给定一个整数N和由3 * N个整数组成的数组arr [] ，任务是在从数组中删除正好N个元素后，找到数组的前半部分和后半部分之间的最大差值。
例子：
Input: N = 2, arr[] = {3, 1, 4, 1, 5, 9}
Output: 1
Explanation:
Removal of arr[1] and arr[5] from the array maximizes the difference = (3 + 4) – (1 + 5) = 7 – 6 = 1.
Input: N = 1, arr[] = {1, 2, 3}
Output: -1
方法：
请按照以下步骤解决问题
- 从数组的开头开始遍历数组，并从数组的开头继续更新最大N个元素的总和。
- 同样，从数组末尾继续更新最小的N个元素的总和。
- 遍历这些和并计算每个点的差异并更新获得的最大差异。
- 打印获得的最大差异。
下面是上述方法的实现：
C++
// C++ Program to implement
// the above approach
#include
using namespace std;
// Function to print the maximum difference
// possible between the two halves of the array
long long FindMaxDif(vector a, int m)
{
int n = m / 3;
vector l(m + 5), r(m + 5);
// Stores n maximum values from the start
multiset s;
for (int i = 1; i <= m; i++) {
// Insert first n elements
if (i <= n) {
// Update sum of largest n
// elements from left
l[i] = a[i - 1] + l[i - 1];
s.insert(a[i - 1]);
}
// For the remaining elements
else {
l[i] = l[i - 1];
// Obtain minimum value
// in the set
long long d = *s.begin();
// Insert only if it is greater
// than minimum value
if (a[i - 1] > d) {
// Update sum from left
l[i] -= d;
l[i] += a[i - 1];
// Remove the minimum
s.erase(s.find(d));
// Insert the current element
s.insert(a[i - 1]);
}
}
}
// Clear the set
s.clear();
// Store n minimum elements from the end
for (int i = m; i >= 1; i--) {
// Insert the last n elements
if (i >= m - n + 1) {
// Update sum of smallest n
// elements from right
r[i] = a[i - 1] + r[i + 1];
s.insert(a[i - 1]);
}
// For the remaining elements
else {
r[i] = r[i + 1];
// Obtain the minimum
long long d = *s.rbegin();
// Insert only if it is smaller
// than maximum value
if (a[i - 1] < d) {
// Update sum from right
r[i] -= d;
r[i] += a[i - 1];
// Remove the minimum
s.erase(s.find(d));
// Insert the new element
s.insert(a[i - 1]);
}
}
}
long long ans = -9e18L;
for (int i = n; i <= m - n; i++) {
// Compare the difference and
// store the maximum
ans = max(ans, l[i] - r[i + 1]);
}
// Return the maximum
// possible difference
return ans;
}
// Driver Code
int main()
{
vector vtr = { 3, 1, 4, 1, 5, 9 };
int n = vtr.size();
cout << FindMaxDif(vtr, n);
return 0;
} Python3
# Python3 Program to implement
# the above approach
# Function to print the maximum difference
# possible between the two halves of the array
def FindMaxDif(a, m) :
n = m // 3
l = [0] * (m + 5)
r = [0] * (m + 5)
# Stores n maximum values from the start
s = []
for i in range(1, m + 1) :
# Insert first n elements
if (i <= n) :
# Update sum of largest n
# elements from left
l[i] = a[i - 1] + l[i - 1]
s.append(a[i - 1])
# For the remaining elements
else :
l[i] = l[i - 1]
# Obtain minimum value
# in the set
s.sort()
d = s[0]
# Insert only if it is greater
# than minimum value
if (a[i - 1] > d) :
# Update sum from left
l[i] -= d
l[i] += a[i - 1]
# Remove the minimum
s.remove(d)
# Insert the current element
s.append(a[i - 1])
# Clear the set
s.clear()
# Store n minimum elements from the end
for i in range(m, 0, -1) :
# Insert the last n elements
if (i >= m - n + 1) :
# Update sum of smallest n
# elements from right
r[i] = a[i - 1] + r[i + 1]
s.append(a[i - 1])
# For the remaining elements
else :
r[i] = r[i + 1]
s.sort()
# Obtain the minimum
d = s[-1]
# Insert only if it is smaller
# than maximum value
if (a[i - 1] < d) :
# Update sum from right
r[i] -= d
r[i] += a[i - 1]
# Remove the minimum
s.remove(d)
# Insert the new element
s.append(a[i - 1])
ans = -9e18
for i in range(n, m - n + 1) :
# Compare the difference and
# store the maximum
ans = max(ans, l[i] - r[i + 1])
# Return the maximum
# possible difference
return ans
# Driver code
vtr = [ 3, 1, 4, 1, 5, 9 ]
n = len(vtr)
print(FindMaxDif(vtr, n))
# This code is contributed by divyesh072019C#
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to print the maximum difference
// possible between the two halves of the array
static long FindMaxDif(List a, int m)
{
int n = m / 3;
long[] l = new long[m + 5];
long[] r = new long[m + 5];
// Stores n maximum values from the start
List s = new List();
for(int i = 1; i <= m; i++)
{
// Insert first n elements
if (i <= n)
{
// Update sum of largest n
// elements from left
l[i] = a[i - 1] + l[i - 1];
s.Add(a[i - 1]);
}
// For the remaining elements
else
{
l[i] = l[i - 1];
s.Sort();
// Obtain minimum value
// in the set
long d = s[0];
// Insert only if it is greater
// than minimum value
if (a[i - 1] > d)
{
// Update sum from left
l[i] -= d;
l[i] += a[i - 1];
// Remove the minimum
s.Remove(d);
// Insert the current element
s.Add(a[i - 1]);
}
}
}
// Clear the set
s.Clear();
// Store n minimum elements from the end
for(int i = m; i >= 1; i--)
{
// Insert the last n elements
if (i >= m - n + 1)
{
// Update sum of smallest n
// elements from right
r[i] = a[i - 1] + r[i + 1];
s.Add(a[i - 1]);
}
// For the remaining elements
else
{
r[i] = r[i + 1];
s.Sort();
// Obtain the minimum
long d = s[s.Count - 1];
// Insert only if it is smaller
// than maximum value
if (a[i - 1] < d)
{
// Update sum from right
r[i] -= d;
r[i] += a[i - 1];
// Remove the minimum
s.Remove(d);
// Insert the new element
s.Add(a[i - 1]);
}
}
}
long ans = (long)(-9e18);
for(int i = n; i <= m - n; i++)
{
// Compare the difference and
// store the maximum
ans = Math.Max(ans, l[i] - r[i + 1]);
}
// Return the maximum
// possible difference
return ans;
}
// Driver Code
static void Main()
{
List vtr = new List(
new long[]{ 3, 1, 4, 1, 5, 9 });
int n = vtr.Count;
Console.Write(FindMaxDif(vtr, n));
}
}
// This code is contributed by divyeshrabadiya07 输出：
1时间复杂度： O(NlogN)
辅助空间： O(N)