Reduce the array by deleting elements which are greater than all elements to its left

Given an array arr[] of N integers, the task is to delete the element from the given array if element to it’s left is smaller than it. Keep on deleting the elements from the array until no element has a smaller adjacent left element. Print the resultant array after above operation.

Examples: 

Input: arr[] = {2, 4, 1, 3, 4} 
Output: 2 1 
Explanation: 
Since 4 is greater than 2 remove 4, and arr become {2, 1, 3, 4}. 
Now 3 is greater than 1 so remove 3 and arr become {2, 1, 4}. 
Now 4 is greater than 1 so remove 4 and arr become {2, 1}. 
Now no elements satisfy the removing criteria so the resultant array is {2, 1}.

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

Approach: The idea is to use the concept of Merge Sort.



  1. Divide the input array into sub-arrays till the size of each sub-array becomes 1.
  2. Start merging the element.
  3. While merging, delete elements from the left subarray till it’s rightmost element, which have a value greater than the leftmost element of the right subarray.
  4. Repeat the above steps in each merging step such all elements with value smaller to it’s left have been deleted.
  5. Finally print the resultant array.

Below is the implementation of the above approach:

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;
 
// Function to implement merging of arr[]
vector<int> merge(vector<int> x, vector<int> y)
{
    for(auto i : y)
    {
        if (x[x.size() - 1] > i)
            x.push_back(i);
    }
    return x;
}
 
// Function to delete all elements which
// satisfy the condition A[i] > A[i-1]
vector<int> mergeDel(vector<int> l)
{
     
    // Divide array into its subarray
    if (l.size() == 1)
        return l;
 
    int m = l.size() / 2;
     
    vector<int> temp1 = {l.begin() + 0,
                         l.begin() + m};
    vector<int> temp2 = {l.begin() + m,
                         l.end()};
     
    // Getting back merged array with all 
    // its right element greater than
    // left one.
    return merge(mergeDel(temp1),
                 mergeDel(temp2));
}
 
// Driver Code
int main()
{
     
    // Given array arr[]
    vector<int> arr({ 5, 4, 3, 2, 1 });
    vector<int> ans = mergeDel(arr);
     
    cout << "[ ";
    for(auto x: ans)
        cout << x << ", ";
         
    cout << "]";
}
 
// This code is contributed by SURENDRA_GANGWAR
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program for the above approach
import java.util.ArrayList;
import java.util.Arrays;
 
class GFG{
     
// Function to implement merging of arr[]
static ArrayList<Integer> merge(ArrayList<Integer> x,
                                ArrayList<Integer> y)
{
    for(Integer i : y)
    {
        if (x.get(x.size() - 1) > i)
            x.add(i);
    }
    return x;
}
 
// Function to delete all elements which
// satisfy the condition A[i] > A[i-1]
static ArrayList<Integer> mergeDel(ArrayList<Integer> l)
{
     
    // Divide array into its subarray
    if (l.size() == 1)
        return l;
 
    int m = l.size() / 2;
 
    ArrayList<Integer> temp1 = new ArrayList<>(
        l.subList(0, m));
    ArrayList<Integer> temp2 = new ArrayList<>(
        l.subList(m, l.size()));
 
    // Getting back merged array with all
    // its right element greater than
    // left one.
    return merge(mergeDel(temp1), mergeDel(temp2));
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given array arr[]
    Integer[] ar = { 5, 4, 3, 2, 1 };
     
    ArrayList<Integer> arr = new ArrayList<>(
        Arrays.asList(ar));
    ArrayList<Integer> ans = mergeDel(arr);
     
    System.out.print("[ ");
    for(Integer x : ans)
        System.out.print(x + ", ");
         
    System.out.println("]");
}
}
 
// This code is contributed by sanjeev2552
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program for the above approach
 
# Function to delete all elements which
# satisfy the condition A[i] > A[i-1]
def mergeDel(l):
 
    # Divide array into its subarray
    if len(l) == 1:
 
        return l
 
    m = int( len(l) / 2)
 
    # Getting back merged array with all
    # its right element greater than left one.
    return merge(mergeDel(l[ 0 : m ]),
                 mergeDel(l[ m : len(l)]) )
 
 
# Function to implement merging of arr[]
def merge(x, y):
 
    for i in y:
 
        if x[-1] > i :
 
            x = x + [i]
 
    return x
   
# Driver Code
 
# Function defination for main()
def main():
 
# Given array arr[]
    arr = [5, 4, 3, 2, 1]
    print(mergeDel(arr))
 
main()
chevron_right

Output: 
[5, 4, 3, 2, 1]








 

Time Complexity: O(N*log N) 
Auxiliary Space: O(1)
 




Recommended Posts:


Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : SURENDRA_GANGWAR, sanjeev2552

Article Tags :
Practice Tags :