Given an array arr, the task is to replace each element of the array with the element that appears after it and replace the last element with -1.
Examples:
Input: arr[] = {5, 1, 3, 2, 4}
Output: 1 3 2 4 -1
Input: arr[] = {6, 8, 32, 12, 14, 10, 25 }
Output: 8 32 12 14 10 25 -1
Approach: Traverse the array from 0 to n-2 and update arr[i] = arr[i+1]. In the end, set a[n-1] = -1 and print the contents of the updated array.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void updateArray(int arr[], int n)
{
for (int i = 0; i <= n - 2; i++)
arr[i] = arr[i + 1];
arr[n - 1] = -1;
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
}
int main()
{
int arr[] = { 5, 1, 3, 2, 4 };
int N = sizeof(arr) / sizeof(arr[0]);
updateArray(arr, N);
return 0;
}
|
Java
class GFG
{
static void updateArray(int arr[], int n)
{
for (int i = 0; i <= n - 2; i++)
arr[i] = arr[i + 1];
arr[n - 1] = -1;
for (int i = 0; i < n; i++)
System.out.print(arr[i] + " ");
}
public static void main(String []args)
{
int arr[] = { 5, 1, 3, 2, 4 } ;
int N = arr.length ;
updateArray(arr, N);
}
}
|
Python3
def updateArray(arr, n):
for i in range (n - 1):
arr[i] = arr[i + 1]
arr[n - 1] = -1
for i in range( n):
print (arr[i], end = " ")
if __name__ == "__main__":
arr = [ 5, 1, 3, 2, 4 ]
N = len(arr)
updateArray(arr, N)
|
C#
using System;
class GFG
{
static void updateArray(int[] arr, int n)
{
for (int i = 0; i <= n - 2; i++)
arr[i] = arr[i + 1];
arr[n - 1] = -1;
for (int i = 0; i < n; i++)
Console.Write(arr[i] + " ");
}
public static void Main()
{
int[] arr = { 5, 1, 3, 2, 4 } ;
int N = arr.Length ;
updateArray(arr, N);
}
}
|
PHP
<?php
function updateArray(&$arr, $n)
{
for ($i = 0; $i <= $n - 2; $i++)
$arr[$i] = $arr[$i + 1];
$arr[$n - 1] = -1;
for ($i = 0; $i < $n; $i++)
{
echo ($arr[$i]);
echo (" ");
}
}
$arr = array(5, 1, 3, 2, 4 );
$N = sizeof($arr);
updateArray($arr, $N);
?>
|
Javascript
<script>
function updateArray(arr, n)
{
for (let i = 0; i <= n - 2; i++)
arr[i] = arr[i + 1];
arr[n - 1] = -1;
for (let i = 0; i < n; i++)
document.write(arr[i] + " ");
}
let arr = [ 5, 1, 3, 2, 4 ];
let N = arr.length;
updateArray(arr, N);
</script>
|
C
#include <stdio.h>
void updateArray(int arr[], int n)
{
for (int i = 0; i <= n - 2; i++)
arr[i] = arr[i + 1];
arr[n - 1] = -1;
for (int i = 0; i < n; i++)
printf("%d ",arr[i]);
}
int main()
{
int arr[] = { 5, 1, 3, 2, 4 };
int N = sizeof(arr) / sizeof(arr[0]);
updateArray(arr, N);
return 0;
}
|
Complexity Analysis:
- Time Complexity: O(n)
- Auxiliary Space: O(1)
Approach 2: Using Stack:
The approach to solving this problem is to use a stack to keep track of the largest element seen so far while traversing the array from right to left. We initialize the stack with the last element of the array since there are no larger elements to the right of it.
Below is the step-by-step algorithm for the above approach:
- Initialize a stack and put the last element of the array into it.
- We iterate over the remaining elements of the array in reverse order. For each element, we check if it is greater than the top element of the stack. If it is, we pop the top element of the stack until we find an element that is greater than the current element or until the stack is empty. The element at the top of the stack after this process is the nearest larger element to the right of the current element. We then replace the current element in the array with the top element of the stack.
- We continue this process for all the elements in the array. At the end, we replace the last element of the array with -1 since there are no larger elements to the right of it.
- Finally, we print the modified array to the console.
Below is the code for the above approach:
C++
#include <iostream>
#include <stack>
void updateArray(int arr[], int n) {
std::stack<int> st;
st.push(arr[n - 1]);
for (int i = n - 2; i >= 0; --i) {
int curr = arr[i];
arr[i] = st.top();
while (!st.empty() && curr > st.top()) {
st.pop();
}
st.push(curr);
}
arr[n - 1] = -1;
for (int i = 0; i < n; ++i) {
std::cout << arr[i] << " ";
}
}
int main() {
int arr[] = {5, 1, 3, 2, 4};
int N = sizeof(arr) / sizeof(arr[0]);
updateArray(arr, N);
return 0;
}
|
Java
import java.util.Stack;
class Main {
public static void updateArray(int[] arr, int n) {
Stack<Integer> st = new Stack<>();
st.push(arr[n - 1]);
for (int i = n - 2; i >= 0; --i) {
int curr = arr[i];
arr[i] = st.peek();
while (!st.empty() && curr > st.peek()) {
st.pop();
}
st.push(curr);
}
arr[n - 1] = -1;
for (int i = 0; i < n; ++i) {
System.out.print(arr[i] + " ");
}
}
public static void main(String[] args) {
int[] arr = {5, 1, 3, 2, 4};
int N = arr.length;
updateArray(arr, N);
}
}
|
Python
def update_array(arr, n):
stack = []
stack.append(arr[n - 1])
for i in range(n - 2, -1, -1):
curr = arr[i]
arr[i] = stack[-1]
while stack and curr > stack[-1]:
stack.pop()
stack.append(curr)
arr[n - 1] = -1
for i in range(n):
print(arr[i])
arr = [5, 1, 3, 2, 4]
N = len(arr)
update_array(arr, N)
|
Javascript
function update_array(arr, n) {
let stack = [];
stack.push(arr[n - 1]);
for (let i = n - 2; i >= 0; i--) {
let curr = arr[i];
arr[i] = stack[stack.length - 1];
while (stack.length && curr > stack[stack.length - 1]) {
stack.pop();
}
stack.push(curr);
}
arr[n - 1] = -1;
for (let i = 0; i < n; i++) {
console.log(arr[i]);
}
}
let arr = [5, 1, 3, 2, 4];
let N = arr.length;
update_array(arr, N);
|
C#
using System;
using System.Collections;
class GFG
{
static void updateArray(int[] arr, int n)
{
Stack st = new Stack();
st.Push(arr[n - 1]);
for (int i = n - 2; i >= 0; --i) {
int curr = arr[i];
arr[i] = Convert.ToInt32(st.Peek());
while ((st.Count!=0) && (curr > Convert.ToInt32(st.Peek()))) {
st.Pop();
}
st.Push(curr);
}
arr[n - 1] = -1;
for (int i = 0; i < n; i++)
Console.Write(arr[i] + " ");
}
public static void Main()
{
int[] arr = { 5, 1, 3, 2, 4 } ;
int N = arr.Length ;
updateArray(arr, N);
}
}
|
Complexity Analysis:
Time Complexity: O(n)
Auxiliary Space: O(n)