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)); |
2
Time Complexity: O(NlogN)
Auxiliary Space: O(N)
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