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 or equal element exists or not. If all elements in the right part of the array are greater than it, then print this element.
Steps for implementation-
- Traverse the input array/vector to pick elements one by one
- Run an inner loop for every element from its next element to the end
- Now if our element is greater than or equal to any element on its right then break the inner loop
- If the inner loop is broken then leave that element else print that element
Code-
C++
// C++ program to find all elements in array// that are smaller than all elements// to their right.#include <bits/stdc++.h>using namespace std;// Function to print all elements which are// smaller than all elements present// to their rightvoid FindDesiredElements(vector<int>& arr){ // Size of input vector int n = arr.size(); // Pick element one by one for (int i = 0; i < n; i++) { int j = i + 1; // Check all elements on its right while (j < n) { // If that element is greater than or equal to // any element on its right then break the inner // loop if (arr[i] >= arr[j]) { break; } j++; } // If that inner loop is not broken then print that // element if (j == n) { cout << arr[i] << " "; } }}// Driver Codeint main(){ vector<int> arr = { 6, 14, 13, 21, 17, 19 }; FindDesiredElements(arr); return 0;} |
Java
import java.util.ArrayList;import java.util.List;class GFG { // Function to print all elements which are // smaller than all elements present // to their right static void findDesiredElements(List<Integer> arr) { // Size of input list int n = arr.size(); // Pick element one by one for (int i = 0; i < n; i++) { int j = i + 1; // Check all elements on its right while (j < n) { // If that element is greater than or equal to // any element on its right then break the inner // loop if (arr.get(i) >= arr.get(j)) { break; } j++; } // If that inner loop is not broken then print that // element if (j == n) { System.out.print(arr.get(i) + " "); } } } // Driver Code public static void main(String[] args) { List<Integer> arr = new ArrayList<>(); arr.add(6); arr.add(14); arr.add(13); arr.add(21); arr.add(17); arr.add(19); findDesiredElements(arr); }} |
Python3
def find_desired_elements(arr): # Size of input list n = len(arr) # Pick element one by one for i in range(n): j = i + 1 # Check all elements on its right while j < n: # If that element is greater than or equal to # any element on its right then break the inner # loop if arr[i] >= arr[j]: break j += 1 # If that inner loop is not broken then print that # element if j == n: print(arr[i], end=' ')# Driver Codeif __name__ == '__main__': arr = [6, 14, 13, 21, 17, 19] find_desired_elements(arr) |
C#
using System;using System.Collections.Generic;class Program { static void Main(string[] args) { List<int> arr = new List<int> { 6, 14, 13, 21, 17, 19 }; FindDesiredElements(arr); } static void FindDesiredElements(List<int> arr) { // Size of input list int n = arr.Count; // Pick element one by one for (int i = 0; i < n; i++) { int j = i + 1; // Check all elements on its right while (j < n) { // If that element is greater than or equal to // any element on its right then break the inner // loop if (arr[i] >= arr[j]) { break; } j++; } // If that inner loop is not broken then print that // element if (j == n) { Console.Write(arr[i] + " "); } } }} |
Javascript
function findDesiredElements(arr) { // Size of input array var n = arr.length; // Pick element one by one for (var i = 0; i < n; i++) { var j = i + 1; // Check all elements on its right while (j < n) { // If that element is greater than or equal to // any element on its right, then break the inner loop if (arr[i] >= arr[j]) { break; } j++; } // If the inner loop is not broken, then print that element if (j == n) { process.stdout.write(arr[i] + " "); } }}// Driver Codevar arr = [6, 14, 13, 21, 17, 19];findDesiredElements(arr); |
Output-
6 13 17 19
Time complexity: O(N*N), because of two nested for loops
Auxiliary Space: O(1),because no extra space has been used
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 rightvoid 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 Codeint 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 rightdef 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 Codearr = []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 rightvoid 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 Codeint 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 approachimport 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 rightdef 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 Codearr = [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 rightusing 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)
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