# 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 quartilefrom 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:

## C++

 `#include ``#include ` `using` `namespace` `std;` `// Function to give ``// index of the median``int` `median(``int` `a[], ``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(a, 0, n - 1);` `    ``// Median of first half``    ``int` `Q1;``    ``if` `(n % 2 == 0)``        ``Q1 = a[median(a, 0, mid_index)];``    ``else``        ``Q1 = a[median(a, 0, mid_index - 1)];` `    ``// Median of second half``    ``int` `Q3;``    ``if` `(n % 2 == 0)``        ``Q3 = a[median(a, mid_index + 1, n - 1)];``    ``else``        ``Q3 = a[median(a, 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

 `// 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` `a[], ``                    ``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(a, ``0``, n - ``1``);` `        ``// Median of first half``        ``int` `Q1;``        ``if` `(n % ``2` `== ``0``)``            ``Q1 = a[median(a, ``0``, mid_index)];``        ``else``            ``Q1 = a[median(a, ``0``, mid_index - ``1``)];` `        ``// Median of second half``        ``int` `Q3;``        ``if` `(n % ``2` `== ``0``)``            ``Q3 = a[median(a, mid_index + ``1``, n - ``1``)];``        ``else``            ``Q3 = a[median(a, 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));``    ``}``}` `// This code is contributed ``// by anuj_67.`

## Python3

 `# Python3 program to find IQR of ``# a data set` `# Function to give index of the median``def` `median(a, 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(a, ``0``, n ``-` `1``)` `    ``# Median of first half``    ``Q1 ``=` `a[median(a, ``0``, mid_index)]` `    ``# Median of second half``    ``if` `n ``%` `2` `=``=` `0``:``        ``Q3 ``=` `a[median(a, mid_index ``+` `1``, n ``-` `1``)]``    ``else``:``        ``Q3 ``=` `a[median(a, 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))` `# This code is contributed by``# Sanjit_Prasad`

## C#

 `// C# program to find ``// IQR of a data set``using` `System;` `class` `GFG``{``    ` `    ``// Function to give ``    ``// index of the median``    ``static` `int` `median(``int` `[]a, ``                      ``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(a, 0, n - 1);` `        ``// Median of first half``        ``int` `Q1 = a[median(a, 0, mid_index)];` `        ``// Median of second half``        ``int` `Q3;``        ``if` `(n % 2 == 0)``            ``Q3 = a[median(a, mid_index + 1, n - 1)];``        ``else``            ``Q3 = a[median(a, 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

 `// JavaScript program to find ``// IQR of a data set` `// Function to give ``// index of the median``function` `median(a, 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(a, 0, n - 1);` `    ``// Median of first half``    ``let Q1 = a[median(a, 0, mid_index)];` `    ``// Median of second half``    ``let Q3;``    ``if` `(n % 2 === 0)``        ``Q3 = a[median(a, mid_index + 1, n - 1)];``    ``else``        ``Q3 = a[median(a, 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));`

## PHP

 ``

Output
```13
```

Time Complexity: O(1)
Auxiliary Space: O(1)

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