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📜  移除N个元素后,最大化数组两半之和之间的差

📅  最后修改于: 2021-04-27 21:41:15             🧑  作者: Mango

给定一个整数N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后半部分之间的最大差值。

例子:

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 divyesh072019


C#
// 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)