Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Find all Array elements that are smaller than all elements to their right

  • Difficulty Level : Medium
  • Last Updated : 23 May, 2022

Given an array arr[] containing N positive integers. The task is to find all the elements which are smaller than all the elements to their right.

Examples:

Input: arr[] = {6, 14, 13, 21, 17, 19}
Output: [6, 13, 17, 19]
Explanation: All the elements in the output are following the condition.

Input: arr[] = {10, 3, 4, 8, 7}
Output: [3, 4, 7]

 

Naive approach: This approach uses two loops. For each element, traverse the array to its right and check if any smaller element exists or not. If all elements in the right part of the array are greater than it, then print this element. 

Time complexity: O(N*N)
Auxiliary Space: O(1)

Efficient approach: In the efficient approach, the idea is to use a Stack. Follow the steps mentioned below:

  • Iterate the array from the beginning of the array.
  • For every element in the array, pop all the elements present in the stack that are greater than it and then push it into the stack.
  • If no element is greater than it, then the current element is an answer.
  • At last, the stack remains with elements that are smaller than all elements present to their right.

Below is the code implementation of the above approach.

C++




// C++ program to find all elements in array
// that are smaller than all elements
// to their right.
#include <iostream>
#include <stack>
#include <vector>
using namespace std;
 
// Function to print all elements which are
// smaller than all elements present
// to their right
void FindDesiredElements(vector<int> const& arr)
{
    // Create an empty stack
    stack<int> stk;
 
    // Do for each element
    for (int i : arr) {
        // Pop all the elements that
        // are greater than the
        // current element
        while (!stk.empty() && stk.top() > i) {
            stk.pop();
        }
 
        // Push current element into the stack
        stk.push(i);
    }
 
    // Print all elements in the stack
    while (!stk.empty()) {
        cout << stk.top() << " ";
        stk.pop();
    }
}
 
// Driver Code
int main()
{
    vector<int> arr = { 6, 14, 13, 21, 17, 19 };
 
    FindDesiredElements(arr);
    return 0;
}

Java




// Java program to find all elements in array
// that are smaller than all elements
// to their right.
import java.util.ArrayList;
import java.util.Stack;
 
class GFG {
 
  // Function to print all elements which are
  // smaller than all elements present
  // to their right
  static void FindDesiredElements(ArrayList<Integer> arr)
  {
 
    // Create an empty stack
    Stack<Integer> stk = new Stack<Integer>();
 
    // Do for each element
    for (int i : arr)
    {
       
      // Pop all the elements that
      // are greater than the
      // current element
      while (!stk.empty() && stk.peek() > i) {
        stk.pop();
      }
 
      // Push current element into the stack
      stk.push(i);
    }
 
    // Print all elements in the stack
    while (!stk.empty()) {
      System.out.print(stk.peek() + " ");
      stk.pop();
    }
  }
 
  // Driver Code
  public static void main(String args[])
  {
    ArrayList<Integer> arr = new ArrayList<Integer>();
    arr.add(6);
    arr.add(14);
    arr.add(13);
    arr.add(21);
    arr.add(17);
    arr.add(19);
 
    FindDesiredElements(arr);
  }
}
 
// This code is contributed by saurabh_jaiswal.

Python3




# python program to find all elements in array
# that are smaller than all elements
# to their right.
 
# Function to print all elements which are
# smaller than all elements present
# to their right
def FindDesiredElements(arr):
 
    # Create an empty stack
    stk = []
 
    # Do for each element
    for i in arr :
 
        # Pop all the elements that
        # are greater than the
        # current element
        while (len(stk)!=0 and stk[len(stk)-1] > i):
            stk.pop()
 
        # Push current element into the stack
        stk.append(i)
 
    # Print all elements in the stack
    while (len(stk) != 0):
        print(stk[len(stk)-1],end = " ")
        stk.pop()
 
# Driver Code
arr = []
arr.append(6)
arr.append(14)
arr.append(13)
arr.append(21)
arr.append(17)
arr.append(19)
 
FindDesiredElements(arr)
 
# This code is contributed by shinjanpatra

C#




// C# program to find all elements in array
// that are smaller than all elements
// to their right.
using System;
using System.Collections.Generic;
public class GFG {
 
  // Function to print all elements which are
  // smaller than all elements present
  // to their right
  static void FindDesiredElements(List<int> arr) {
 
    // Create an empty stack
    Stack<int> stk = new Stack<int>();
 
    // Do for each element
    foreach (int i in arr) {
 
      // Pop all the elements that
      // are greater than the
      // current element
      while (stk.Count!=0 && stk.Peek() > i) {
        stk.Pop();
      }
 
      // Push current element into the stack
      stk.Push(i);
    }
 
    // Print all elements in the stack
    while (stk.Count>0) {
      Console.Write(stk.Peek() + " ");
      stk.Pop();
    }
  }
 
  // Driver Code
  public static void Main(String []args) {
    List<int> arr = new List<int>();
    arr.Add(6);
    arr.Add(14);
    arr.Add(13);
    arr.Add(21);
    arr.Add(17);
    arr.Add(19);
 
    FindDesiredElements(arr);
  }
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
// javascript program to find all elements in array
// that are smaller than all elements
// to their right.
 
    // Function to print all elements which are
    // smaller than all elements present
    // to their right
    function FindDesiredElements( arr) {
 
        // Create an empty stack
        var stk = [];
 
        // Do for each element
        for (var i of arr) {
 
            // Pop all the elements that
            // are greater than the
            // current element
            while (stk.length!=0 && stk[stk.length-1] > i) {
                stk.pop();
            }
 
            // Push current element into the stack
            stk.push(i);
        }
 
        // Print all elements in the stack
        while (stk.length != 0) {
            document.write(stk[stk.length-1] + " ");
            stk.pop();
        }
    }
 
    // Driver Code
        var arr = [];
        arr.push(6);
        arr.push(14);
        arr.push(13);
        arr.push(21);
        arr.push(17);
        arr.push(19);
 
        FindDesiredElements(arr);
 
// This code is contributed by Rajput-Ji
</script>
Output
19 17 13 6 

 Time complexity: O(N)
Auxiliary Space: O(N)

Space Optimized approach: The idea is to traverse the array from right to left (reverse order) and maintain an auxiliary variable that stores the minimum element found so far. This approach will neglect the use of stack. Follow the below steps:

  • Start iterating from the end of the array.
  • If the current element is smaller than the minimum so far, then the element is found. Update the value storing minimum value so far. Print all such values as the answer.

 Below is the implementation of the above approach.

C++




// C++ program to print all elements which are
// smaller than all elements present to their right
 
#include <iostream>
#include <limits.h>
using namespace std;
 
// Function to print all elements which are
// smaller than all elements
// present to their right
void FindDesiredElements(int arr[], int n)
{
    int min_so_far = INT_MAX;
 
    // Traverse the array from right to left
    for (int j = n - 1; j >= 0; j--) {
        // If the current element is greater
        //  than the maximum so far, print it
        // and update `max_so_far`
        if (arr[j] <= min_so_far) {
            min_so_far = arr[j];
            cout << arr[j] << " ";
        }
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 6, 14, 13, 21, 17, 19 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    FindDesiredElements(arr, N);
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
class GFG {
 
  // Function to print all elements which are
  // smaller than all elements
  // present to their right
  static void FindDesiredElements(int[] arr, int n)
  {
    int min_so_far = Integer.MAX_VALUE;
 
    // Traverse the array from right to left
    for (int j = n - 1; j >= 0; j--)
    {
 
      // If the current element is greater
      //  than the maximum so far, print it
      // and update `max_so_far`
      if (arr[j] <= min_so_far) {
        min_so_far = arr[j];
        System.out.print(arr[j] + " ");
      }
    }
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int[] arr = { 6, 14, 13, 21, 17, 19 };
    int N = arr.length;
 
    FindDesiredElements(arr, N);
 
  }
}
 
// This code is contributed by code_hunt.

Python3




# Python code for the above approach
 
# Function to print all elements which are
# smaller than all elements
# present to their right
def FindDesiredElements(arr, n):
    min_so_far = 10 ** 9;
 
    # Traverse the array from right to left
    for j in range(n - 1, -1, -1):
     
        # If the current element is greater
        #  than the maximum so far, print it
        # and update `max_so_far`
        if (arr[j] <= min_so_far):
            min_so_far = arr[j];
            print(arr[j], end=" ")
 
# Driver Code
arr = [6, 14, 13, 21, 17, 19];
N = len(arr)
 
FindDesiredElements(arr, N);
 
# This code is contributed by Saurabh Jaiswal

C#




// C# program to print all elements which are
// smaller than all elements present to their right
 
using System;
class GFG {
    // Function to print all elements which are
    // smaller than all elements
    // present to their right
    static void FindDesiredElements(int[] arr, int n)
    {
        int min_so_far = Int32.MaxValue;
 
        // Traverse the array from right to left
        for (int j = n - 1; j >= 0; j--) {
            // If the current element is greater
            //  than the maximum so far, print it
            // and update `max_so_far`
            if (arr[j] <= min_so_far) {
                min_so_far = arr[j];
                Console.Write(arr[j] + " ");
            }
        }
    }
 
    // Driver Code
    public static void Main()
    {
        int[] arr = { 6, 14, 13, 21, 17, 19 };
        int N = arr.Length;
 
        FindDesiredElements(arr, N);
    }
}

Javascript




<script>
      // JavaScript code for the above approach
 
      // Function to print all elements which are
      // smaller than all elements
      // present to their right
      function FindDesiredElements(arr, n)
      {
          let min_so_far = Number.MAX_VALUE;
 
          // Traverse the array from right to left
          for (let j = n - 1; j >= 0; j--)
          {
           
              // If the current element is greater
              //  than the maximum so far, print it
              // and update `max_so_far`
              if (arr[j] <= min_so_far) {
                  min_so_far = arr[j];
                  document.write(arr[j] + " ");
              }
          }
      }
 
      // Driver Code
      let arr = [6, 14, 13, 21, 17, 19];
      let N = arr.length;
 
      FindDesiredElements(arr, N);
 
// This code is contributed by Potta Lokesh
  </script>
Output
19 17 13 6 

 Time complexity: O(N)
Auxiliary Space: O(1)


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!