Data Modelling & AI Data Structure & Algorithm

Program to find weighted median of a given array

21 June 2025

1

Given two arrays arr[] of N integers and W[] of N weights where W[i] is the weight for the element arr[i]. The task is to find the weighted median of the given array.

Note: The sum of the weight of all elements will always be 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:

$\sum _{i=1}^{k-1}W_{i}\leq 1/2 \;and\; \sum _{i=k+1}^{N}W_{i}\leq 1/2$
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:
$\sum _{i=1}^{k-1}W_{i}< 1/2 \;and\; \sum _{i=k+1}^{N}W_{i}= 1/2$
The upper weighted median for element arr[k] which satisfies the following:
$\sum _{i=1}^{k-1}W_{i}= 1/2 \;and\; \sum _{i=k+1}^{N}W_{i}< 1/2$

Examples:

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.

Approach: Follow the steps below to solve the given problem:

Now to find the median of the array arr[] in increasing order with their respective order of weight shouldn’t be changed.
So, create a set of pairs where the first element of the pair will be arr[i] and the second element of the pair will be its corresponding weights W[i].
Then sort the set of Pairs according to the arr[] values.
If the number of pairs is odd, then find the weighted median as:
- Traverse over the set of pairs and compute sum by adding weights.
- When the sum becomes greater than 0.5 print the arr[i] value of that Pair.
But, if the number of pairs is even, then find both lower and upper weighted medians:
- For the lower median traverse over the set pairs from the left and compute sum by adding weights.
- When the sum becomes greater than or equal to 0.5 print the arr[i] value of that Pair.
- For the upper median traverse over the set pairs from the right and compute sum by adding weights.
- When the sum becomes greater than or equal to 0.5 print the arr[i] value of that Pair.

Below is the implementation of the above approach:

C++14

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate weighted median
void weightedMedian(vector<int> arr,
                    vector<float> W)
{
     
    // Store pr of arr[i] and W[i]
    vector<pair<int, float>> 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<int> arr = { 4, 1, 3, 2 };
     
    // Given weights W[]
    vector<float> 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<Pair>
{
  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<Integer> arr,
                           Vector<Double> W)
{
  // Store pr of arr[i] and W[i]
  Vector<Pair> 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<Integer> arr = new Vector<>();
  arr.add(4);
  arr.add(1);
  arr.add(3);
  arr.add(2);
 
  // Given weights W[]
  Vector<Double> 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<Tuple<int, 
               float>> pr = new List<Tuple<int,
                                           float>>();
  
    for(int index = 0; index < arr.Length; index++)
        pr.Add(new Tuple<int, float>(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<int, float> 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<int, float> 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

Javascript

<script>
// Javascript program for the
// above approach
 
// Function to calculate weighted median
function weightedMedian(arr,W)
{
 
    // Store pr of arr[i] and W[i]
  let pr = [];
  
  for(let index = 0;
      index < arr.length;
      index++)
    pr.push([arr[index],
                    W[index]]);
  
  // Sort the list of pr w.r.t.
  // to their arr[] values
  (pr).sort(function(a,b){return a[1]-b[1];});
  
  // If N is odd
  if (arr.length % 2 != 0)
  {
    // Traverse the set pr
    // from left to right
    let sums = 0;
    for(let element=0;element< pr.length;element++)
    {
      // Update sums
      sums += pr[element][1];
  
      // If sum becomes > 0.5
      if (sums > 0.5)
        document.write(
               "The Weighted Median is element " +
                pr[element][0] + "<br>");
    }
  }
  
  // If N is even
  else
  {
    // For lower median traverse
    // the set pr from left
    let sums = 0;
    for(let element=0;element< pr.length;element++)
    {
      // Update sums
      sums += pr[element][1];
  
      // When sum >= 0.5
      if (sums <= 0.5)
      {
        document.write(
               "Lower Weighted Median is element " +
                 pr[element][0] + "<br>");
        break;
      }
    }
  
    // For upper median traverse
    // the set pr from right
    sums = 0;
    for(let index = pr.length - 1;
            index >= 0; index--)
    {
      let element = pr[index][0];
      let weight = pr[index][1];
  
      // Update sums
      sums += weight;
  
      // When sum >= 0.5
      if (sums >= 0.5)
      {
        document.write(
               "Upper Weighted Median is element " +
                element);
        break;
      }
    }
  }
}
 
// Driver Code
// Given array arr[]
let arr = [];
arr.push(4);
arr.push(1);
arr.push(3);
arr.push(2);
 
// Given weights W[]
let W =  [];
W.push(0.25);
W.push(0.49);
W.push(0.25);
W.push(0.01);
 
// Function Call
weightedMedian(arr, W);
 
// This code is contributed by patel2127
</script>

Output:

Lower Weighted Median is element 2
Upper Weighted Median is element 3

Time Complexity: O(N log N)
Auxiliary Space: O(N)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!

Program to find weighted median of a given array

C++14

Java

Python3

C#

Javascript

From Word2Vec to LLM2Vec: How to Choose the Right Embedding Model for RAG

How to Debug Slow Search Requests in Milvus

When Context Engineering Is Done Right, Hallucinations Can Be the Spark of AI Creativity

LEAVE A REPLY Cancel reply

Most Popular

Google TV could borrow this Netflix feature to ensure you’re still watching

Gemini has a known vulnerability, and Google is leaving it alone

Google’s October 2025 update for Pixels is here, but it’s a light one

It wasn’t even on your radar, but this 12.6-inch AMOLED Android tablet is now the best deal around

EDITOR PICKS

Google TV could borrow this Netflix feature to ensure you’re still watching

Gemini has a known vulnerability, and Google is leaving it alone

Google’s October 2025 update for Pixels is here, but it’s a light one

POPULAR POSTS

Google TV could borrow this Netflix feature to ensure you’re still watching

Gemini has a known vulnerability, and Google is leaving it alone

Google’s October 2025 update for Pixels is here, but it’s a light one

POPULAR CATEGORY

ABOUT US

FOLLOW US