给定正整数的数组arr [] 。任务是反转子数组，以使偶数位置的元素总数最小，并打印最小总数。
注意：仅执行一次移动。子数组可能不会反转。
例子：
Input: arr[] = {1, 2, 3, 4, 5}
Output: 7
Explanation:
Sum of elements at even positions initially = arr[0] + arr[2] + arr[4] = 1 + 3 + 5 = 9
On reversing the subarray from position [1, 4], the array becomes: {1, 5, 4, 3, 2}
Now the sum of elements at even positions = arr[0] + arr[2] + arr[4] = 1 + 4 + 2 = 7, which is the minimum sum.
Input: arr[] = {0, 1, 4, 3}
Output: 1
Explanation:
Sum of elements at even positions initially = arr[0] + arr[2] = 0 + 4 = 4
On reversing the subarray from position [1, 2], the array becomes: {0, 4, 1, 3}
Now the sum of elements at even positions = arr[0] + arr[2] = 0 + 1 = 1, which is the minimum sum.
天真的方法：想法是应用蛮力方法并生成所有子数组，并在偶数位置检查元素的总和。打印总数中的最小值。
时间复杂度： O(N 3 )
辅助空间： O(N)
高效方法：这个想法是要观察数组arr []的以下要点：
- 反转奇数长度的子数组不会更改总和，因为偶数索引处的所有元素都将再次变为偶数索引。
- 反转偶数长度的子数组将使索引位置处的所有元素均到达偶数索引，而偶数索引处的元素均变为奇数索引。
- 仅当将奇数索引元素置于偶数索引元素时，元素的总和才会更改。假设索引1的元素可以转到索引0或索引2。
- 从偶数索引转换为索引2到4的另一个偶数索引示例时，它将在第一种情况下覆盖；如果从奇数索引到索引1至4的偶数索引示例，则将覆盖第二种情况。
以下是基于上述观察结果的方法步骤：
- 因此，想法是仅找到偶数索引元素的总和，并初始化两个数组，假设v1和v2使得如果第一个索引元素变为0，v1将保留更改说明，而如果第一个索引元素变为v2将保留更改说明至2。
- 获取两个值中的最小值，然后检查它是否小于0。如果是，则将其添加到答案中，最后返回答案。
下面是上述方法的实现：
C++
// C++ implementation to reverse a subarray
// of the given array to minimize the
// sum of elements at even position
#include
#define N 5
using namespace std;
// Function that will give
// the max negative value
int after_rev(vector v)
{
int mini = 0, count = 0;
for (int i = 0; i < v.size(); i++) {
count += v[i];
// Check for count
// graeter than 0
// as we require only
// negative solution
if (count > 0)
count = 0;
if (mini > count)
mini = count;
}
return mini;
}
// Function to print the minimum sum
void print(int arr[N])
{
int sum = 0;
// Taking sum of only
// even index elements
for (int i = 0; i < N; i += 2)
sum += arr[i];
// Initialize two vectors v1, v2
vector v1, v2;
// v1 will keep accout for change
// if 1th index element goes to 0
for (int i = 0; i + 1 < N; i += 2)
v1.push_back(arr[i + 1] - arr[i]);
// v2 will keep account for change
// if 1th index element goes to 2
for (int i = 1; i + 1 < N; i += 2)
v2.push_back(arr[i] - arr[i + 1]);
// Get the max negative value
int change = min(after_rev(v1),
after_rev(v2));
if (change < 0)
sum += change;
cout << sum << endl;
}
// Driver code
int main()
{
int arr[N] = { 0, 1, 4, 3 };
print(arr);
return 0;
} Java
// Java implementation to reverse a subarray
// of the given array to minimize the
// sum of elements at even position
import java.util.*;
class GFG{
static final int N = 5;
// Function that will give
// the max negative value
static int after_rev(Vector v)
{
int mini = 0, count = 0;
for(int i = 0; i < v.size(); i++)
{
count += v.get(i);
// Check for count graeter
// than 0 as we require only
// negative solution
if (count > 0)
count = 0;
if (mini > count)
mini = count;
}
return mini;
}
// Function to print the minimum sum
static void print(int arr[])
{
int sum = 0;
// Taking sum of only
// even index elements
for(int i = 0; i < N; i += 2)
sum += arr[i];
// Initialize two vectors v1, v2
Vector v1, v2;
v1 = new Vector();
v2 = new Vector();
// v1 will keep accout for change
// if 1th index element goes to 0
for(int i = 0; i + 1 < N; i += 2)
v1.add(arr[i + 1] - arr[i]);
// v2 will keep account for change
// if 1th index element goes to 2
for(int i = 1; i + 1 < N; i += 2)
v2.add(arr[i] - arr[i + 1]);
// Get the max negative value
int change = Math.min(after_rev(v1),
after_rev(v2));
if (change < 0)
sum += change;
System.out.print(sum + "\n");
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 0, 1, 4, 3, 0 };
print(arr);
}
}
// This code is contributed by 29AjayKumar Python3
# Python3 implementation to reverse
# a subarray of the given array to
# minimize the sum of elements at
# even position
# Function that will give
# the max negative value
def after_rev(v):
mini = 0
count = 0
for i in range(len(v)):
count += v[i]
# Check for count graeter
# than 0 as we require only
# negative solution
if(count > 0):
count = 0
if(mini > count):
mini = count
return mini
# Function to print the
# minimum sum
def print_f(arr):
sum = 0
# Taking sum of only
# even index elements
for i in range(0, len(arr), 2):
sum += arr[i]
# Initialize two vectors v1, v2
v1, v2 = [], []
# v1 will keep accout for change
# if 1th index element goes to 0
i = 1
while i + 1 < len(arr):
v1.append(arr[i + 1] - arr[i])
i += 2
# v2 will keep account for change
# if 1th index element goes to 2
i = 1
while i + 1 < len(arr):
v2.append(arr[i] - arr[i + 1])
i += 2
# Get the max negative value
change = min(after_rev(v1),
after_rev(v2))
if(change < 0):
sum += change
print(sum)
# Driver code
if __name__ == '__main__':
arr = [ 0, 1, 4, 3 ]
print_f(arr)
# This code is contributed by Shivam SinghC#
// C# implementation to reverse a subarray
// of the given array to minimize the
// sum of elements at even position
using System;
using System.Collections.Generic;
class GFG{
static readonly int N = 5;
// Function that will give
// the max negative value
static int after_rev(List v)
{
int mini = 0, count = 0;
for(int i = 0; i < v.Count; i++)
{
count += v[i];
// Check for count graeter
// than 0 as we require only
// negative solution
if (count > 0)
count = 0;
if (mini > count)
mini = count;
}
return mini;
}
// Function to print the minimum sum
static void print(int []arr)
{
int sum = 0;
// Taking sum of only
// even index elements
for(int i = 0; i < N; i += 2)
sum += arr[i];
// Initialize two vectors v1, v2
List v1, v2;
v1 = new List();
v2 = new List();
// v1 will keep accout for change
// if 1th index element goes to 0
for(int i = 0; i + 1 < N; i += 2)
v1.Add(arr[i + 1] - arr[i]);
// v2 will keep account for change
// if 1th index element goes to 2
for(int i = 1; i + 1 < N; i += 2)
v2.Add(arr[i] - arr[i + 1]);
// Get the max negative value
int change = Math.Min(after_rev(v1),
after_rev(v2));
if (change < 0)
sum += change;
Console.Write(sum + "\n");
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 0, 1, 4, 3, 0 };
print(arr);
}
}
// This code is contributed by sapnasingh4991 输出：
1
时间复杂度： O(N)
辅助空间： O(N)