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++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to print the array after removing // k consecutive elements such that the sum // of the remaining elements is maximized void maxSumArr( int arr[], int n, int k)
{ int cur = 0, index = 0;
// Find the sum of first k elements
for ( int i = 0; i < k; i++)
cur += arr[i];
// To store the minimum sum of k
// consecutive elements of the array
int min = cur;
for ( int i = 0; i < n - k; i++) {
// Calculating sum of next k elements
cur = cur - arr[i] + arr[i + k];
// Update the minimum sum so far and the
// index of the first element
if (cur < min) {
cur = min;
index = i + 1;
}
}
// Printing result
for ( int i = 0; i < index; i++)
cout << arr[i] << " " ;
for ( int i = index + k; i < n; i++)
cout << arr[i] << " " ;
} // Driver code 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 implementation of the approach class GFG {
// Function to print the array after removing
// k consecutive elements such that the sum
// of the remaining elements is maximized
static void maxSumArr( int arr[], int n, int k)
{
int cur = 0 , index = 0 ;
// Find the sum of first k elements
for ( int i = 0 ; i < k; i++)
cur += arr[i];
// To store the minimum sum of k
// consecutive elements of the array
int min = cur;
for ( int i = 0 ; i < n - k; i++) {
// Calculating sum of next k elements
cur = cur - arr[i] + arr[i + k];
// Update the minimum sum so far and the
// index of the first element
if (cur < min) {
cur = min;
index = i + 1 ;
}
}
// Printing result
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] + " " );
}
// Driver code
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);
}
} |
# Python3 implementation of the approach # Function to print the array after removing # k consecutive elements such that the sum # of the remaining elements is maximized def maxSumArr(arr, n, k):
cur = 0
index = 0
# Find the sum of first k elements
for i in range (k):
cur + = arr[i]
# To store the minimum sum of k
# consecutive elements of the array
min = cur
for i in range (n - k):
# Calculating sum of next k elements
cur = cur - arr[i] + arr[i + k]
# Update the minimum sum so far and the
# index of the first element
if (cur < min ):
cur = min
index = i + 1
# Printing result
for i in range (index):
print (arr[i], end = " " )
i = index + k
while i < n:
print (arr[i], end = " " )
i + = 1
# Driver code arr = [ - 1 , 1 , 2 , - 3 , 2 , 2 ]
n = len (arr)
k = 3
maxSumArr(arr, n, k) |
// C# implementation of the above approach using System;
class GFG {
// Function to print the array after removing
// k consecutive elements such that the sum
// of the remaining elements is maximized
static void maxSumArr( int [] arr, int n, int k)
{
int cur = 0, index = 0;
// Find the sum of first k elements
for ( int i = 0; i < k; i++)
cur = cur + arr[i];
// To store the minimum sum of k
// consecutive elements of the array
int min = cur;
for ( int i = 0; i < n - k; i++) {
// Calculating sum of next k elements
cur = (cur - arr[i]) + (arr[i + k]);
// Update the minimum sum so far and the
// index of the first element
if (cur < min) {
cur = min;
index = i + 1;
}
}
// Printing result
for ( int i = 0; i < index; i++)
Console.Write(arr[i] + " " );
for ( int i = index + k; i < n; i++)
Console.Write(arr[i] + " " );
}
// Driver code
static public void Main()
{
int [] arr = { -1, 1, 2, -3, 2, 2 };
int n = arr.Length;
int k = 3;
maxSumArr(arr, n, k);
}
} // This code is contributed by ajit.. |
<script> // Javascript implementation of the above approach
// Function to print the array after removing
// k consecutive elements such that the sum
// of the remaining elements is maximized
function maxSumArr(arr, n, k)
{
let cur = 0, index = 0;
// Find the sum of first k elements
for (let i = 0; i < k; i++)
cur = cur + arr[i];
// To store the minimum sum of k
// consecutive elements of the array
let min = cur;
for (let i = 0; i < n - k; i++)
{
// Calculating sum of next k elements
cur = (cur - arr[i]) + (arr[i + k]);
// Update the minimum sum so far and the
// index of the first element
if (cur < min)
{
cur = min;
index = i + 1;
}
}
// Printing result
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> |
-1 2 2
Time Complexity: O(n)
Auxiliary space: O(1)