Minimize length of an array by repeatedly removing elements that are smaller than the next element
Last Updated :
19 Jul, 2023
Given an array arr[] consisting of N integers, the task is to repeatedly remove elements that are smaller than the next element.
Examples:
Input: arr[] = {20, 10, 25, 30, 40}
Output: 40
Explanation:
Below are the operations that are performed:
- Current array: arr[] = {20, 10, 25, 30, 40}
Currently, arr[1](= 10) is less than arr[2](= 25). Therefore, removing arr[1] modifies the array to {20, 25, 30, 40}.
- Currently arr[0](= 20) is less than arr[1](= 25). Therefore, removing arr[0] modifies the array to {25, 30, 40}.
- Currently, arr[0](= 25) is less than arr[1](= 30). Therefore, removing arr[0] modifies the array to {30, 40}.
- Currently, arr[0](= 30) is less than arr[1](= 40). Therefore, removing arr[0] modifies the array to {40}.
Therefore, the remaining array is arr[] = {40}.
Input: arr[] = {2, 5, 1, 0}
Output: 5 1 0
Naive Approach: The simplest approach to solve the problem is to traverse the array and remove the element if there are any greater elements in the range [i + 1, N – 1].
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach – Iterative: The above approach can be optimized by traversing the array from the end and keeping track of the largest element found and deleting the current element if it is lesser than the largest element. Follow the steps below to solve the problem:
- Initialize a vector, say res, to store the resultant array.
- Initialize a variable, say maxi as INT_MIN, to store the maximum value from the end.
- Traverse the array arr[] in a reverse manner and perform the following steps:
- If the value of arr[i] is less than the maxi then continue.
- Otherwise, push the element arr[i] in res and update the value of maxi to the maximum of maxi and arr[i].
- Reverse the vector res.
- After completing the above steps, print the vector res as the result.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void DeleteSmaller(int arr[], int N)
{
int maxi = INT_MIN;
vector<int> res;
for (int i = N - 1; i >= 0; i--) {
if (arr[i] < maxi)
continue;
res.push_back(arr[i]);
maxi = max(maxi, arr[i]);
}
reverse(res.begin(), res.end());
for (auto x : res)
cout << x << " ";
}
int main()
{
int arr[] = { 2, 5, 1, 0 };
int N = sizeof(arr) / sizeof(arr[0]);
DeleteSmaller(arr, N);
return 0;
}
|
Java
import java.util.*;
class GFG {
static void DeleteSmaller(int arr[], int N)
{
int maxi = Integer.MIN_VALUE;
List<Integer> res = new ArrayList<Integer>();
for (int i = N - 1; i >= 0; i--) {
if (arr[i] < maxi)
continue;
res.add(arr[i]);
maxi = Math.max(maxi, arr[i]);
}
Collections.reverse(res);
for(int i = 0; i < res.size(); i++)
System.out.println(res.get(i) + " ");
}
public static void main(String[] args)
{
int arr[] = { 2, 5, 1, 0 };
int N = arr.length;
DeleteSmaller(arr, N);
}
}
|
Python3
def DeleteSmaller(arr, N):
maxi =-10**9
res = []
for i in range(N - 1, -1, -1):
if (arr[i] < maxi):
continue
res.append(arr[i])
maxi = max(maxi, arr[i])
res = res[::-1]
print(*res)
if __name__ == '__main__':
arr = [2, 5, 1, 0]
N = len(arr)
DeleteSmaller(arr, N)
|
C#
using System;
using System.Collections.Generic;
class GFG {
static void DeleteSmaller(int[] arr, int N)
{
int maxi = Int32.MinValue;
List<int> res = new List<int>();
for (int i = N - 1; i >= 0; i--) {
if (arr[i] < maxi)
continue;
res.Add(arr[i]);
maxi = Math.Max(maxi, arr[i]);
}
res.Reverse();
foreach(int x in res)
Console.Write(x + " ");
}
public static void Main()
{
int[] arr = { 2, 5, 1, 0 };
int N = arr.Length;
DeleteSmaller(arr, N);
}
}
|
Javascript
<script>
function DeleteSmaller(arr, N)
{
let maxi = Number.MIN_SAFE_INTEGER;
let res = [];
for (let i = N - 1; i >= 0; i--) {
if (arr[i] < maxi)
continue;
res.push(arr[i]);
maxi = Math.max(maxi, arr[i]);
}
res.reverse();
for (let x of res)
document.write(x + " ");
}
let arr = [2, 5, 1, 0];
let N = arr.length;
DeleteSmaller(arr, N);
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(N)
Efficient Approach – Recursive: The above approach can also be implemented using recursion. Follow the steps below to solve the problem:
- Initialize a vector, say res to store the resultant array.
- Define a recursive function, say RecursiveDeleteFunction(int i) to delete element smaller than next element:
- If the value of i is equal to N then return INT_MIN.
- Call recursively call the function RecursiveDeleteFunction(i+1) and store the returned value in a variable say curr.
- If the value of arr[i] is at least curr then push arr[i] in vector res and update the value of curr as the maximum of curr and arr[i].
- Now, return the curr.
- Call the function RecursiveDeleteFunction(0) to remove the smaller elements to next elements and then, reverse the vector res.
- After completing the above steps, print the vector res as the result.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int RecursiveDeleteFunction(
int arr[], int N, int i,
vector<int>& res)
{
if (i == N)
return INT_MIN;
int curr = RecursiveDeleteFunction(
arr, N, i + 1, res);
if (arr[i] >= curr) {
res.push_back(arr[i]);
curr = max(arr[i], curr);
}
return curr;
}
void DeleteSmaller(int arr[], int N)
{
int maxi = INT_MIN;
vector<int> res;
RecursiveDeleteFunction(arr, N, 0, res);
reverse(res.begin(), res.end());
for (auto x : res)
cout << x << " ";
}
int main()
{
int arr[] = { 2, 5, 1, 0 };
int N = sizeof(arr) / sizeof(arr[0]);
DeleteSmaller(arr, N);
return 0;
}
|
Java
import java.util.*;
public class GFG
{
static int RecursiveDeleteFunction(
int []arr, int N, int i, ArrayList<Integer> res)
{
if (i == N)
return Integer.MIN_VALUE;
int curr = RecursiveDeleteFunction(
arr, N, i + 1, res);
if (arr[i] >= curr) {
res.add(arr[i]);
curr = Math.max(arr[i], curr);
}
return curr;
}
static void DeleteSmaller(int []arr, int N)
{
int maxi = Integer.MIN_VALUE;
ArrayList<Integer>
res = new ArrayList<Integer>();
RecursiveDeleteFunction(arr, N, 0, res);
Collections.reverse(res);
for (int i = 0; i < res.size(); i++)
System.out.print(res.get(i) + " ");
}
public static void main(String args[])
{
int []arr = { 2, 5, 1, 0 };
int N = arr.length;
DeleteSmaller(arr, N);
}
}
|
Python3
def RecursiveDeleteFunction(arr, N, i, res) :
if (i == N) :
return -10**9
curr = RecursiveDeleteFunction(
arr, N, i + 1, res)
if (arr[i] >= curr) :
res.append(arr[i])
curr = max(arr[i], curr)
return curr
def DeleteSmaller(arr, N) :
maxi = -10**9
res = []
RecursiveDeleteFunction(arr, N, 0, res)
res = res[::-1]
for x in res :
print(x, end = " ")
if __name__ == '__main__':
arr = [2, 5, 1, 0]
N = len(arr)
DeleteSmaller(arr, N)
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static int RecursiveDeleteFunction(
int []arr, int N, int i, List<int> res)
{
if (i == N)
return Int32.MinValue;
int curr = RecursiveDeleteFunction(
arr, N, i + 1, res);
if (arr[i] >= curr) {
res.Add(arr[i]);
curr = Math.Max(arr[i], curr);
}
return curr;
}
static void DeleteSmaller(int []arr, int N)
{
int maxi = Int32.MinValue;
List<int> res = new List<int>();
RecursiveDeleteFunction(arr, N, 0, res);
res.Reverse();
for (int i = 0; i < res.Count; i++)
Console.Write(res[i] + " ");
}
public static void Main()
{
int []arr = { 2, 5, 1, 0 };
int N = arr.Length;
DeleteSmaller(arr, N);
}
}
|
Javascript
<script>
function RecursiveDeleteFunction(
arr, N, i,
res) {
if (i == N)
return -999999999;
let curr = RecursiveDeleteFunction(
arr, N, i + 1, res);
if (arr[i] >= curr) {
res.push(arr[i]);
curr = Math.max(arr[i], curr);
}
return curr;
}
function DeleteSmaller(arr, N) {
let maxi = Number.MIN_VALUE;
let res = [];
RecursiveDeleteFunction(arr, N, 0, res);
res.reverse();
document.write(res);
}
let arr = [2, 5, 1, 0];
let N = arr.length;
DeleteSmaller(arr, N);
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(N)
Optimized Solution(using stack data structure)
Approach- We can use a stack data structure.First iterate through the array from left to right, and for each element, we will compare it with the top element of the stack. If the current element is greater than the top element, we will pop the top element from the stack until either the stack becomes empty or the top element is greater than the current element. Finally, we will push the current element onto the stack.
Implementation of the above approach
C++
#include <iostream>
#include <stack>
#include <vector>
#include <algorithm> // add this line
using namespace std;
vector<int> removeElements(vector<int>& arr) {
stack<int> s;
for (int i = 0; i < arr.size(); i++) {
while (!s.empty() && s.top() < arr[i]) {
s.pop();
}
s.push(arr[i]);
}
vector<int> result;
while (!s.empty()) {
result.push_back(s.top());
s.pop();
}
reverse(result.begin(), result.end());
return result;
}
int main() {
vector<int> arr = {2,5,1,0};
vector<int> result = removeElements(arr);
for (int i = 0; i < result.size(); i++) {
cout << result[i] << " ";
}
cout << endl;
return 0;
}
|
Java
import java.util.*;
public class Main {
public static List<Integer> removeElements(List<Integer> arr) {
Stack<Integer> s = new Stack<>();
for (int i = 0; i < arr.size(); i++) {
while (!s.empty() && s.peek() < arr.get(i)) {
s.pop();
}
s.push(arr.get(i));
}
List<Integer> result = new ArrayList<>();
while (!s.empty()) {
result.add(s.peek());
s.pop();
}
Collections.reverse(result);
return result;
}
public static void main(String[] args) {
List<Integer> arr = Arrays.asList(2, 5, 1, 0);
List<Integer> result = removeElements(arr);
for (int i = 0; i < result.size(); i++) {
System.out.print(result.get(i) + " ");
}
System.out.println();
}
}
|
Python
def removeElements(arr):
s = []
for i in range(len(arr)):
while s and s[-1] < arr[i]:
s.pop()
s.append(arr[i])
result = []
while s:
result.append(s.pop())
result.reverse()
return result
arr = [2,5,1,0]
result = removeElements(arr)
print(result)
|
C#
using System;
using System.Collections.Generic;
using System.Linq;
public class GFG
{
public static List<int> RemoveElements(List<int> arr)
{
Stack<int> s = new Stack<int>();
for (int i = 0; i < arr.Count; i++)
{
while (s.Count > 0 && s.Peek() < arr[i])
{
s.Pop();
}
s.Push(arr[i]);
}
List<int> result = new List<int>();
while (s.Count > 0)
{
result.Add(s.Peek());
s.Pop();
}
result.Reverse();
return result;
}
public static void Main()
{
List<int> arr = new List<int> {2, 5, 1, 0};
List<int> result = RemoveElements(arr);
for (int i = 0; i < result.Count; i++)
{
Console.Write(result[i] + " ");
}
Console.WriteLine();
}
}
|
Javascript
function removeElements(arr) {
let s = [];
for (let i = 0; i < arr.length; i++) {
while (s.length > 0 && s[s.length - 1] < arr[i]) {
s.pop();
}
s.push(arr[i]);
}
let result = [];
while (s.length > 0) {
result.push(s[s.length - 1]);
s.pop();
}
result.reverse();
return result;
}
let arr = [2, 5, 1, 0];
let result = removeElements(arr);
for (let i = 0; i < result.length; i++) {
console.log(result[i] + " ");
}
console.log();
|
Time complexity: O(n), where n is the length of the input array ‘arr’
The auxiliary space:O(n), because in the worst case scenario the entire input array may need to be stored in the stack before any elements are removed.
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