Remove elements to make array satisfy arr[ i+1] < arr[i] for each valid i

Given an array arr[] of non-negative integers. We have to delete elements from this array such that arr[i + 1] > arr[j] for each valid i and this will be counted as one step. We have to apply the same operations until the array has become strictly decreasing. Now the task is to count the number of steps required to get the desired array.

Examples:

Input: arr[] = {6, 5, 8, 4, 7, 10, 9}
Output: 2
Initially 8, 7 and 10 do not satisfy the condition
so they all are deleted in the first step
and the array becomes {6, 5, 4, 9}
In the next step 9 gets deleted and
the array becomes {6, 5, 4} which is strictly decreasing.

Input: arr[] = {1, 2, 3, 4, 5}
Output: 1

Approach: The idea is to keep the indices of only required elements that are to be checked against a particular element. Thus, we use a vector to store only the required indices. We insert every index at the back and then remove the indices from back if the following condition is satisfied.

arr[vect.back()] ≥ val[i]

We take another array in which we update the no of steps particular element takes to delete.
If status[i] = -1 then element is not to be deleted, 0 denotes first step and so on…. That’s why we will add 1 to the answer.
While popping the indices, we repeatedly update the status of elements. If all indices are popped i.e. vect.size() = 0 then this element is not to be deleted so change its status to -1.

Below is the implementation of the above approach:

CPP

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
int status[100000];
  
// Function to return the required
// number of steps
int countSteps(int* val, int n)
{
    int sol = 0;
    vector<int> vec(1, 0);
    status[0] = -1;
  
    // Compute the number of steps
    for (int i = 1; i < n; ++i) {
  
        // Current status is to
        // delete in first step
        status[i] = 0;
  
        // Pop the indices while
        // condition is satisfied
        while (vec.size() > 0
               && val[vec.back()] >= val[i]) {
  
            // Inserting the correct
            // step no to delete
            status[i] = max(status[i],
                            status[vec.back()] + 1);
            vec.pop_back();
        }
        if (vec.size() == 0) {
  
            // Status changed to not delete
            status[i] = -1;
        }
  
        // Pushing a new index in the vector
        vec.push_back(i);
  
        // Build the solution from
        // smaller to larger size
        sol = max(sol, status[i] + 1);
    }
    return sol;
}
  
// Driver code
int main()
{
    int val[] = { 6, 5, 8, 4, 7, 10, 9 };
    int n = sizeof(val) / sizeof(val[0]);
  
    cout << countSteps(val, n);
  
    return 0;
}

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Java

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// A Java implementation of the approach
import java.util.*;
  
class GFG 
{
      
static int []status = new int[100000];
  
// Function to return the required
// number of steps
static int countSteps(int[]val, int n)
{
    int sol = 0;
    Vector<Integer> vec = new Vector<>(1);
    vec.add(0);
    status[0] = -1;
  
    // Compute the number of steps
    for (int i = 1; i < n; ++i) 
    {
  
        // Current status is to
        // delete in first step
        status[i] = 0;
  
        // Pop the indices while
        // condition is satisfied
        while (vec.size() > 0
            && val[vec.lastElement()] >= val[i])
        {
  
            // Inserting the correct
            // step no to delete
            status[i] = Math.max(status[i],
                            status[vec.lastElement()] + 1);
            vec.remove(vec.lastElement());
        }
        if (vec.isEmpty())
        {
  
            // Status changed to not delete
            status[i] = -1;
        }
  
        // Pushing a new index in the vector
        vec.add(i);
  
        // Build the solution from
        // smaller to larger size
        sol = Math.max(sol, status[i] + 1);
    }
    return sol;
}
  
// Driver code
public static void main(String[] args) 
{
    int val[] = { 6, 5, 8, 4, 7, 10, 9 };
    int n = val.length;
  
    System.out.println(countSteps(val, n));
}
}
  
/* This code contributed by PrinciRaj1992 */

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Python3

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# Python3 implementation of the approach 
  
status = [0]*100000
  
# Function to return the required 
# number of steps 
def countSteps(val, n) :
      
    sol = 0
    vec = [1, 0]; 
    status[0] = -1
  
    # Compute the number of steps 
    for i in range(n) :
  
        # Current status is to 
        # delete in first step 
        status[i] = 0
  
        # Pop the indices while 
        # condition is satisfied 
        while (len(vec) > 0
            and val[vec[len(vec)-1]] >= val[i]) : 
  
            # Inserting the correct 
            # step no to delete 
            status[i] = max(status[i], 
                            status[len(vec)-1] + 1); 
            vec.pop(); 
          
        if (len(vec) == 0) :
  
            # Status changed to not delete 
            status[i] = -1
          
  
        # Pushing a new index in the vector 
        vec.append(i); 
  
        # Build the solution from 
        # smaller to larger size 
        sol = max(sol, status[i] + 1); 
      
    return sol; 
  
  
# Driver code 
if __name__ == "__main__"
  
    val = [ 6, 5, 8, 4, 7, 10, 9 ]; 
    n = len(val); 
  
    print(countSteps(val, n)); 
      
# This code is contributed by AnkitRai01

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C#

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// A C# implementation of the approach 
using System;
using System.Collections.Generic;
  
class GFG 
      
static int []status = new int[100000]; 
  
// Function to return the required 
// number of steps 
static int countSteps(int[]val, int n) 
    int sol = 0; 
    List<int> vec = new List<int>(1); 
    vec.Add(0); 
    status[0] = -1; 
  
    // Compute the number of steps 
    for (int i = 1; i < n; ++i) 
    
  
        // Current status is to 
        // delete in first step 
        status[i] = 0; 
  
        // Pop the indices while 
        // condition is satisfied 
        while (vec.Count > 0
            && val[vec[vec.Count-1]] >= val[i]) 
        
  
            // Inserting the correct 
            // step no to delete 
            status[i] = Math.Max(status[i], 
                            status[vec[vec.Count-1]] + 1); 
            vec.Remove(vec[vec.Count-1]); 
        
        if (vec.Count == 0) 
        
  
            // Status changed to not delete 
            status[i] = -1; 
        
  
        // Pushing a new index in the vector 
        vec.Add(i); 
  
        // Build the solution from 
        // smaller to larger size 
        sol = Math.Max(sol, status[i] + 1); 
    
    return sol; 
  
// Driver code 
public static void Main(String[] args) 
    int []val = { 6, 5, 8, 4, 7, 10, 9 }; 
    int n = val.Length; 
  
    Console.WriteLine(countSteps(val, n)); 
  
// This code contributed by Rajput-Ji

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Output:

2


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