给定长度为N的数组arr [] ，任务是找到所有数组元素对的差的中值。
例子：
Input: arr[] = {1, 2, 3, 4}
Output: 1
Explanation:
The difference of all pairs from the given array are {2 – 1, 3 – 2, 4 – 3, 3 – 1, 4 – 2, 4 – 1} = {1, 1, 1, 2, 2, 3}.
Since the array contains 6 elements, the median is the element at index 3 of the difference array.
Therefore, the answer is 1.
Input: arr[] = {1, 3, 5}
Output: 2
Explanation: The difference array is {2, 2, 4}. Therefore, the median is 2.
天真的方法：任务是从给定的数组中生成所有可能的对，并计算arr []数组中每对的差并将它们存储在diff []数组中。排序diff []并找到中间元素。
时间复杂度： O(N 2 * log(N 2 ))
辅助空间： O(N 2 )
高效的方法：可以使用二进制搜索和排序来优化上述方法。请按照以下步骤解决问题：
- 对给定的数组进行排序。
- 初始化low = 0和high = arr [N-1] -arr [0] 。
- 计算中间等于(低+高)/ 2。
- 计算小于中点的差数。如果超过差值阵列的中值化指数，[小区(N *(N – 1)/ 2)]，然后更新高达中间- 1。否则，将低位更新为mid +1 。
- 重复上述步骤，直到低和高相等。
下面是上述方法的实现：
C++
// C++ Program to implement
// the above approach
#include
#define ll long long
using namespace std;
// Function check if mid can be median
// index of the difference array
bool possible(ll mid, vector& a)
{
// Size of the array
ll n = a.size();
// Total possible no of pair
// possible
ll total = (n * (n - 1)) / 2;
// The index of the element in the
// differece of all pairs
// from the array
ll need = (total + 1) / 2;
ll count = 0;
ll start = 0, end = 1;
// Count the number of pairs
// having difference <= mid
while (end < n) {
if (a[end] - a[start] <= mid) {
end++;
}
else {
count += (end - start - 1);
start++;
}
}
// If the difference between end
// and first element is less then
// or equal to mid
if (end == n && start < end
&& a[end - 1] - a[start] <= mid) {
ll t = end - start - 1;
count += (t * (t + 1) / 2);
}
// Checking for the no of element less than
// or equal to mid is greater than median or
// not
if (count >= need)
return true;
else
return false;
}
// Function to calculate the median
// of differences of all pairs
// from the array
ll findMedian(vector& a)
{
// Size of the array
ll n = a.size();
// Initialising the low and high
ll low = 0, high = a[n - 1] - a[0];
// Binary search
while (low <= high) {
// Calculate mid
ll mid = (low + high) / 2;
// If mid can be the median
// of the array
if (possible(mid, a))
high = mid - 1;
else
low = mid + 1;
}
// Returning the median of the
// differences of pairs from
// the array
return high + 1;
}
// Driver Code
int main()
{
vector a = { 1, 7, 5, 2 };
sort(a.begin(), a.end());
cout << findMedian(a) << endl;
} Java
// Java program to implement
// the above approach
import java.util.*;
class GFG{
// Function check if mid can be median
// index of the difference array
static boolean possible(long mid, int[] a)
{
// Size of the array
long n = a.length;
// Total possible no of pair
// possible
long total = (n * (n - 1)) / 2;
// The index of the element in the
// differece of all pairs
// from the array
long need = (total + 1) / 2;
long count = 0;
long start = 0, end = 1;
// Count the number of pairs
// having difference <= mid
while (end < n)
{
if (a[(int)end] - a[(int)start] <= mid)
{
end++;
}
else
{
count += (end - start - 1);
start++;
}
}
// If the difference between end
// and first element is less then
// or equal to mid
if (end == n && start < end &&
a[(int)end - 1] - a[(int)start] <= mid)
{
long t = end - start - 1;
count += (t * (t + 1) / 2);
}
// Checking for the no of element less than
// or equal to mid is greater than median or
// not
if (count >= need)
return true;
else
return false;
}
// Function to calculate the median
// of differences of all pairs
// from the array
static long findMedian(int[] a)
{
// Size of the array
long n = a.length;
// Initialising the low and high
long low = 0, high = a[(int)n - 1] - a[0];
// Binary search
while (low <= high)
{
// Calculate mid
long mid = (low + high) / 2;
// If mid can be the median
// of the array
if (possible(mid, a))
high = mid - 1;
else
low = mid + 1;
}
// Returning the median of the
// differences of pairs from
// the array
return high + 1;
}
// Driver code
public static void main (String[] args)
{
int[] a = { 1, 7, 5, 2 };
Arrays.sort(a);
System.out.println(findMedian(a));
}
}
// This code is contributed by offbeatPython3
# Python3 program to implement
# the above approach
# Function check if mid can be median
# index of the difference array
def possible(mid, a):
# Size of the array
n = len(a);
# Total possible no of pair
# possible
total = (n * (n - 1)) // 2;
# The index of the element in the
# differece of all pairs
# from the array
need = (total + 1) // 2;
count = 0;
start = 0; end = 1;
# Count the number of pairs
# having difference <= mid
while (end < n):
if (a[end] - a[start] <= mid):
end += 1;
else:
count += (end - start - 1);
start += 1;
# If the difference between end
# and first element is less then
# or equal to mid
if (end == n and start < end and
a[end - 1] - a[start] <= mid):
t = end - start - 1;
count += (t * (t + 1) // 2);
# Checking for the no of element
# less than or equal to mid is
# greater than median or not
if (count >= need):
return True;
else:
return False;
# Function to calculate the median
# of differences of all pairs
# from the array
def findMedian(a):
# Size of the array
n = len(a);
# Initialising the low and high
low = 0; high = a[n - 1] - a[0];
# Binary search
while (low <= high):
# Calculate mid
mid = (low + high) // 2;
# If mid can be the median
# of the array
if (possible(mid, a)):
high = mid - 1;
else :
low = mid + 1;
# Returning the median of the
# differences of pairs from
# the array
return high + 1;
# Driver Code
if __name__ == "__main__" :
a = [ 1, 7, 5, 2 ];
a.sort()
print(findMedian(a));
# This code is contributed by AnkitRai01C#
// C# program to implement
// the above approach
using System;
class GFG{
// Function check if mid can be median
// index of the difference array
static bool possible(long mid, int[] a)
{
// Size of the array
long n = a.Length;
// Total possible no of pair
// possible
long total = (n * (n - 1)) / 2;
// The index of the element in the
// differece of all pairs
// from the array
long need = (total + 1) / 2;
long count = 0;
long start = 0, end = 1;
// Count the number of pairs
// having difference <= mid
while (end < n)
{
if (a[(int)end] - a[(int)start] <= mid)
{
end++;
}
else
{
count += (end - start - 1);
start++;
}
}
// If the difference between end
// and first element is less then
// or equal to mid
if (end == n && start < end &&
a[(int)end - 1] - a[(int)start] <= mid)
{
long t = end - start - 1;
count += (t * (t + 1) / 2);
}
// Checking for the no of element less than
// or equal to mid is greater than median or
// not
if (count >= need)
return true;
else
return false;
}
// Function to calculate the median
// of differences of all pairs
// from the array
static long findMedian(int[] a)
{
// Size of the array
long n = a.Length;
// Initialising the low and high
long low = 0, high = a[(int)n - 1] - a[0];
// Binary search
while (low <= high)
{
// Calculate mid
long mid = (low + high) / 2;
// If mid can be the median
// of the array
if (possible(mid, a))
high = mid - 1;
else
low = mid + 1;
}
// Returning the median of the
// differences of pairs from
// the array
return high + 1;
}
// Driver code
public static void Main (string[] args)
{
int[] a = { 1, 7, 5, 2 };
Array.Sort(a);
Console.Write(findMedian(a));
}
}
// This code is contributed by rutvik_56输出：
3
时间复杂度： O(N * log(M))，其中N是元素数，M是数组元素对之间的最大差。
辅助空间： O(1)