找到对 i, j 使得 |A[i]−A[j]|与它们与任何 Array 元素的差异之和相同

给定一个具有N个非负整数的数组A[] ，找到一对索引i 和 j ，使得它们之间的绝对差值等于与任何其他数组元素的差值之和，即| A[i] - A[k] | + | A[k]− A[j] | = | A[i] - A[j] |，其中k可以是任何索引。

例子：

Input: N = 3, A[] = {2, 7, 5}
Output: 0 1
Explanation:
For k = 0: 
|A[0] – A[0]| + |A[0] – A[1]|
= |2 − 2| + |2 – 7| = 0 + 5 = 5
= |A[0] – A[1]|
For k = 1:
|A[0] – A[1]| + |A[1] – A[1]|
= |2 − 7| + |7 – 7| = 5 + 0 = 5
= |A[0] – A[1]|
For k = 2:
|A[0] – A[2]| + |A[2] – A[1]|
= |2 − 5| + |5 – 7| = 3 + 2 = 5
= |A[0] – A[1]|

Input: N = 4, arr[] = {5, 9, 1, 3}
Output: 1 2
Explanation: 
For k = 0: 
|A[1] – A[0]| + |A[0] – A[2]|
= |9 − 5| + |5 – 1| = 4 + 4 = 8
= |A[1] – A[2]|
For k = 1:
|A[1] – A[1]| + |A[1] – A[2]|
= |9 − 9| + |9 – 1| = 0 + 8 = 8
= |A[1] – A[2]|
For k = 2:
|A[1] – A[2]| + |A[2] – A[2]|
= |9 − 1| + |1 – 1| = 8 + 0 = 8
= |A[1] – A[2]|
For k = 3:
|A[1] – A[3]| + |A[3] – A[2]|
= |9 − 3| + |3 – 1| = 6 + 2 = 8
= |A[1] – A[2]|

方法：这个问题可以通过以下数学观察来解决：

Depending on the relation between A[i], A[j] and A[k], the inequality can be written in the following 4 ways:

When A[i] ≥ A[k] ≥ A[j]:
A[i] − A[k]  + A[k]− A[j] = A[i] − A[j]
=> A[i] – A[j] =  A[i] − A[j] 

When A[k] ≥ A[i], A[k] ≥ A[j]:
A[k] − A[i]  + A[k]− A[j] = |A[i] − A[j]|
=> 2*A[k] – A[i] – A[j] =  |A[i] − A[j]| 

When A[i] ≥ A[k], A[j] ≥ A[k]:
A[i] − A[k]  – A[k]+ A[j] = |A[i] − A[j]|
=> A[i] + A[j] – 2*A[k] =  |A[i] − A[j]| 

When A[j] ≥ A[k] ≥ A[i]:
– A[i] + A[k]  – A[k] + A[j] = – A[i] + A[j]
=> A[j] – A[i] =  A[j] − A[i].

From the above equations, it is clear that if value of A[i] and A[j] are not the extreme values of the array then the probability of the equation being satisfied depends on the value of A[k] and will not hold true when A[k] lies outside the range of [A[i], A[j]].

根据以上观察，很明显 A[i] 和 A[j] 的值应该是数组元素中的最大值和最小值。请按照以下步骤解决问题：

• 从k = 0 到 N-1遍历数组：
  • 如果 A[k] 大于 A[i]，则更新最大元素的索引（比如i ）。
  • 如果 A[k] 小于 A[j]，则更新最小元素（比如j ）的索引。
• 返回对(i, j)作为答案。

下面是上述方法的实现。

C++

// C++ program for aboe approach
 
#include 
using namespace std;
 
// Function to find the pair
pair findPair(int N,
                        vector vec)
{
    // Maximum element of the array
    int maxi = *max_element(vec.begin(),
                            vec.end());
 
    // Minimum element of the array
    int mini = *min_element(vec.begin(),
                            vec.end());
 
    int idx1 = 0;
    int idx2 = 0;
 
    // Loop to find index of maximum element
    for (int i = 0; i < N; i++) {
        if (vec[i] == maxi)
            idx1 = i;
    }
 
    // Loop to find index of minimum element
    for (int i = 0; i < N; i++) {
        if (vec[i] == mini)
            idx2 = i;
    }
 
    return { idx2, idx1 };
}
 
// Driver code
int main()
{
    int N = 3;
    vector vec{ 2, 7, 5 };
 
    // Function call
    pair ans = findPair(N, vec);
    cout << ans.first << " " << ans.second;
    return 0;
}

Java

// Java program for above approach
import java.io.*;
 
class GFG
{
 
  // Function to find the pair
  public static int[] findPair(int N, int vec[])
  {
 
    // Maximum element of the array
    int maxi = Integer.MIN_VALUE;
    for (int i = 0; i < N; i++) {
      maxi = Math.max(maxi, vec[i]);
    }
    // Minimum element of the array
    int mini = Integer.MAX_VALUE;
    for (int i = 0; i < N; i++) {
      mini = Math.min(mini, vec[i]);
    }
 
    int idx1 = 0;
    int idx2 = 0;
 
    // Loop to find index of maximum element
    for (int i = 0; i < N; i++) {
      if (vec[i] == maxi)
        idx1 = i;
    }
 
    // Loop to find index of minimum element
    for (int i = 0; i < N; i++) {
      if (vec[i] == mini)
        idx2 = i;
    }
    int res[] = { idx2, idx1 };
 
    return res;
  }
  public static void main(String[] args)
  {
    int N = 3;
    int vec[] = { 2, 7, 5 };
 
    // Function call
    int ans[] = findPair(N, vec);
    System.out.print(ans[0] + " " + ans[1]);
  }
}
 
// This code is contributed by Rohit Pradhan

Python3

# Python program for aboe approach
 
# Function to find the pair
def findPair(N, vec):
 
    # Maximum element of the array
    maxi = max(vec)
 
    # Minimum element of the array
    mini = min(vec)
 
    idx1 = 0
    idx2 = 0
 
    # Loop to find index of maximum element
    for i in range(N):
        if (vec[i] == maxi):
            idx1 = i
 
# Loop to find index of minimum element
    for i in range(N):
        if (vec[i] == mini):
            idx2 = i
 
    return [idx2, idx1]
 
# Driver code
N = 3
vec = [2, 7, 5]
 
# Function call
ans = findPair(N, vec)
print(f"{ans[0]} {ans[1]}")
 
# This code is contributed by shinjanpatra

C#

// C# program for above approach
using System;
 
public class GFG{
 
  // Function to find the pair
  public static int[] findPair(int N, int[] vec)
  {
 
    // Maximum element of the array
    int maxi = Int32.MinValue;
    for (int i = 0; i < N; i++) {
      maxi = Math.Max(maxi, vec[i]);
    }
    // Minimum element of the array
    int mini = Int32.MaxValue;
    for (int i = 0; i < N; i++) {
      mini = Math.Min(mini, vec[i]);
    }
 
    int idx1 = 0;
    int idx2 = 0;
 
    // Loop to find index of maximum element
    for (int i = 0; i < N; i++) {
      if (vec[i] == maxi)
        idx1 = i;
    }
 
    // Loop to find index of minimum element
    for (int i = 0; i < N; i++) {
      if (vec[i] == mini)
        idx2 = i;
    }
    int[] res = { idx2, idx1 };
 
    return res;
  }
    static public void Main (){
 
        int N = 3;
        int[] vec = { 2, 7, 5 };
 
        // Function call
        int[] ans = findPair(N, vec);
        Console.Write(ans[0] + " " + ans[1]);
    }
}
 
// This code is contributed by hrithikgarg03188.

Javascript

    // JavaScript program for aboe approach
 
    // Function to find the pair
    const findPair = (N, vec) => {
     
        // Maximum element of the array
        let maxi = Math.max(...vec);
 
        // Minimum element of the array
        let mini = Math.min(...vec);
 
        let idx1 = 0;
        let idx2 = 0;
 
        // Loop to find index of maximum element
        for (let i = 0; i < N; i++) {
            if (vec[i] == maxi)
                idx1 = i;
        }
 
        // Loop to find index of minimum element
        for (let i = 0; i < N; i++) {
            if (vec[i] == mini)
                idx2 = i;
        }
 
        return [idx2, idx1];
    }
 
    // Driver code
    let N = 3;
    let vec = [2, 7, 5];
 
    // Function call
    let ans = findPair(N, vec);
    document.write(`${ans[0]} ${ans[1]}`);
 
// This code is contributed by rakeshsahni
 

输出

0 1

时间复杂度： O(N)
辅助空间： O(1)