给定两个包含N 个整数的数组arr[]和包含N 个权重的W[] ，其中W[i]是元素arr[i]的权重。任务是找到给定数组的加权中位数。
注意：所有元素的权重总和始终为 1。
Let the array arr[] be arranged in increasing order with their corresponding weights.
If N is odd, then there is only one weighted median say arr[k] which satisfies the below property:
If N is even, then there are two weighted medians, i.e., lower and upper weighted median.
The lower weighted median for element arr[k] which satisfies the following:
The upper weighted median for element arr[k] which satisfies the following:
例子：
Input: arr={5, 1, 3, 2, 4}, W=[0.25, 0.15, 0.2, 0.1, 0.3]
Output: The weighted median is element 4
Explanation:
Here the number of element is odd, so there is only one weighted median because at K = 3 the above condition is satisfied.
The cumulative weights on each side of element 4 is 0.45 and 0.25.
Input: arr=[4, 1, 3, 2], W=[0.25, 0.49, 0.25, 0.01]
Output:
The lower weighted median is element 2
The upper weighted median is element 3
Explanation:
Here there are an even number of elements, so there are two weighted medians.
Lower weighted median is at K = 2 because at K = 2 the above condition is satisfied with cumulative weight on each side of element 2 is 0.49 and 0.5.
Upper weighted median is at K = 3 because at K = 3 the above condition is satisfied with cumulative weight on each side of element 3 is 0.5 and 0.25.
方法：按照以下步骤解决给定的问题：
- 现在要以递增的顺序查找数组arr[]的中位数，它们各自的权重顺序不应改变。
- 因此，创建一组对，其中该对的第一个元素将是arr[i] ，该对的第二个元素将是其相应的权重W[i] 。
- 然后根据arr[]值对 Pairs 集进行排序。
- 如果对数为奇数，则求加权中位数为：
- 遍历对集并通过添加权重计算总和。
- 当总和大于 0.5 时，打印该对的arr[i]值。
- 但是，如果对的数量是偶数，则找到下加权中位数和上加权中位数：
- 对于较低的中位数，从左侧遍历集合对并通过添加权重来计算总和。
- 当总和大于或等于 0.5 时，打印该对的arr[i]值。
- 对于上中位数，从右侧遍历集合对并通过添加权重计算总和。
- 当总和大于或等于 0.5 时，打印该对的arr[i]值。
下面是上述方法的实现：
C++14
// C++ program for the above approach
#include
using namespace std;
// Function to calculate weighted median
void weightedMedian(vector arr,
vector W)
{
// Store pr of arr[i] and W[i]
vector> pr;
for(int index = 0;
index < arr.size();
index++)
pr.push_back({arr[index],
W[index]});
// Sort the list of pr w.r.t.
// to their arr[] values
sort(pr.begin(), pr.end());
// If N is odd
if (arr.size() % 2 != 0)
{
// Traverse the set pr
// from left to right
float sums = 0;
for(auto element : pr)
{
// Update sums
sums += element.second;
// If sum becomes > 0.5
if (sums > 0.5)
cout << "The Weighted Median is element "
<< element.first << endl;
}
}
// If N is even
else
{
// For lower median traverse
// the set pr from left
float sums = 0;
for(auto element : pr)
{
// Update sums
sums += element.second;
// When sum >= 0.5
if (sums >= 0.5)
{
cout << "Lower Weighted Median is element "
<< element.first << endl;
break;
}
}
// For upper median traverse
// the set pr from right
sums = 0;
for(int index = pr.size() - 1;
index >= 0;
index--)
{
int element = pr[index].first;
float weight = pr[index].second;
// Update sums
sums += weight;
// When sum >= 0.5
if (sums >= 0.5)
{
cout << "Upper Weighted Median is element "
<< element;
break;
}
}
}
}
// Driver Code
int main()
{
// Given array arr[]
vector arr = { 4, 1, 3, 2 };
// Given weights W[]
vector W = { 0.25, 0.49, 0.25, 0.01 };
// Function Call
weightedMedian(arr, W);
}
// This code is contributed by mohit kumar 29 Java
// Java program for the
// above approach
import java.util.*;
class GFG{
static class Pair implements Comparable
{
int first;
double second;
Pair(int f, double s)
{
first = f;
second = s;
}
@Override
public int compareTo(Pair o)
{
if(this.second > o.second)
return 1;
else if(this.second == o.second)
return 0;
return -1;
}
}
// Function to calculate weighted median
static void weightedMedian(Vector arr,
Vector W)
{
// Store pr of arr[i] and W[i]
Vector pr = new Vector<>();
for(int index = 0;
index < arr.size();
index++)
pr.add(new Pair(arr.get(index),
W.get(index)));
// Sort the list of pr w.r.t.
// to their arr[] values
Collections.sort(pr);
// If N is odd
if (arr.size() % 2 != 0)
{
// Traverse the set pr
// from left to right
float sums = 0;
for(Pair element : pr)
{
// Update sums
sums += element.second;
// If sum becomes > 0.5
if (sums > 0.5)
System.out.print(
"The Weighted Median is element " +
element.first + "\n");
}
}
// If N is even
else
{
// For lower median traverse
// the set pr from left
double sums = 0;
for(Pair element : pr)
{
// Update sums
sums += element.second;
// When sum >= 0.5
if (sums <= 0.5)
{
System.out.print(
"Lower Weighted Median is element " +
element.first + "\n");
break;
}
}
// For upper median traverse
// the set pr from right
sums = 0;
for(int index = pr.size() - 1;
index >= 0; index--)
{
int element = pr.get(index).first;
double weight = pr.get(index).second;
// Update sums
sums += weight;
// When sum >= 0.5
if (sums >= 0.5)
{
System.out.print(
"Upper Weighted Median is element " +
element);
break;
}
}
}
}
// Driver Code
public static void main(String[] args)
{
// Given array arr[]
Vector arr = new Vector<>();
arr.add(4);
arr.add(1);
arr.add(3);
arr.add(2);
// Given weights W[]
Vector W = new Vector<>();
W.add(0.25);
W.add(0.49);
W.add(0.25);
W.add(0.01);
// Function Call
weightedMedian(arr, W);
}
}
// This code is contributed by gauravrajput1 Python3
# Python3 program for the above approach
# Function to calculate weighted median
def weightedMedian(arr, W):
# Store pairs of arr[i] and W[i]
pairs = []
for index in range(len(arr)):
pairs.append([arr[index], W[index]])
# Sort the list of pairs w.r.t.
# to their arr[] values
pairs.sort(key = lambda p: p[0])
# If N is odd
if len(arr) % 2 != 0:
# Traverse the set pairs
# from left to right
sums = 0
for element, weight in pairs:
# Update sums
sums += weight
# If sum becomes > 0.5
if sums > 0.5:
print("The Weighted Median", end = ' ')
print("is element {}".format(element))
# If N is even
else:
# For lower median traverse
# the set pairs from left
sums = 0
for element, weight in pairs:
# Update sums
sums += weight
# When sum >= 0.5
if sums >= 0.5:
print("Lower Weighted Median", end = ' ')
print("is element {}".format(element))
break
# For upper median traverse
# the set pairs from right
sums = 0
for index in range(len(pairs)-1, -1, -1):
element = pairs[index][0]
weight = pairs[index][1]
# Update sums
sums += weight
# When sum >= 0.5
if sums >= 0.5:
print("Upper Weighted Median", end = ' ')
print("is element {}".format(element))
break
# Driver Code
if __name__ == "__main__":
# Given array arr[]
arr = [4, 1, 3, 2]
# Given weights W[]
W = [0.25, 0.49, 0.25, 0.01]
# Function Call
weightedMedian(arr, W)C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to calculate weighted median
static void weightedMedian(int[] arr,
float[] W)
{
// Store pr of arr[i] and W[i]
List> pr = new List>();
for(int index = 0; index < arr.Length; index++)
pr.Add(new Tuple(arr[index], W[index]));
// Sort the list of pr w.r.t.
// to their arr[] values
pr.Sort();
// If N is odd
if (arr.Length % 2 != 0)
{
// Traverse the set pr
// from left to right
float sums = 0;
foreach(Tuple element in pr)
{
// Update sums
sums += element.Item2;
// If sum becomes > 0.5
if (sums > 0.5)
Console.WriteLine("The Weighted Median " +
"is element " + element.Item1);
}
}
// If N is even
else
{
// For lower median traverse
// the set pr from left
float sums = 0;
foreach(Tuple element in pr)
{
// Update sums
sums += element.Item2;
// When sum >= 0.5
if (sums >= 0.5)
{
Console.WriteLine("Lower Weighted Median " +
"is element " + element.Item1);
break;
}
}
// For upper median traverse
// the set pr from right
sums = 0;
for(int index = pr.Count - 1; index >= 0; index--)
{
int element = pr[index].Item1;
float weight = pr[index].Item2;
// Update sums
sums += weight;
// When sum >= 0.5
if (sums >= 0.5)
{
Console.Write("Upper Weighted Median " +
"is element " + element);
break;
}
}
}
}
// Driver code
static void Main()
{
// Given array arr[]
int[] arr = { 4, 1, 3, 2 };
// Given weights W[]
float[] W = { 0.25f, 0.49f, 0.25f, 0.01f };
// Function Call
weightedMedian(arr, W);
}
}
// This code is contributed by divyeshrabadiya07 输出：
Lower Weighted Median is element 2
Upper Weighted Median is element 3时间复杂度： O(N log N)
辅助空间： O(N)
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