Maximize first element of Array by deleting first or adding a previously deleted element
Last Updated :
08 Apr, 2022
Given an array arr[] of size N, and an integer K, the task is to maximize the first element of the array in K operations where in each operation:
- If the array is not empty, remove the topmost element of the array.
- Add any one of the previously removed element back at the starting of the array.
Examples:
Input: arr[] = [5, 2, 2, 4, 0, 6], K = 4
Output: 5
Explanation: The 4 operations are as:
- Remove the topmost element = 5. The arr becomes [2, 2, 4, 0, 6].
- Remove the topmost element = 2. The arr becomes [2, 4, 0, 6].
- Remove the topmost element = 2. The arr becomes [4, 0, 6].
- Add 5 back onto the arr. The arr becomes [5, 4, 0, 6].
Here 5 is the largest answer possible after 4 moves.
Input: arr[] = [2], K = 1
Output: -1
Explanation: Only one move can be applied and in the first move.
The only option is to remove the first element of the arr[].
If that is done the array becomes empty. So answer is -1
Approach: This problem can be solved with the help of the Greedy approach based on the following idea:
In first K-1 operations, the K-1 value of the starting can be removed. So currently at Kth node. Now in the last operation there are two possible choice:
- Either remove the current starting node (optimal if the value of the (K+1)th node is greater than the largest amongst first K-1 already removed elements)
- Add the largest from the already removed K-1 elements (optimal when the (K+1)th node has less value than this largest one)
Follow the illustration shown below for a better understanding.
Illustration:
For example arr[] = {5, 2, 2, 4, 0, 6}, K = 4
1st Operation:
=> Remove 5. arr[] = {2, 2, 4, 0, 6}
=> maximum = 5, K = 4 – 1 = 3
2nd Operation:
=> Remove 2. arr[] = {2, 4, 0, 6}
=> maximum = max (5, 2) = 5, K = 3 – 1 = 2
3rd Operation:
=> Remove 2. arr[] = {4, 0, 6}
=> maximum = max (5, 2) = 5, K = 2 – 1 = 1
4th Operation:
=> Here the current 2nd element i.e. 0 is less than 5.
=> So add 5 back in the array. arr[] = {5, 4, 0, 6}
=> maximum = max (5, 0) = 5, K = 1 – 1 = 0
Therefore the maximum possible first element is 5.
Follow the steps to solve the problem:
- If K = 0, then return first node value.
- If K = 1, then return the second node value(if any) else return -1 (because after K operations the list does not exist).
- If the size of the linked list is one then in every odd operation (i.e. 1, 3, 5, . . . ), return -1, else return the first node value (because if performed odd operation then array will become empty).
- If K > 2, then:
- Traverse first K-1 nodes and find out the maximum value.
- Compare that maximum value with the (K+1)th node value.
- If (K+1)th value is greater than the previous maximum value, update it with (K+1)th Node value. Otherwise, don’t update the maximum value.
- Return the maximum value.
Below is the implementation of the above approach :
C++
#include <iostream>
using namespace std;
int maximumTopMost( int arr[], int k, int N){
if (N == 1 and k % 2 != 0)
return -1;
int ans = -1;
for ( int i = 0; i < min(N, k - 1); i++)
ans = max(ans, arr[i]);
if (k < N)
ans = max(ans, arr[k]);
return ans;
}
int main() {
int arr[] = {5, 2, 2, 4, 0, 6};
int N = 6;
int K = 4;
cout <<(maximumTopMost(arr, K, N));
return 0;
}
|
Java
class GFG {
static int maximumTopMost( int [] arr, int k, int N)
{
if (N == 1 && k % 2 != 0 )
return - 1 ;
int ans = - 1 ;
for ( int i = 0 ; i < Math.min(N, k - 1 ); i++)
ans = Math.max(ans, arr[i]);
if (k < N)
ans = Math.max(ans, arr[k]);
return ans;
}
public static void main(String[] args)
{
int [] arr = { 5 , 2 , 2 , 4 , 0 , 6 };
int N = 6 ;
int K = 4 ;
System.out.println(maximumTopMost(arr, K, N));
}
}
|
Python3
def maximumTopMost(arr, k):
if len (arr) = = 1 and k % 2 ! = 0 :
return - 1
ans = - 1
for i in range ( min ( len (arr), k - 1 )):
ans = max (ans, arr[i])
if k < len (arr):
ans = max (ans, arr[k])
return ans
if __name__ = = "__main__" :
arr = [ 5 , 2 , 2 , 4 , 0 , 6 ]
K = 4
print (maximumTopMost(arr, K))
|
C#
using System;
class GFG {
static int maximumTopMost( int [] arr, int k, int N)
{
if (N == 1 && k % 2 != 0)
return -1;
int ans = -1;
for ( int i = 0; i < Math.Min(N, k - 1); i++)
ans = Math.Max(ans, arr[i]);
if (k < N)
ans = Math.Max(ans, arr[k]);
return ans;
}
public static void Main()
{
int [] arr = { 5, 2, 2, 4, 0, 6 };
int N = 6;
int K = 4;
Console.Write(maximumTopMost(arr, K, N));
}
}
|
Javascript
<script>
const maximumTopMost = (arr, k) => {
if (arr.length == 1 && k % 2 != 0)
return -1;
let ans = -1;
for (let i = 0; i < Math.min(arr.length, k - 1); ++i)
ans = Math.max(ans, arr[i]);
if (k < arr.length)
ans = Math.max(ans, arr[k]);
return ans;
}
let arr = [5, 2, 2, 4, 0, 6];
let K = 4;
document.write(maximumTopMost(arr, K));
</script>
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Time Complexity: O(N)
Auxiliary Space: O(1)
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