Maximum sum after K consecutive deletions
Last Updated :
24 Oct, 2022
Given an array arr[] of size N and an integer K, the task is to delete K continuous elements from the array such that the sum of the remaining element is maximum. Here we need to print the remaining elements of the array.
Examples:
Input: arr[] = {-1, 1, 2, -3, 2, 2}, K = 3
Output: -1 2 2
Delete 1, 2, -3 and the sum of the remaining
elements will be 3 which is maximum possible.
Input: arr[] = {1, 2, -3, 4, 5}, K = 1
Output: 1 2 4 5
Approach: Calculate the sum of k-consecutive elements and remove the elements with the minimum sum. Print the rest of the elements of the array.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void maxSumArr( int arr[], int n, int k)
{
int cur = 0, index = 0;
for ( int i = 0; i < k; i++)
cur += arr[i];
int min = cur;
for ( int i = 0; i < n - k; i++) {
cur = cur - arr[i] + arr[i + k];
if (cur < min) {
cur = min;
index = i + 1;
}
}
for ( int i = 0; i < index; i++)
cout << arr[i] << " " ;
for ( int i = index + k; i < n; i++)
cout << arr[i] << " " ;
}
int main()
{
int arr[] = { -1, 1, 2, -3, 2, 2 };
int n = sizeof (arr) / sizeof (arr[0]);
int k = 3;
maxSumArr(arr, n, k);
return 0;
}
|
Java
class GFG {
static void maxSumArr( int arr[], int n, int k)
{
int cur = 0 , index = 0 ;
for ( int i = 0 ; i < k; i++)
cur += arr[i];
int min = cur;
for ( int i = 0 ; i < n - k; i++) {
cur = cur - arr[i] + arr[i + k];
if (cur < min) {
cur = min;
index = i + 1 ;
}
}
for ( int i = 0 ; i < index; i++)
System.out.print(arr[i] + " " );
for ( int i = index + k; i < n; i++)
System.out.print(arr[i] + " " );
}
public static void main(String[] args)
{
int arr[] = { - 1 , 1 , 2 , - 3 , 2 , 2 };
int n = arr.length;
int k = 3 ;
maxSumArr(arr, n, k);
}
}
|
Python
def maxSumArr(arr, n, k):
cur = 0
index = 0
for i in range (k):
cur + = arr[i]
min = cur
for i in range (n - k):
cur = cur - arr[i] + arr[i + k]
if (cur < min ):
cur = min
index = i + 1
for i in range (index):
print (arr[i], end = " " )
i = index + k
while i < n:
print (arr[i], end = " " )
i + = 1
arr = [ - 1 , 1 , 2 , - 3 , 2 , 2 ]
n = len (arr)
k = 3
maxSumArr(arr, n, k)
|
C#
using System;
class GFG {
static void maxSumArr( int [] arr, int n, int k)
{
int cur = 0, index = 0;
for ( int i = 0; i < k; i++)
cur = cur + arr[i];
int min = cur;
for ( int i = 0; i < n - k; i++) {
cur = (cur - arr[i]) + (arr[i + k]);
if (cur < min) {
cur = min;
index = i + 1;
}
}
for ( int i = 0; i < index; i++)
Console.Write(arr[i] + " " );
for ( int i = index + k; i < n; i++)
Console.Write(arr[i] + " " );
}
static public void Main()
{
int [] arr = { -1, 1, 2, -3, 2, 2 };
int n = arr.Length;
int k = 3;
maxSumArr(arr, n, k);
}
}
|
Javascript
<script>
function maxSumArr(arr, n, k)
{
let cur = 0, index = 0;
for (let i = 0; i < k; i++)
cur = cur + arr[i];
let min = cur;
for (let i = 0; i < n - k; i++)
{
cur = (cur - arr[i]) + (arr[i + k]);
if (cur < min)
{
cur = min;
index = i + 1;
}
}
for (let i = 0; i < index; i++)
document.write(arr[i] + " " );
for (let i = index + k; i < n; i++)
document.write(arr[i] + " " );
}
let arr = [ -1, 1, 2, -3, 2, 2 ];
let n = arr.length;
let k = 3;
maxSumArr(arr, n, k);
</script>
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Time Complexity: O(n)
Auxiliary space: O(1)
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