给定一个整数k和n个元素的整数数组arr [] ，任务是在修改后的数组中找到最大的子数组总和(通过重复给定数组k次来形成)。例如，如果arr [] = {1，2}且k = 3，则修改后的数组将为{1,2,1,2,1,2,2}。
例子：
Input: arr[] = {1, 2}, k = 3
Output: 9
Modified array will be {1, 2, 1, 2, 1, 2}
And the maximum sub-array sum will be 1 + 2 + 1 + 2 + 1 + 2 = 9
Input: arr[] = {1, -2, 1}, k = 5
Output: 2
一个简单的解决方案是创建一个大小为n * k的数组，然后运行Kadane算法查找最大子数组和。时间复杂度为O(n * k) ，辅助空间为O(n * k) 。
更好的解决方案是计算数组arr []的总和并将其存储在sum中。
- 如果sum <0，则不管K，通过两次串联该数组来计算一个数组的最大子数组和。例如，取arr [] = {1，-4，1}且k =5。该数组的总和小于0。因此，将两个数组并置后可以找到该数组的最大子数组总和。仅与K的值无关，即b [] = {1，-4，1，1，-4，1}和最大子数组总和= 1 + 1 = 2
。
- 如果sum> 0，则最大子数组将包括上一步中计算的最大和(该数组被串联两次)+数组的其余重复(k – 2)也可以包括，因为它们的和大于0将贡献最大的总和。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to concatenate array
void arrayConcatenate(int *arr, int *b,
int k,int len)
{
// Array b will be formed by concatenation
// array a exactly k times
int j = 0;
while (k > 0)
{
for (int i = 0; i < len; i++)
{
b[j++] = arr[i];
}
k--;
}
}
// Function to return the maximum
// subarray sum of arr[]
long maxSubArrSum(int *a,int len)
{
int size = len;
int max_so_far = INT_MIN;
long max_ending_here = 0;
for (int i = 0; i < size; i++)
{
max_ending_here = max_ending_here + a[i];
if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
if (max_ending_here < 0)
max_ending_here = 0;
}
return max_so_far;
}
// Function to return the maximum sub-array
// sum of the modified array
long maxSubKSum(int *arr, int k,int len)
{
int arrSum = 0;
long maxSubArrSum1 = 0;
int b[(2 * len)]={0};
// Concatenating the array 2 times
arrayConcatenate(arr, b, 2,len);
// Finding the sum of the array
for (int i = 0; i < len; i++)
arrSum += arr[i];
// If sum is less than zero
if (arrSum < 0)
maxSubArrSum1 = maxSubArrSum(b,2*len);
// If sum is greater than zero
else
maxSubArrSum1 = maxSubArrSum(b,2*len) +
(k - 2) * arrSum;
return maxSubArrSum1;
}
// Driver code
int main()
{
int arr[] = { 1, -2, 1 };
int arrlen=sizeof(arr)/sizeof(arr[0]);
int k = 5;
cout << maxSubKSum(arr, k,arrlen) << endl;
return 0;
}
// This code is contributed by mits Java
// Java implementation of the approach
public class GFG {
// Function to concatenate array
static void arrayConcatenate(int arr[], int b[],
int k)
{
// Array b will be formed by concatenation
// array a exactly k times
int j = 0;
while (k > 0) {
for (int i = 0; i < arr.length; i++) {
b[j++] = arr[i];
}
k--;
}
}
// Function to return the maximum
// subarray sum of arr[]
static int maxSubArrSum(int a[])
{
int size = a.length;
int max_so_far = Integer.MIN_VALUE,
max_ending_here = 0;
for (int i = 0; i < size; i++) {
max_ending_here = max_ending_here + a[i];
if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
if (max_ending_here < 0)
max_ending_here = 0;
}
return max_so_far;
}
// Function to return the maximum sub-array
// sum of the modified array
static long maxSubKSum(int arr[], int k)
{
int arrSum = 0;
long maxSubArrSum = 0;
int b[] = new int[(2 * arr.length)];
// Concatenating the array 2 times
arrayConcatenate(arr, b, 2);
// Finding the sum of the array
for (int i = 0; i < arr.length; i++)
arrSum += arr[i];
// If sum is less than zero
if (arrSum < 0)
maxSubArrSum = maxSubArrSum(b);
// If sum is greater than zero
else
maxSubArrSum = maxSubArrSum(b) +
(k - 2) * arrSum;
return maxSubArrSum;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, -2, 1 };
int k = 5;
System.out.println(maxSubKSum(arr, k));
}
}
Below is the implementation of the above approach:Python
# Python approach to this problem
# A python module where element
# are added to list k times
def MaxsumArrKtimes(c, ktimes):
# Store element in list d k times
d = c * ktimes
# two variable which can keep
# track of maximum sum seen
# so far and maximum sum ended.
maxsofar = -99999
maxending = 0
for i in d:
maxending = maxending + i
if maxsofar < maxending:
maxsofar = maxending
if maxending < 0:
maxending = 0
return maxsofar
# Get the Maximum sum of element
print(MaxsumArrKtimes([1, -2, 1], 5))
# This code is contributed by AnupGaurav.C#
// C# implementation of the approach
using System;
class GFG
{
// Function to concatenate array
static void arrayConcatenate(int []arr,
int []b, int k)
{
// Array b will be formed by concatenation
// array a exactly k times
int j = 0;
while (k > 0)
{
for (int i = 0; i < arr.Length; i++)
{
b[j++] = arr[i];
}
k--;
}
}
// Function to return the maximum
// subarray sum of arr[]
static int maxSubArrSum(int []a)
{
int size = a.Length;
int max_so_far = int.MinValue,
max_ending_here = 0;
for (int i = 0; i < size; i++)
{
max_ending_here = max_ending_here + a[i];
if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
if (max_ending_here < 0)
max_ending_here = 0;
}
return max_so_far;
}
// Function to return the maximum sub-array
// sum of the modified array
static long maxSubKSum(int []arr, int k)
{
int arrSum = 0;
long maxSubArrsum = 0;
int []b = new int[(2 * arr.Length)];
// Concatenating the array 2 times
arrayConcatenate(arr, b, 2);
// Finding the sum of the array
for (int i = 0; i < arr.Length; i++)
arrSum += arr[i];
// If sum is less than zero
if (arrSum < 0)
maxSubArrsum = maxSubArrSum(b);
// If sum is greater than zero
else
maxSubArrsum = maxSubArrSum(b) +
(k - 2) * arrSum;
return maxSubArrsum;
}
// Driver code
public static void Main()
{
int []arr = { 1, -2, 1 };
int k = 5;
Console.WriteLine(maxSubKSum(arr, k));
}
}
// This code is contributed by RyugaPHP
0)
{
for ($i = 0; $i < sizeof($arr); $i++)
{
$b[$j++] = $arr[$i];
}
$k--;
}
}
// Function to return the maximum
// subarray sum of arr[]
function maxSubArrSum(&$a)
{
$size = sizeof($a);
$max_so_far = 0;
$max_ending_here = 0;
for ($i = 0; $i < $size; $i++)
{
$max_ending_here = $max_ending_here + $a[$i];
if ($max_so_far < $max_ending_here)
$max_so_far = $max_ending_here;
if ($max_ending_here < 0)
$max_ending_here = 0;
}
return $max_so_far;
}
// Function to return the maximum sub-array
// sum of the modified array
function maxSubKSum(&$arr,$k)
{
$arrSum = 0;
$maxSubArrSum = 0;
$b = array_fill(0,(2 * sizeof($arr)),NULL);
// Concatenating the array 2 times
arrayConcatenate($arr, $b, 2);
// Finding the sum of the array
for ($i = 0; $i < sizeof($arr); $i++)
$arrSum += $arr[$i];
// If sum is less than zero
if ($arrSum < 0)
$maxSubArrSum = maxSubArrSum($b);
// If sum is greater than zero
else
$maxSubArrSum = maxSubArrSum($b) +
($k - 2) * $arrSum;
return $maxSubArrSum;
}
// Driver code
$arr = array(1, -2, 1 );
$k = 5;
echo maxSubKSum($arr, $k);
// This code is contributed by Ita_c.
?>输出：
2
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