给定一个由N个整数组成的数组arr [] ，任务是使一个元素的绝对差之和与数组中其余元素的差最大。
例子：
Input: arr[] = {1, 2, 4, 7}
Output: 6
Explanation:
For i = 1, |1 – 2| + |1 – 4| + |1 – 7| = 1 + 3 + 6 =10.
For i = 2, |2 – 1| + |2 – 4| + |2 – 7| = 1 + 2 + 5 = 8.
For i = 3, |4 – 1| + |4 – 2| + |4 – 7| = 3 + 2 + 3 = 8.
For i = 4, |7 – 1| + |7 – 2| + |7 – 4| = 6 + 5 + 3 = 14.
Maximum=14, Minimum=8.
Therefore, the maximum difference = 14 – 8 = 6.
Input: arr[] = {2, 1, 5, 4, 3}
Output: 4
天真的方法：最简单的想法是遍历数组，对于每个数组元素，使用嵌套循环遍历数组，然后计算并存储其与其余数组的绝对差之和。在计算时，请跟踪获得的最大和最小总和。最后，打印最大和最小和之间的差。
时间复杂度： O(N 2 )
辅助空间： O(1)
高效方法：为了优化上述方法，该思想基于以下观察：在排序数组中，对于任何索引i ，其左侧的元素将较小，而其右侧的元素将较大。可以使用以下公式计算此排序数组中任何元素arr [i]的绝对差之和：
(Number of elements to its left)*(arr[i]) – Sum of elements to its left + Sum of elements to its right – (Number of elements to its right)*(arr[i]))
请按照以下步骤解决问题：
- 初始化totalSum作为0来存储所有的总和的阵列和leftSum的元素作为0存储在任何索引的左元素的总和。
- 初始化两个变量， Max为INT_MIN ， Min为INT_MAX 。
- 以升序对数组arr []进行排序。
- 使用变量i遍历数组arr []并执行以下操作：
- 使用公式Sum =(i * arr [i])– leftSum + totalSum –(((N – i – 1)* arr [i]) ，将arr [i]与其余元素的绝对差之和存储。
- 将Max更新为Max和Sum的最大值。
- 将Min更新为Min和Sum的最小值。
- 完成上述步骤后，将“最大”和“最小”的值打印出来。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to maximize difference of
// the sum of absolute difference of
// an element with the rest of the
// elements in the array
void findMaxDifference(int arr[], int n)
{
// Sort the array in ascending order
sort(arr, arr + n);
// Stores prefix sum at any instant
int Leftsum = 0;
// Store the total array sum
int Totalsum = 0;
// Initialize minimum and maximum
// absolute difference
int Min = INT_MAX, Max = INT_MIN;
// Traverse the array to find
// the total array sum
for (int i = 0; i < n; i++)
Totalsum += arr[i];
// Traverse the array arr[]
for (int i = 0; i < n; i++) {
// Store the number of
// elements to its left
int leftNumbers = i;
// Store the number of
// elements to its right
int rightNumbers = n - i - 1;
// Update the sum of elements
// on its left
Totalsum = Totalsum - arr[i];
// Store the absolute difference sum
int sum = (leftNumbers * arr[i])
- Leftsum
+ Totalsum
- (rightNumbers * arr[i]);
// Update the Minimum
Min = min(Min, sum);
// Update the Maximum
Max = max(Max, sum);
// Update sum of elements
// on its left
Leftsum += arr[i];
}
// Print the result
cout << Max - Min;
}
// Driven Code
int main()
{
int arr[] = { 1, 2, 4, 7 };
int N = sizeof(arr) / sizeof(arr[0]);
findMaxDifference(arr, N);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG
{
// Function to maximize difference of
// the sum of absolute difference of
// an element with the rest of the
// elements in the array
static void findMaxDifference(int arr[], int n)
{
// Sort the array in ascending order
Arrays.sort(arr);
// Stores prefix sum at any instant
int Leftsum = 0;
// Store the total array sum
int Totalsum = 0;
// Initialize minimum and maximum
// absolute difference
int Min = Integer.MAX_VALUE, Max = Integer.MIN_VALUE;
// Traverse the array to find
// the total array sum
for (int i = 0; i < n; i++)
Totalsum += arr[i];
// Traverse the array arr[]
for (int i = 0; i < n; i++)
{
// Store the number of
// elements to its left
int leftNumbers = i;
// Store the number of
// elements to its right
int rightNumbers = n - i - 1;
// Update the sum of elements
// on its left
Totalsum = Totalsum - arr[i];
// Store the absolute difference sum
int sum = (leftNumbers * arr[i])
- Leftsum
+ Totalsum
- (rightNumbers * arr[i]);
// Update the Minimum
Min = Math.min(Min, sum);
// Update the Maximum
Max = Math.max(Max, sum);
// Update sum of elements
// on its left
Leftsum += arr[i];
}
// Print the result
System.out.print(Max - Min);
}
// Driven Code
public static void main(String[] args)
{
int arr[] = { 1, 2, 4, 7 };
int N = arr.length;
findMaxDifference(arr, N);
}
}
// This code is contributed by 29AjayKumarPython3
# Python3 program for the above approach
# Function to maximize difference of
# the sum of absolute difference of
# an element with the rest of the
# elements in the array
def findMaxDifference(arr, n):
# Sort the array in ascending order
arr = sorted(arr)
# Stores prefix sum at any instant
Leftsum = 0
# Store the total array sum
Totalsum = 0
# Initialize minimum and maximum
# absolute difference
Min, Max = 10**8, -10**8
# Traverse the array to find
# the total array sum
for i in range(n):
Totalsum += arr[i]
# Traverse the array arr[]
for i in range(n):
# Store the number of
# elements to its left
leftNumbers = i
# Store the number of
# elements to its right
rightNumbers = n - i - 1
# Update the sum of elements
# on its left
Totalsum = Totalsum - arr[i]
# Store the absolute difference sum
sum = (leftNumbers * arr[i])- Leftsum + Totalsum - (rightNumbers * arr[i])
# Update the Minimum
Min = min(Min, sum)
# Update the Maximum
Max = max(Max, sum)
# Update sum of elements
# on its left
Leftsum += arr[i]
# Prthe result
print (Max - Min)
# Driven Code
if __name__ == '__main__':
arr = [1, 2, 4, 7]
N = len(arr)
findMaxDifference(arr, N)
# This code is contributed by mohit kumar 29.C#
// C# Program to implement
// the above approach
using System;
class GFG
{
// Function to maximize difference of
// the sum of absolute difference of
// an element with the rest of the
// elements in the array
static void findMaxDifference(int[] arr, int n)
{
// Sort the array in ascending order
Array.Sort(arr);
// Stores prefix sum at any instant
int Leftsum = 0;
// Store the total array sum
int Totalsum = 0;
// Initialize minimum and maximum
// absolute difference
int Minn = Int32.MaxValue, Maxx = Int32.MinValue;
// Traverse the array to find
// the total array sum
for (int i = 0; i < n; i++)
Totalsum += arr[i];
// Traverse the array arr[]
for (int i = 0; i < n; i++)
{
// Store the number of
// elements to its left
int leftNumbers = i;
// Store the number of
// elements to its right
int rightNumbers = n - i - 1;
// Update the sum of elements
// on its left
Totalsum = Totalsum - arr[i];
// Store the absolute difference sum
int sum = (leftNumbers * arr[i])
- Leftsum
+ Totalsum
- (rightNumbers * arr[i]);
// Update the Minimum
Minn = Math.Min(Minn, sum);
// Update the Maximum
Maxx = Math.Max(Maxx, sum);
// Update sum of elements
// on its left
Leftsum += arr[i];
}
// Print the result
Console.WriteLine(Maxx - Minn);
}
// Driver Code
public static void Main(String[] args)
{
int[] arr = { 1, 2, 4, 7 };
int N = arr.Length;
findMaxDifference(arr, N);
}
}
// This code is contributd by sanjoy_62.输出：
6时间复杂度： O(N * log N)
辅助空间： O(1)