Interquartile Range (IQR)

The quartiles of a ranked set of data values are three points that divide the data into exactly four equal parts, each part comprising quarter data. 

1. Q1 is defined as the middle number between the smallest number and the median of the data set.
2. Q2 is the median of the data.
3. Q3 is the middle value between the median and the highest value of the data set.

The interquartile range IQR tells us the range 
where the bulk of the values lie. The interquartile 
range is calculated by subtracting the first quartile
from the third quartile. 
IQR = Q3 - Q1

Uses 
1. Unlike range, IQR tells where the majority of data lies and is thus preferred over range. 
2. IQR can be used to identify outliers in a data set. 
3. Gives the central tendency of the data. 

Examples: 

Input : 1, 19, 7, 6, 5, 9, 12, 27, 18, 2, 15
Output : 13
The data set after being sorted is
1, 2, 5, 6, 7, 9, 12, 15, 18, 19, 27
As mentioned above Q2 is the median of the data.
Hence Q2 = 9
Q1 is the median of lower half, taking Q2 as pivot.
So Q1 = 5
Q3 is the median of upper half talking Q2 as pivot.
So Q3 = 18
Therefore IQR for given data=Q3-Q1=18-5=13

Input : 1, 3, 4, 5, 5, 6, 7, 11
Output : 3

Below is the implementation of the above approach:

#include <iostream>
#include <algorithm>

using namespace std;

// Function to give 
// index of the median
int median(int l, int r){
    int n = r - l + 1;
    n = (n + 1) / 2 - 1;
    return n + l;
}

// Function to 
// calculate IQR
int IQR(int a[], int n){
    sort(a, a + n);

    // Index of median 
    // of entire data
    int mid_index = median(0, n - 1);

    // Median of first half
    int Q1;
    if (n % 2 == 0)
        Q1 = a[median(0, mid_index)];
    else
        Q1 = a[median(0, mid_index - 1)];

    // Median of second half
    int Q3;
    if (n % 2 == 0)
        Q3 = a[median(mid_index + 1, n - 1)];
    else
        Q3 = a[median(mid_index + 1, n)];

    // IQR calculation
    return (Q3 - Q1);
}

// Driver Code
int main() {
    int a[] = {1, 19, 7, 6, 5, 9, 12, 27, 18, 2, 15};
    int n = sizeof(a) / sizeof(a[0]);
    cout << IQR(a, n) << endl;
    return 0;
}

// Java program to find 
// IQR of a data set
import java.io.*;
import java.util.*;

class GFG{
    
    // Function to give 
    // index of the median
    static int median(int l, int r){
        int n = r - l + 1;
        n = (n + 1) / 2 - 1;
        return n + l;
    }

    // Function to 
    // calculate IQR
    static int IQR(int [] a, int n){
        Arrays.sort(a);

        // Index of median 
        // of entire data
        int mid_index = median(0, n - 1);

        // Median of first half
        int Q1;
        if (n % 2 == 0)
            Q1 = a[median(0, mid_index)];
        else
            Q1 = a[median(0, mid_index - 1)];

        // Median of second half
        int Q3;
        if (n % 2 == 0)
            Q3 = a[median(mid_index + 1, n - 1)];
        else
            Q3 = a[median(mid_index + 1, n)];

        // IQR calculation
        return (Q3 - Q1);
    }

    // Driver Code
    public static void main (String[] args) {
        int []a = {1, 19, 7, 6, 5, 9, 
                12, 27, 18, 2, 15};
        int n = a.length;
        System.out.println(IQR(a, n));
    }
}

# Python3 program to find IQR of 
# a data set

# Function to give index of the median
def median(l, r):
    n = r - l + 1
    n = (n + 1) // 2 - 1
    return n + l

# Function to calculate IQR
def IQR(a, n):

    a.sort()

    # Index of median of entire data
    mid_index = median(0, n - 1)

    # Median of first half
    Q1 = a[median(0, mid_index)]

    # Median of second half
    if n % 2 == 0:
        Q3 = a[median(mid_index + 1, n - 1)]
    else:
        Q3 = a[median(mid_index + 1, n - 1)]

    # IQR calculation
    return (Q3 - Q1)

# Driver Function
if __name__=='__main__':
    a = [1, 19, 7, 6, 5, 9, 12, 27, 18, 2, 15]
    n = len(a)
    print(IQR(a, n))

// C# program to find 
// IQR of a data set
using System;

class GFG{
    
    // Function to give 
    // index of the median
    static int median(int l, int r){
        int n = r - l + 1;
        n = (n + 1) / 2 - 1;
        return n + l;
    }

    // Function to 
    // calculate IQR
    static int IQR(int [] a, int n){
        Array.Sort(a);

        // Index of median 
        // of entire data
        int mid_index = median(0, n - 1);

        // Median of first half
        int Q1 = a[median(0, mid_index)];

        // Median of second half
        int Q3;
        if (n % 2 == 0)
            Q3 = a[median(mid_index + 1, n - 1)];
        else
            Q3 = a[median(mid_index + 1, n)];

        // IQR calculation
        return (Q3 - Q1);
    }

    // Driver Code
    public static void Main () {
        int []a = {1, 19, 7, 6, 5, 9, 
                12, 27, 18, 2, 15};
        int n = a.Length;
        Console.WriteLine(IQR(a, n));
    }
}

// This code is contributed 
// by anuj_67.

// JavaScript program to find 
// IQR of a data set

// Function to give 
// index of the median
function median(l, r) {
    let n = r - l + 1;
    n = Math.floor((n + 1) / 2) - 1;
    return n + l;
}

// Function to 
// calculate IQR
function IQR(a, n) {
    a.sort((x, y) => x - y);

    // Index of median 
    // of entire data
    let mid_index = median(0, n - 1);

    // Median of first half
    let Q1 = a[median(0, mid_index)];

    // Median of second half
    let Q3;
    if (n % 2 === 0)
        Q3 = a[median(mid_index + 1, n - 1)];
    else
        Q3 = a[median(mid_index + 1, n)];

    // IQR calculation
    return Q3 - Q1;
}

// Driver Code
let a = [1, 19, 7, 6, 5, 9, 12, 27, 18, 2, 15];
let n = a.length;
console.log(IQR(a, n));

13

Time Complexity: O(n * log(n)), due to sorting of array.
Auxiliary Space: O(1)