给定N个整数的数组arr [] 。任务是首先找到最大子阵列总和，然后从子阵列中移除最多一个元素。如果存在多个子数组，且子数组的总和最大，则最多移除单个元素，以使移除后的最大总和最大化。任务是使移除后获得的总和最大化。
注意：您必须首先找到最大子数组总和，然后在必要时从该子数组中删除该元素。同样在移除后，子阵列的大小至少应为1。
例子：
Input: arr[] = {1, 2, 3, -2, 3}
Output: 9
The maximum sub-array sum is given by the sub-array {2, 3, -2, 3}
Hence, we can remove -2 to further maximize the sub-array sum.
Input: arr[] = {-1, -2}
Output: -1
The maximum sub-array sum is from the sub-array {-1} and no removal is required.
方法：使用Kadane算法查找最大子数组和。找到总和后，请重新应用Kadane的算法，以稍作更改再次找到最大和。在循环中使用两个额外的变量cnt和mini 。变量cnt对子数组中的元素数进行计数，并且mini将最小值存储在所有子数组中，这些子数组的和与最大子数组的总和相同。如果由此获得的最小元素小于0 ，那么只有我们从子数组中删除一个元素，否则我们就不这样做。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the maximum sub-array sum
int maxSubArraySum(int a[], int size)
{
// Initialized
int max_so_far = INT_MIN, max_ending_here = 0;
// Traverse in the array
for (int i = 0; i < size; i++) {
// Increase the sum
max_ending_here = max_ending_here + a[i];
// If sub-array sum is more than the previous
if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
// If sum is negative
if (max_ending_here < 0)
max_ending_here = 0;
}
return max_so_far;
}
// Function that returns the maximum sub-array sum
// after removing an element from the same sub-array
int maximizeSum(int a[], int n)
{
int cnt = 0;
int mini = INT_MAX;
int minSubarray = INT_MAX;
// Maximum sub-array sum using Kadane's Algorithm
int sum = maxSubArraySum(a, n);
int max_so_far = INT_MIN, max_ending_here = 0;
// Re-apply Kadane's with minor changes
for (int i = 0; i < n; i++) {
// Increase the sum
max_ending_here = max_ending_here + a[i];
cnt++;
minSubarray = min(a[i], minSubarray);
// If sub-array sum is greater than the previous
if (sum == max_ending_here) {
// If elements are 0, no removal
if (cnt == 1)
mini = min(mini, 0);
// If elements are more, then store
// the minimum value in the sub-array
// obtained till now
else
mini = min(mini, minSubarray);
}
// If sum is negative
if (max_ending_here < 0) {
// Re-initialize everything
max_ending_here = 0;
cnt = 0;
minSubarray = INT_MAX;
}
}
return sum - mini;
}
// Driver code
int main()
{
int a[] = { 1, 2, 3, -2, 3 };
int n = sizeof(a) / sizeof(a[0]);
cout << maximizeSum(a, n);
return 0;
} Java
// Java implementation of the approach
class GFG
{
// Function to return the maximum sub-array sum
static int maxSubArraySum(int a[], int size)
{
// Initialized
int max_so_far = Integer.MIN_VALUE,
max_ending_here = 0;
// Traverse in the array
for (int i = 0; i < size; i++)
{
// Increase the sum
max_ending_here = max_ending_here + a[i];
// If sub-array sum is more than the previous
if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
// If sum is negative
if (max_ending_here < 0)
max_ending_here = 0;
}
return max_so_far;
}
// Function that returns the maximum sub-array sum
// after removing an element from the same sub-array
static int maximizeSum(int a[], int n)
{
int cnt = 0;
int mini = Integer.MAX_VALUE;
int minSubarray = Integer.MAX_VALUE;
// Maximum sub-array sum
// using Kadane's Algorithm
int sum = maxSubArraySum(a, n);
int max_so_far = Integer.MIN_VALUE,
max_ending_here = 0;
// Re-apply Kadane's with minor changes
for (int i = 0; i < n; i++)
{
// Increase the sum
max_ending_here = max_ending_here + a[i];
cnt++;
minSubarray = Math.min(a[i], minSubarray);
// If sub-array sum is greater than the previous
if (sum == max_ending_here)
{
// If elements are 0, no removal
if (cnt == 1)
mini = Math.min(mini, 0);
// If elements are more, then store
// the minimum value in the sub-array
// obtained till now
else
mini = Math.min(mini, minSubarray);
}
// If sum is negative
if (max_ending_here < 0)
{
// Re-initialize everything
max_ending_here = 0;
cnt = 0;
minSubarray = Integer.MAX_VALUE;
}
}
return sum - mini;
}
// Driver code
public static void main(String[] args)
{
int a[] = { 1, 2, 3, -2, 3 };
int n = a.length;
System.out.println(maximizeSum(a, n));
}
}
// This code is contributed by Code_MechPython3
# Python3 implementation of the approach
import sys;
# Function to return the maximum sub-array sum
def maxSubArraySum(a, size) :
# Initialized
max_so_far = -(sys.maxsize - 1);
max_ending_here = 0;
# Traverse in the array
for i in range(size) :
# Increase the sum
max_ending_here = max_ending_here + a[i];
# If sub-array sum is more than the previous
if (max_so_far < max_ending_here) :
max_so_far = max_ending_here;
# If sum is negative
if (max_ending_here < 0) :
max_ending_here = 0;
return max_so_far;
# Function that returns the maximum
# sub-array sum after removing an
# element from the same sub-array
def maximizeSum(a, n) :
cnt = 0;
mini = sys.maxsize;
minSubarray = sys.maxsize;
# Maximum sub-array sum using
# Kadane's Algorithm
sum = maxSubArraySum(a, n);
max_so_far = -(sys.maxsize - 1);
max_ending_here = 0;
# Re-apply Kadane's with minor changes
for i in range(n) :
# Increase the sum
max_ending_here = max_ending_here + a[i];
cnt += 1;
minSubarray = min(a[i], minSubarray);
# If sub-array sum is greater
# than the previous
if (sum == max_ending_here) :
# If elements are 0, no removal
if (cnt == 1) :
mini = min(mini, 0);
# If elements are more, then store
# the minimum value in the sub-array
# obtained till now
else :
mini = min(mini, minSubarray);
# If sum is negative
if (max_ending_here < 0) :
# Re-initialize everything
max_ending_here = 0;
cnt = 0;
minSubarray = sys.maxsize;
return sum - mini;
# Driver code
if __name__ == "__main__" :
a = [ 1, 2, 3, -2, 3 ];
n = len(a)
print(maximizeSum(a, n));
# This code is contributed by RyugaC#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the maximum sub-array sum
static int maxSubArraySum(int []a, int size)
{
// Initialized
int max_so_far = int.MinValue,
max_ending_here = 0;
// Traverse in the array
for (int i = 0; i < size; i++)
{
// Increase the sum
max_ending_here = max_ending_here + a[i];
// If sub-array sum is more than the previous
if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
// If sum is negative
if (max_ending_here < 0)
max_ending_here = 0;
}
return max_so_far;
}
// Function that returns the maximum sub-array sum
// after removing an element from the same sub-array
static int maximizeSum(int []a, int n)
{
int cnt = 0;
int mini = int.MaxValue;
int minSubarray = int.MaxValue;
// Maximum sub-array sum
// using Kadane's Algorithm
int sum = maxSubArraySum(a, n);
int max_so_far = int.MinValue,
max_ending_here = 0;
// Re-apply Kadane's with minor changes
for (int i = 0; i < n; i++)
{
// Increase the sum
max_ending_here = max_ending_here + a[i];
cnt++;
minSubarray = Math.Min(a[i], minSubarray);
// If sub-array sum is greater than the previous
if (sum == max_ending_here)
{
// If elements are 0, no removal
if (cnt == 1)
mini = Math.Min(mini, 0);
// If elements are more, then store
// the minimum value in the sub-array
// obtained till now
else
mini = Math.Min(mini, minSubarray);
}
// If sum is negative
if (max_ending_here < 0)
{
// Re-initialize everything
max_ending_here = 0;
cnt = 0;
minSubarray = int.MaxValue;
}
}
return sum - mini;
}
// Driver code
public static void Main(String[] args)
{
int []a = { 1, 2, 3, -2, 3 };
int n = a.Length;
Console.WriteLine(maximizeSum(a, n));
}
}
// This code has been contributed by 29AjayKumarPHP
输出：
9