
                stack.Push(A[i]);
            }
        }

        // The size of the stack represents the minimum number
       // of increasing subsequences
      // that can be deleted to empty the array
        return stack.Count;
    }

    static void Main()
    {
        // Example usage
        int[] A = { 2, 1, 4, 5, 3 };
        int N = A.Length;

        // Function Call
        Console.WriteLine(FindMinOp(A, N));
    }
}

Javascript

function findMinOp(A, N) {
  // Declaring priority queue
  let pq = [];

  // Iterating from 0 to N
  for (let i = 0; i < N; i++) {
    // Insert current element in priority queue
    pq.push(A[i]);
    // Convert array to min heap using a comparator function
    pq.sort((a, b) => a - b);

    // If it is the first element in the sorted
    // priority queue, skip the current iteration
    if (pq.indexOf(A[i]) === 0) {
      continue;
    }

    // Delete that previous element from the priority queue
    pq.splice(pq.indexOf(A[i - 1]), 1);
  }

  // Return the final answer which is the size of the priority queue
  return pq.length;
}

// Driver code
let A = [2, 1, 4, 5, 3];
let N = A.length;

// Function call
console.log(findMinOp(A, N));

Output

2

Time Complexity: O(NlogN)
Auxiliary Space: O(N)