Find maximum sum subarray of given size

Here given code implementation process.

``````//C Program
//Find maximum sum subarray of given size
#include <stdio.h>

//Print array element
void print_array(int arr[], int size)
{
printf("\n");
for (int i = 0; i < size; i++)
{
printf(" %d ", arr[i]);
}
}
//Find the maximum sum subarray of given length k
//Assuming that the length of subarray k is greater than 2
void max_k_subarray(int arr[], int size, int k)
{
if (size < 2 || size < k || k < 2)
{
//When find invalid inputs
//Here can be three possibilities
//When array contains less than 2 elements, so subarray is not possible
//when array size are less than given subarray length
//Last is when given subarray have less than 2 elements
return;
}
//Set the initial resultant of sub array
int start = 0, last = k - 1;
int sum = 0;
int result = 0;
for (int i = 0; i < size; ++i)
{
if (k > i)
{
//Get the first subarray sum
result += arr[i];
sum = result;
}
else
{
//When possible of more than 1 subarray of given length k
//Remove first element of pervious subarray
result = result - arr[i - k];
result = result + arr[i];
if (result > sum)
{
//Modify the resultant index
start = i - (k - 1);
last = i;
//Set new max sum
sum = result;
}
}
}
//Display array elements
print_array(arr, size);
//Display result
printf("\nMax sum subarray of given length [%d] is %d, are exist between index [%d] to [%d].\n", k, sum, start, last);
}
int main()
{
// Define the array elements
int arr1[] =  {6,8,1,-3,15,5,-4};

// Find the size
int size = sizeof(arr1)/sizeof(arr1[0]);

max_k_subarray(arr1,size,3);

// Define the array elements
int arr2[] ={6,10,-7,2,9,-2,5,1};

// Find the size
size = sizeof(arr2)/sizeof(arr2[0]);

max_k_subarray(arr2,size,4);
return 0;
}``````

Output

`````` 6  8  1  -3  15  5  -4
Max sum subarray of given length [3] is 17, are exist between index [3] to [5].

6  10  -7  2  9  -2  5  1
Max sum subarray of given length [4] is 14, are exist between index [1] to [4].``````
``````/*
Java Program
Find maximum sum subarray of given size
*/
class MyArray
{
//Print array element
void print_array(int[] arr, int size)
{
System.out.print("\n");
for (int i = 0; i < size; i++)
{
System.out.print(" " + arr[i] + " ");
}
}
//Find the maximum sum subarray of given length k
//Assuming that the length of subarray k is greater than 2
void max_k_subarray(int[] arr, int size, int k)
{
if (size < 2 || size < k || k < 2)
{
//When find invalid inputs
//Here can be three possibilities
//When array contains less than 2 elements, so subarray is not possible
//when array size are less than given subarray length
//Last is when given subarray have less than 2 elements
return;
}
//Set the initial resultant of sub array
int start = 0, last = k - 1;
int sum = 0;
int result = 0;
for (int i = 0; i < size; ++i)
{
if (k > i)
{
//Get the first subarray sum
result += arr[i];
sum = result;
}
else
{
//When possible of more than 1 subarray of given length k
//Remove first element of pervious subarray
result = result - arr[i - k];
result = result + arr[i];
if (result > sum)
{
//Modify the resultant index
start = i - (k - 1);
last = i;
//Set new max sum
sum = result;
}
}
}
//Display array elements
print_array(arr, size);
System.out.print("\nMax sum subarray of given length [" + k + "] is " + sum + ", are exist between index [" + start + "] to [" + last + "].\n");
}
public static void main(String[] args)
{
MyArray obj = new MyArray();
// Define the array elements
int[] arr1 = {
6,
8,
1,
-3,
15,
5,
-4
};
// Find the size
int size = arr1.length;
obj.max_k_subarray(arr1, size, 3);
// Define the array elements
int[] arr2 = {
6,
10,
-7,
2,
9,
-2,
5,
1
};
// Find the size
size = arr2.length;
obj.max_k_subarray(arr2, size, 4);
}
}``````

Output

`````` 6  8  1  -3  15  5  -4
Max sum subarray of given length [3] is 17, are exist between index [3] to [5].

6  10  -7  2  9  -2  5  1
Max sum subarray of given length [4] is 14, are exist between index [1] to [4].``````
``````/*
C++ Program
Find maximum sum subarray of given size
*/
#include<iostream>

using namespace std;
class MyArray
{
public:
//Print array element
void print_array(int arr[], int size)
{
cout << "\n";
for (int i = 0; i < size; i++)
{
cout << " " << arr[i] << " ";
}
}
//Find the maximum sum subarray of given length k
//Assuming that the length of subarray k is greater than 2
void max_k_subarray(int arr[], int size, int k)
{
if (size < 2 || size < k || k < 2)
{
//When find invalid inputs
//Here can be three possibilities
//When array contains less than 2 elements, so subarray is not possible
//when array size are less than given subarray length
//Last is when given subarray have less than 2 elements
return;
}
//Set the initial resultant of sub array
int start = 0, last = k - 1;
int sum = 0;
int result = 0;
for (int i = 0; i < size; ++i)
{
if (k > i)
{
//Get the first subarray sum
result += arr[i];
sum = result;
}
else
{
//When possible of more than 1 subarray of given length k
//Remove first element of pervious subarray
result = result - arr[i - k];
result = result + arr[i];
if (result > sum)
{
//Modify the resultant index
start = i - (k - 1);
last = i;
//Set new max sum
sum = result;
}
}
}
//Display array elements
this->print_array(arr, size);
cout << "\nMax sum subarray of given length [" << k << "] is " << sum << ", are exist between index [" << start << "] to [" << last << "].\n";
}
};
int main()
{
MyArray obj;
int arr1[] = {
6,
8,
1,
-3,
15,
5,
-4
};
// Find the size
int size = sizeof(arr1) / sizeof(arr1[0]);
obj.max_k_subarray(arr1, size, 3);
int arr2[] = {
6,
10,
-7,
2,
9,
-2,
5,
1
};
// Find the size
size = sizeof(arr2) / sizeof(arr2[0]);
obj.max_k_subarray(arr2, size, 4);
return 0;
}``````

Output

`````` 6  8  1  -3  15  5  -4
Max sum subarray of given length [3] is 17, are exist between index [3] to [5].

6  10  -7  2  9  -2  5  1
Max sum subarray of given length [4] is 14, are exist between index [1] to [4].``````
``````/*
C# Program
Find maximum sum subarray of given size
*/
using System;
class MyArray
{
//Print array element
void print_array(int[] arr, int size)
{
Console.Write("\n");
for (int i = 0; i < size; i++)
{
Console.Write(" " + arr[i] + " ");
}
}
//Assuming that the length of subarray k is greater than 2
void max_k_subarray(int[] arr, int size, int k)
{
if (size < 2 || size < k || k < 2)
{
return;
}
//Set the initial resultant of sub array
int start = 0, last = k - 1;
int sum = 0;
int result = 0;
for (int i = 0; i < size; i++)
{
if (k > i)
{
//Get the first subarray sum
result += arr[i];
sum = result;
}
else
{
//Remove first element of pervious subarray
result = result - arr[i - k];
result = result + arr[i];
if (result > sum)
{
//Modify the resultant index
start = i - (k - 1);
last = i;
//Set new max sum
sum = result;
}
}
}
//Display array elements
print_array(arr, size);
Console.Write("\nMax sum subarray of given length [" + k + "] is " + sum + ", are exist between index [" + start + "] to [" + last + "].\n");
}
public static void Main(String[] args)
{
MyArray obj = new MyArray();
int[] arr1 = {
6,
8,
1,
-3,
15,
5,
-4
};
// Find the size
int size = arr1.Length;
obj.max_k_subarray(arr1, size, 3);
int[] arr2 = {
6,
10,
-7,
2,
9,
-2,
5,
1
};
// Find the size
size = arr2.Length;
obj.max_k_subarray(arr2, size, 4);
}
}``````

Output

`````` 6  8  1  -3  15  5  -4
Max sum subarray of given length [3] is 17, are exist between index [3] to [5].

6  10  -7  2  9  -2  5  1
Max sum subarray of given length [4] is 14, are exist between index [1] to [4].``````
``````<?php
/*
Php Program
Find maximum sum subarray of given size
*/
class MyArray
{
//Print array element
function print_array( \$arr, \$size)
{
echo "\n";
for (\$i = 0; \$i < \$size; \$i++)
{
echo " ". \$arr[\$i] ." ";
}
}
//Assuming that the length of subarray k is greater than 2
function max_k_subarray( & \$arr, \$size, \$k)
{
if (\$size < 2 || \$size < \$k || \$k < 2)
{
return;
}
//Set the initial resultant of sub array
\$start = 0;
\$last = \$k - 1;
\$sum = 0;
\$result = 0;
for (\$i = 0; \$i < \$size; ++\$i)
{
if (\$k > \$i)
{
//Get the first subarray sum
\$result += \$arr[\$i];
\$sum = \$result;
}
else
{
//Remove first element of pervious subarray
\$result = \$result - \$arr[\$i - \$k];
\$result = \$result + \$arr[\$i];
if (\$result > \$sum)
{
//Modify the resultant index
\$start = \$i - (\$k - 1);
\$last = \$i;
//Set new max sum
\$sum = \$result;
}
}
}
//Display array elements
\$this->print_array(\$arr, \$size);
echo "\nMax sum subarray of given length [". \$k ."] is ". \$sum .", are exist between index [". \$start ."] to [". \$last ."].\n";
}
}

function main()
{
\$obj = new MyArray();
// Define the array elements
\$arr1 = array(6, 8, 1, -3, 15, 5, -4);
// Find the size
\$size = count(\$arr1);
\$obj->max_k_subarray(\$arr1, \$size, 3);
// Define the array elements
\$arr2 = array(6, 10, -7, 2, 9, -2, 5, 1);
// Find the size
\$size = count(\$arr2);
\$obj->max_k_subarray(\$arr2, \$size, 4);
}
main();``````

Output

`````` 6  8  1  -3  15  5  -4
Max sum subarray of given length [3] is 17, are exist between index [3] to [5].

6  10  -7  2  9  -2  5  1
Max sum subarray of given length [4] is 14, are exist between index [1] to [4].``````
``````/*
Node Js Program
Find maximum sum subarray of given size
*/
class MyArray
{
//Print array element
print_array(arr, size)
{
process.stdout.write("\n");
for (var i = 0; i < size; i++)
{
process.stdout.write(" " + arr[i] + " ");
}
}
//Find the maximum sum subarray of given length k
//Assuming that the length of subarray k is greater than 2
max_k_subarray(arr, size, k)
{
if (size < 2 || size < k || k < 2)
{
//When find invalid inputs
//Here can be three possibilities
//When array contains less than 2 elements, so subarray is not possible
//when array size are less than given subarray length
//Last is when given subarray have less than 2 elements
return;
}
//Set the initial resultant of sub array
var start = 0;
var last = k - 1;
var sum = 0;
var result = 0;
for (var i = 0; i < size; ++i)
{
if (k > i)
{
//Get the first subarray sum
result += arr[i];
sum = result;
}
else
{
//When possible of more than 1 subarray of given length k
//Remove first element of pervious subarray
result = result - arr[i - k];
result = result + arr[i];
if (result > sum)
{
//Modify the resultant index
start = i - (k - 1);
last = i;
//Set new max sum
sum = result;
}
}
}
//Display array elements
this.print_array(arr, size);
process.stdout.write("\nMax sum subarray of given length [" + k + "] is " + sum + ", are exist between index [" + start + "] to [" + last + "].\n");
}
}

function main()
{
var obj = new MyArray();
// Define the array elements
var arr1 = [6, 8, 1, -3, 15, 5, -4];
// Find the size
var size = arr1.length;
obj.max_k_subarray(arr1, size, 3);
// Define the array elements
var arr2 = [6, 10, -7, 2, 9, -2, 5, 1];
// Find the size
size = arr2.length;
obj.max_k_subarray(arr2, size, 4);
}
main();``````

Output

`````` 6  8  1  -3  15  5  -4
Max sum subarray of given length [3] is 17, are exist between index [3] to [5].

6  10  -7  2  9  -2  5  1
Max sum subarray of given length [4] is 14, are exist between index [1] to [4].``````
``````#   Python 3 Program
#   Find maximum sum subarray of given size

class MyArray :
# Print array element
def print_array(self, arr, size) :
print("\n", end = "")
i = 0
while (i < size) :
print(" ", arr[i] ," ", end = "")
i += 1

# Assuming that the length of subarray k is greater than 2
# Find the maximum sum subarray of given length k
def max_k_subarray(self, arr, size, k) :
if (size < 2 or size < k or k < 2) :
# Last is when given subarray have less than 2 elements
# when array size are less than given subarray length
# When array contains less than 2 elements, so subarray is not possible
# Here can be three possibilities
# When find invalid inputs
return

# Set the initial resultant of sub array
start = 0
last = k - 1
sum = 0
result = 0
i = 0
while (i < size) :
if (k > i) :
# Get the first subarray sum
result += arr[i]
sum = result
else :
# Remove first element of pervious subarray
# When possible of more than 1 subarray of given length k
result = result - arr[i - k]
result = result + arr[i]
if (result > sum) :
# Modify the resultant index
start = i - (k - 1)
last = i
# Set new max sum
sum = result

i += 1

# Display array elements
self.print_array(arr, size)
print("\nMax sum subarray of given length [", k ,"] is ", sum ,", are exist between index [", start ,"] to [", last ,"].\n", end = "")

def main() :
obj = MyArray()
#  Define the array elements
arr1 = [6, 8, 1, -3, 15, 5, -4]
#  Find the size
size = len(arr1)
obj.max_k_subarray(arr1, size, 3)
#  Define the array elements
arr2 = [6, 10, -7, 2, 9, -2, 5, 1]
#  Find the size
size = len(arr2)
obj.max_k_subarray(arr2, size, 4)

if __name__ == "__main__": main()``````

Output

``````  6    8    1    -3    15    5    -4
Max sum subarray of given length [ 3 ] is  17 , are exist between index [ 3 ] to [ 5 ].

6    10    -7    2    9    -2    5    1
Max sum subarray of given length [ 4 ] is  14 , are exist between index [ 1 ] to [ 4 ].``````
``````#   Ruby Program
#   Find maximum sum subarray of given size

class MyArray

# Print array element
def print_array(arr, size)

print("\n")
i = 0
while (i < size)

print(" ", arr[i] ," ")
i += 1
end
end
# Assuming that the length of subarray k is greater than 2
# Find the maximum sum subarray of given length k
def max_k_subarray(arr, size, k)

if (size < 2 || size < k || k < 2)

# Last is when given subarray have less than 2 elements
# when array size are less than given subarray length
# When array contains less than 2 elements, so subarray is not possible
# Here can be three possibilities
# When find invalid inputs
return
end
# Set the initial resultant of sub array
start = 0
last = k - 1
sum = 0
result = 0
i = 0
while (i < size)

if (k > i)

# Get the first subarray sum
result += arr[i]
sum = result
else

# Remove first element of pervious subarray
# When possible of more than 1 subarray of given length k
result = result - arr[i - k]
result = result + arr[i]
if (result > sum)

# Modify the resultant index
start = i - (k - 1)
last = i
# Set new max sum
sum = result
end
end
i += 1
end
# Display array elements
self.print_array(arr, size)
print("\nMax sum subarray of given length [", k ,"] is ", sum ,", are exist between index [", start ,"] to [", last ,"].\n")
end
end
def main()

obj = MyArray.new()
#  Define the array elements
arr1 = [6, 8, 1, -3, 15, 5, -4]
#  Find the size
size = arr1.length
obj.max_k_subarray(arr1, size, 3)
#  Define the array elements
arr2 = [6, 10, -7, 2, 9, -2, 5, 1]
#  Find the size
size = arr2.length
obj.max_k_subarray(arr2, size, 4)
end
main()``````

Output

`````` 6  8  1  -3  15  5  -4
Max sum subarray of given length [3] is 17, are exist between index [3] to [5].

6  10  -7  2  9  -2  5  1
Max sum subarray of given length [4] is 14, are exist between index [1] to [4].
``````
``````/*
Scala Program
Find maximum sum subarray of given size
*/
class MyArray
{
//Print array element
def print_array(arr: Array[Int], size: Int): Unit = {
print("\n");
var i: Int = 0;
while (i < size)
{
print(" " + arr(i) + " ");
i += 1;
}
}
//Assuming that the length of subarray k is greater than 2
//Find the maximum sum subarray of given length k
def max_k_subarray(arr: Array[Int], size: Int, k: Int): Unit = {
if (size < 2 || size < k || k < 2)
{
//Last is when given subarray have less than 2 elements
//when array size are less than given subarray length
//When array contains less than 2 elements, so subarray is not possible
//Here can be three possibilities
//When find invalid inputs
return;
}
//Set the initial resultant of sub array
var start: Int = 0;
var last: Int = k - 1;
var sum: Int = 0;
var result: Int = 0;
var i: Int = 0;
while (i < size)
{
if (k > i)
{
//Get the first subarray sum
result += arr(i);
sum = result;
}
else
{
//Remove first element of pervious subarray
//When possible of more than 1 subarray of given length k
result = result - arr(i - k);
result = result + arr(i);
if (result > sum)
{
//Modify the resultant index
start = i - (k - 1);
last = i;
//Set new max sum
sum = result;
}
}
i += 1;
}
//Display array elements
print_array(arr, size);
print("\nMax sum subarray of given length [" + k + "] is " + sum + ", are exist between index [" + start + "] to [" + last + "].\n");
}
}
object Main
{
def main(args: Array[String]): Unit = {
var obj: MyArray = new MyArray();
// Define the array elements
var arr1: Array[Int] = Array(6, 8, 1, -3, 15, 5, -4);
// Find the size
var size: Int = arr1.length;
obj.max_k_subarray(arr1, size, 3);
// Define the array elements
var arr2: Array[Int] = Array(6, 10, -7, 2, 9, -2, 5, 1);
// Find the size
size = arr2.length;
obj.max_k_subarray(arr2, size, 4);
}
}``````

Output

`````` 6  8  1  -3  15  5  -4
Max sum subarray of given length [3] is 17, are exist between index [3] to [5].

6  10  -7  2  9  -2  5  1
Max sum subarray of given length [4] is 14, are exist between index [1] to [4].``````
``````/*
Swift Program
Find maximum sum subarray of given size
*/
class MyArray
{
//Print array element
func print_array(_ arr: [Int], _ size: Int)
{
print("\n", terminator: "");
var i: Int = 0;
while (i < size)
{
print(" ", arr[i] ," ", terminator: "");
i += 1;
}
}
//Assuming that the length of subarray k is greater than 2
//Find the maximum sum subarray of given length k
func max_k_subarray(_ arr: [Int], _ size: Int, _ k: Int)
{
if (size < 2 || size < k || k < 2)
{
//Last is when given subarray have less than 2 elements
//when array size are less than given subarray length
//When array contains less than 2 elements, so subarray is not possible
//Here can be three possibilities
//When find invalid inputs
return;
}
//Set the initial resultant of sub array
var start: Int = 0;
var last: Int = k - 1;
var sum: Int = 0;
var result: Int = 0;
var i: Int = 0;
while (i < size)
{
if (k > i)
{
//Get the first subarray sum
result += arr[i];
sum = result;
}
else
{
//Remove first element of pervious subarray
//When possible of more than 1 subarray of given length k
result = result - arr[i - k];
result = result + arr[i];
if (result > sum)
{
//Modify the resultant index
start = i - (k - 1);
last = i;
//Set new max sum
sum = result;
}
}
i += 1;
}
//Display array elements
self.print_array(arr, size);
print("\nMax sum subarray of given length [", k ,"] is ", sum ,", are exist between index [", start ,"] to [", last ,"].\n", terminator: "");
}
}
func main()
{
let obj: MyArray = MyArray();
// Define the array elements
let arr1: [Int] = [6, 8, 1, -3, 15, 5, -4];
// Find the size
var size: Int = arr1.count;
obj.max_k_subarray(arr1, size, 3);
// Define the array elements
let arr2: [Int] = [6, 10, -7, 2, 9, -2, 5, 1];
// Find the size
size = arr2.count;
obj.max_k_subarray(arr2, size, 4);
}
main();``````

Output

``````  6    8    1    -3    15    5    -4
Max sum subarray of given length [ 3 ] is  17 , are exist between index [ 3 ] to [ 5 ].

6    10    -7    2    9    -2    5    1
Max sum subarray of given length [ 4 ] is  14 , are exist between index [ 1 ] to [ 4 ].``````

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