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

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 ``using` `namespace` `std;` `// Function to print all elements which are``// smaller than all elements present``// to their right``void` `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 Code``int` `main()``{``    ``vector<``int``> arr = { 6, 14, 13, 21, 17, 19 };``    ``FindDesiredElements(arr);``    ``return` `0;``}`

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 Code``if` `__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] + ``" "``);``            ``}``        ``}``    ``}``}`

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 ``#include ``#include ``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 arr)``  ``{` `    ``// Create an empty stack``    ``Stack stk = ``new` `Stack();` `    ``// 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 arr = ``new` `ArrayList();``    ``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

 ``

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 ``#include ``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

 ``

Output

`19 17 13 6 `

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

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