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:

C++

// 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

Java

// 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

Python3

# 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 definition for main()
def main():

# Given array arr[]
    arr = [5, 4, 3, 2, 1]
    print(mergeDel(arr))

main()

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG{

// Function to implement merging of arr[]
static List<int> merge(List<int> x, List<int> y)
{
    foreach(int i in y)
    {
        if (x[x.Count - 1] > i)
            x.Add(i);
    }
    return x;
}

// Function to delete all elements which
// satisfy the condition A[i] > A[i-1]
static List<int> mergeDel(List<int> l)
{

    // Divide array into its subarray
    if (l.Count == 1)
        return l;

    int m = l.Count / 2;

    List<int> temp1 = l.GetRange(0, m);
    List<int> temp2 = l.GetRange(m, l.Count - m);

    // 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[]
    List<int> arr = new List<int>{ 5, 4, 3, 2, 1 };

    List<int> ans = mergeDel(arr);

    Console.Write("[ ");
    foreach(int x in ans)
        Console.Write(x + ", ");

    Console.Write("]");
}
}

// This code is contributed by chitranayal

Javascript

<script>
// Javascript program for the above approach

// Function to implement merging of arr[]
function  merge(x,y)
{
    for(let i=0;i<y.length;i++)
    {
        if (x[x.length - 1] > y[i])
            x.push(y[i]);
    }
    return x;
}

// Function to delete all elements which
// satisfy the condition A[i] > A[i-1]
function mergeDel(l)
{
    // Divide array into its subarray
    if (l.length == 1)
        return l;

    let m = Math.floor(l.length / 2);

    let temp1 = l.slice(0, m);
    let temp2 = l.slice(m, l.length);

    // Getting back merged array with all
    // its right element greater than
    // left one.
    return merge(mergeDel(temp1), mergeDel(temp2));
}
// Driver Code

// Given array arr[]
let arr=[5, 4, 3, 2, 1];
let ans = mergeDel(arr);
document.write("[ ");
document.write(ans.join(", "));
document.write("]<br>");


// This code is contributed by avanitrachhadiya2155
</script>

Output:

[5, 4, 3, 2, 1]

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