TimSort

Difficulty Level : Medium
Last Updated : 26 Jun, 2021

TimSort is a sorting algorithm based on Insertion Sort and Merge Sort.

A stable sorting algorithm works in O(n Log n) time
Used in Java’s Arrays.sort() as well as Python’s sorted() and sort().
First sort small pieces using Insertion Sort, then merges the pieces using merge of merge sort.

We divide the Array into blocks known as Run. We sort those runs using insertion sort one by one and then merge those runs using the combine function used in merge sort. If the size of the Array is less than run, then Array gets sorted just by using Insertion Sort. The size of the run may vary from 32 to 64 depending upon the size of the array. Note that the merge function performs well when size subarrays are powers of 2. The idea is based on the fact that insertion sort performs well for small arrays.

Details of below implementation:

We consider size of run as 32.
We one by one sort pieces of size equal to run
After sorting individual pieces, we merge them one by one. We double the size of merged subarrays after every iteration.

C++

// C++ program to perform TimSort.
#include<bits/stdc++.h>
using namespace std;
const int RUN = 32;
 
// This function sorts array from left index to
// to right index which is of size atmost RUN
void insertionSort(int arr[], int left, int right)
{
    for (int i = left + 1; i <= right; i++)
    {
        int temp = arr[i];
        int j = i - 1;
        while (j >= left && arr[j] > temp)
        {
            arr[j+1] = arr[j];
            j--;
        }
        arr[j+1] = temp;
    }
}
 
// Merge function merges the sorted runs
void merge(int arr[], int l, int m, int r)
{
     
    // Original array is broken in two parts
    // left and right array
    int len1 = m - l + 1, len2 = r - m;
    int left[len1], right[len2];
    for (int i = 0; i < len1; i++)
        left[i] = arr[l + i];
    for (int i = 0; i < len2; i++)
        right[i] = arr[m + 1 + i];
 
    int i = 0;
    int j = 0;
    int k = l;
 
    // After comparing, we
    // merge those two array
    // in larger sub array
    while (i < len1 && j < len2)
    {
        if (left[i] <= right[j])
        {
            arr[k] = left[i];
            i++;
        }
        else
        {
            arr[k] = right[j];
            j++;
        }
        k++;
    }
 
    // Copy remaining elements of left, if any
    while (i < len1)
    {
        arr[k] = left[i];
        k++;
        i++;
    }
 
    // Copy remaining element of right, if any
    while (j < len2)
    {
        arr[k] = right[j];
        k++;
        j++;
    }
}
 
// Iterative Timsort function to sort the
// array[0...n-1] (similar to merge sort)
void timSort(int arr[], int n)
{
     
    // Sort individual subarrays of size RUN
    for (int i = 0; i < n; i+=RUN)
        insertionSort(arr, i, min((i+RUN-1),
                                    (n-1)));
 
    // Start merging from size RUN (or 32).
    // It will merge
    // to form size 64, then 128, 256
    // and so on ....
    for (int size = RUN; size < n;
                             size = 2*size)
    {
         
        // pick starting point of
        // left sub array. We
        // are going to merge
        // arr[left..left+size-1]
        // and arr[left+size, left+2*size-1]
        // After every merge, we
        // increase left by 2*size
        for (int left = 0; left < n;
                             left += 2*size)
        {
             
            // find ending point of
            // left sub array
            // mid+1 is starting point
            // of right sub array
            int mid = left + size - 1;
            int right = min((left + 2*size - 1),
                                            (n-1));
 
            // merge sub array arr[left.....mid] &
            // arr[mid+1....right]
              if(mid < right)
                merge(arr, left, mid, right);
        }
    }
}
 
// Utility function to print the Array
void printArray(int arr[], int n)
{
    for (int i = 0; i < n; i++)
        printf("%d  ", arr[i]);
    printf("\n");
}
 
// Driver program to test above function
int main()
{
    int arr[] = {-2, 7, 15, -14, 0, 15, 0, 7, -7,
                       -4, -13, 5, 8, -14, 12};
    int n = sizeof(arr)/sizeof(arr[0]);
    printf("Given Array is\n");
    printArray(arr, n);
 
    // Function Call
    timSort(arr, n);
 
    printf("After Sorting Array is\n");
    printArray(arr, n);
    return 0;
}

Java

// Java program to perform TimSort.
class GFG
{
 
    static int MIN_MERGE = 32;
 
    public static int minRunLength(int n)
    {
        assert n >= 0;
       
        // Becomes 1 if any 1 bits are shifted off
        int r = 0;
        while (n >= MIN_MERGE)
        {
            r |= (n & 1);
            n >>= 1;
        }
        return n + r;
    }
 
    // This function sorts array from left index to
    // to right index which is of size atmost RUN
    public static void insertionSort(int[] arr, int left,
                                     int right)
    {
        for (int i = left + 1; i <= right; i++)
        {
            int temp = arr[i];
            int j = i - 1;
            while (j >= left && arr[j] > temp)
            {
                arr[j + 1] = arr[j];
                j--;
            }
            arr[j + 1] = temp;
        }
    }
 
    // Merge function merges the sorted runs
    public static void merge(int[] arr, int l,
                                 int m, int r)
    {
        // Original array is broken in two parts
        // left and right array
        int len1 = m - l + 1, len2 = r - m;
        int[] left = new int[len1];
        int[] right = new int[len2];
        for (int x = 0; x < len1; x++)
        {
            left[x] = arr[l + x];
        }
        for (int x = 0; x < len2; x++)
        {
            right[x] = arr[m + 1 + x];
        }
 
        int i = 0;
        int j = 0;
        int k = l;
 
        // After comparing, we merge those two array
        // in larger sub array
        while (i < len1 && j < len2)
        {
            if (left[i] <= right[j])
            {
                arr[k] = left[i];
                i++;
            }
            else {
                arr[k] = right[j];
                j++;
            }
            k++;
        }
 
        // Copy remaining elements
        // of left, if any
        while (i < len1)
        {
            arr[k] = left[i];
            k++;
            i++;
        }
 
        // Copy remaining element
        // of right, if any
        while (j < len2)
        {
            arr[k] = right[j];
            k++;
            j++;
        }
    }
 
    // Iterative Timsort function to sort the
    // array[0...n-1] (similar to merge sort)
    public static void timSort(int[] arr, int n)
    {
        int minRun = minRunLength(MIN_MERGE);
       
        // Sort individual subarrays of size RUN
        for (int i = 0; i < n; i += minRun)
        {
            insertionSort(arr, i,
                          Math.min((i + MIN_MERGE - 1), (n - 1)));
        }
 
        // Start merging from size
        // RUN (or 32). It will
        // merge to form size 64,
        // then 128, 256 and so on
        // ....
        for (int size = minRun; size < n; size = 2 * size)
        {
 
            // Pick starting point
            // of left sub array. We
            // are going to merge
            // arr[left..left+size-1]
            // and arr[left+size, left+2*size-1]
            // After every merge, we
            // increase left by 2*size
            for (int left = 0; left < n;
                                 left += 2 * size)
            {
 
                // Find ending point of left sub array
                // mid+1 is starting point of right sub
                // array
                int mid = left + size - 1;
                int right = Math.min((left + 2 * size - 1),
                                     (n - 1));
 
                // Merge sub array arr[left.....mid] &
                // arr[mid+1....right]
                  if(mid < right)
                    merge(arr, left, mid, right);
            }
        }
    }
 
    // Utility function to print the Array
    public static void printArray(int[] arr, int n)
    {
        for (int i = 0; i < n; i++) {
            System.out.print(arr[i] + " ");
        }
        System.out.print("\n");
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int[] arr = { -2, 7,  15,  -14, 0, 15,  0, 7,
                      -7, -4, -13, 5,   8, -14, 12 };
        int n = arr.length;
        System.out.println("Given Array is");
        printArray(arr, n);
 
        timSort(arr, n);
 
        System.out.println("After Sorting Array is");
        printArray(arr, n);
    }
}
 
// This code has been contributed by 29AjayKumar

Python3

# Python3 program to perform basic timSort
MIN_MERGE = 32
 
def calcMinRun(n):
    """Returns the minimum length of a
    run from 23 - 64 so that
    the len(array)/minrun is less than or
    equal to a power of 2.
 
    e.g. 1=>1, ..., 63=>63, 64=>32, 65=>33,
    ..., 127=>64, 128=>32, ...
    """
    r = 0
    while n >= MIN_MERGE:
        r |= n & 1
        n >>= 1
    return n + r
 
 
# This function sorts array from left index to
# to right index which is of size atmost RUN
def insertionSort(arr, left, right):
    for i in range(left + 1, right + 1):
        j = i
        while j > left and arr[j] < arr[j - 1]:
            arr[j], arr[j - 1] = arr[j - 1], arr[j]
            j -= 1
 
 
# Merge function merges the sorted runs
def merge(arr, l, m, r):
     
    # original array is broken in two parts
    # left and right array
    len1, len2 = m - l + 1, r - m
    left, right = [], []
    for i in range(0, len1):
        left.append(arr[l + i])
    for i in range(0, len2):
        right.append(arr[m + 1 + i])
 
    i, j, k = 0, 0, l
     
    # after comparing, we merge those two array
    # in larger sub array
    while i < len1 and j < len2:
        if left[i] <= right[j]:
            arr[k] = left[i]
            i += 1
 
        else:
            arr[k] = right[j]
            j += 1
 
        k += 1
 
    # Copy remaining elements of left, if any
    while i < len1:
        arr[k] = left[i]
        k += 1
        i += 1
 
    # Copy remaining element of right, if any
    while j < len2:
        arr[k] = right[j]
        k += 1
        j += 1
 
 
# Iterative Timsort function to sort the
# array[0...n-1] (similar to merge sort)
def timSort(arr):
    n = len(arr)
    minRun = calcMinRun(n)
     
    # Sort individual subarrays of size RUN
    for start in range(0, n, minRun):
        end = min(start + minRun - 1, n - 1)
        insertionSort(arr, start, end)
 
    # Start merging from size RUN (or 32). It will merge
    # to form size 64, then 128, 256 and so on ....
    size = minRun
    while size < n:
         
        # Pick starting point of left sub array. We
        # are going to merge arr[left..left+size-1]
        # and arr[left+size, left+2*size-1]
        # After every merge, we increase left by 2*size
        for left in range(0, n, 2 * size):
 
            # Find ending point of left sub array
            # mid+1 is starting point of right sub array
            mid = min(n - 1, left + size - 1)
            right = min((left + 2 * size - 1), (n - 1))
 
            # Merge sub array arr[left.....mid] &
            # arr[mid+1....right]
            if mid < right:
                merge(arr, left, mid, right)
 
        size = 2 * size
 
 
# Driver program to test above function
if __name__ == "__main__":
 
    arr = [-2, 7, 15, -14, 0, 15, 0,
           7, -7, -4, -13, 5, 8, -14, 12]
 
    print("Given Array is")
    print(arr)
 
    # Function Call
    timSort(arr)
 
    print("After Sorting Array is")
    print(arr)
    # [-14, -14, -13, -7, -4, -2, 0, 0,
           5, 7, 7, 8, 12, 15, 15]

C#

// C# program to perform TimSort.
using System;
  
class GFG
{
    public const int RUN = 32;
     
    // This function sorts array from left index to
    // to right index which is of size atmost RUN
    public static void insertionSort(int[] arr,
                                int left, int right)
    {
        for (int i = left + 1; i <= right; i++)
        {
            int temp = arr[i];
            int j = i - 1;
            while (j >= left && arr[j] > temp)
            {
                arr[j+1] = arr[j];
                j--;
            }
            arr[j+1] = temp;
        }
    }
       
    // merge function merges the sorted runs
    public static void merge(int[] arr, int l,
                                   int m, int r)
    {
        // original array is broken in two parts
        // left and right array
        int len1 = m - l + 1, len2 = r - m;
        int[] left = new int[len1];
        int[] right = new int[len2];
        for (int x = 0; x < len1; x++)
            left[x] = arr[l + x];
        for (int x = 0; x < len2; x++)
            right[x] = arr[m + 1 + x];
       
        int i = 0;
        int j = 0;
        int k = l;
       
        // After comparing, we merge those two array
        // in larger sub array
        while (i < len1 && j < len2)
        {
            if (left[i] <= right[j])
            {
                arr[k] = left[i];
                i++;
            }
            else
            {
                arr[k] = right[j];
                j++;
            }
            k++;
        }
       
        // Copy remaining elements
        // of left, if any
        while (i < len1)
        {
            arr[k] = left[i];
            k++;
            i++;
        }
       
        // Copy remaining element
        // of right, if any
        while (j < len2)
        {
            arr[k] = right[j];
            k++;
            j++;
        }
    }
       
    // Iterative Timsort function to sort the
    // array[0...n-1] (similar to merge sort)
    public static void timSort(int[] arr, int n)
    {
         
        // Sort individual subarrays of size RUN
        for (int i = 0; i < n; i+=RUN)
            insertionSort(arr, i,
                         Math.Min((i+RUN-1), (n-1)));
       
        // Start merging from size RUN (or 32).
        // It will merge
        // to form size 64, then
        // 128, 256 and so on ....
        for (int size = RUN; size < n;
                                 size = 2*size)
        {
             
            // Pick starting point of
            // left sub array. We
            // are going to merge
            // arr[left..left+size-1]
            // and arr[left+size, left+2*size-1]
            // After every merge, we increase
            // left by 2*size
            for (int left = 0; left < n;
                                  left += 2*size)
            {
                 
                // Find ending point of left sub array
                // mid+1 is starting point of
                // right sub array
                int mid = left + size - 1;
                int right = Math.Min((left +
                                    2*size - 1), (n-1));
       
                // Merge sub array arr[left.....mid] &
                // arr[mid+1....right]
                  if(mid < right)
                    merge(arr, left, mid, right);
            }
        }
    }
       
    // Utility function to print the Array
    public static void printArray(int[] arr, int n)
    {
        for (int i = 0; i < n; i++)
            Console.Write(arr[i] + " ");
        Console.Write("\n");
    }
       
    // Driver program to test above function
    public static void Main()
    {
        int[] arr = {-2, 7, 15, -14, 0, 15, 0,
                   7, -7, -4, -13, 5, 8, -14, 12};
        int n = arr.Length;
        Console.Write("Given Array is\n");
        printArray(arr, n);
       
        // Function Call
        timSort(arr, n);
       
        Console.Write("After Sorting Array is\n");
        printArray(arr, n);
    }   
}
 
//This code is contributed by DrRoot_

Javascript

<script>
 
// Javascript program to perform TimSort.
let MIN_MERGE = 32;
 
function minRunLength(n)
{
     
    // Becomes 1 if any 1 bits are shifted off
    let r = 0;
    while (n >= MIN_MERGE)
    {
        r |= (n & 1);
        n >>= 1;
    }
    return n + r;
}
 
// This function sorts array from left index to
// to right index which is of size atmost RUN
function insertionSort(arr,left,right)
{
    for(let i = left + 1; i <= right; i++)
    {
        let temp = arr[i];
        let j = i - 1;
         
        while (j >= left && arr[j] > temp)
        {
            arr[j + 1] = arr[j];
            j--;
        }
        arr[j + 1] = temp;
    }
}
 
// Merge function merges the sorted runs
function merge(arr, l, m, r)
{
     
    // Original array is broken in two parts
    // left and right array
    let len1 = m - l + 1, len2 = r - m;
    let left = new Array(len1);
    let right = new Array(len2);
    for(let x = 0; x < len1; x++)
    {
        left[x] = arr[l + x];
    }
    for(let x = 0; x < len2; x++)
    {
        right[x] = arr[m + 1 + x];
    }
 
    let i = 0;
    let j = 0;
    let k = l;
 
    // After comparing, we merge those two
    // array in larger sub array
    while (i < len1 && j < len2)
    {
        if (left[i] <= right[j])
        {
            arr[k] = left[i];
            i++;
        }
        else
        {
            arr[k] = right[j];
            j++;
        }
        k++;
    }
 
    // Copy remaining elements
    // of left, if any
    while (i < len1)
    {
        arr[k] = left[i];
        k++;
        i++;
    }
 
    // Copy remaining element
    // of right, if any
    while (j < len2)
    {
        arr[k] = right[j];
        k++;
        j++;
    }
}
 
// Iterative Timsort function to sort the
// array[0...n-1] (similar to merge sort)
function  timSort(arr, n)
{
    let minRun = minRunLength(MIN_MERGE);
        
    // Sort individual subarrays of size RUN
    for(let i = 0; i < n; i += minRun)
    {
        insertionSort(arr, i, Math.min(
            (i + MIN_MERGE - 1), (n - 1)));
    }
 
    // Start merging from size
    // RUN (or 32). It will
    // merge to form size 64,
    // then 128, 256 and so on
    // ....
    for(let size = minRun; size < n; size = 2 * size)
    {
         
        // Pick starting point
        // of left sub array. We
        // are going to merge
        // arr[left..left+size-1]
        // and arr[left+size, left+2*size-1]
        // After every merge, we
        // increase left by 2*size
        for(let left = 0; left < n;
                          left += 2 * size)
        {
 
            // Find ending point of left sub array
            // mid+1 is starting point of right sub
            // array
            let mid = left + size - 1;
            let right = Math.min((left + 2 * size - 1),
                                    (n - 1));
 
            // Merge sub array arr[left.....mid] &
            // arr[mid+1....right]
            if(mid < right)
                merge(arr, left, mid, right);
        }
    }
}
 
// Utility function to print the Array
function printArray(arr,n)
{
    for(let i = 0; i < n; i++)
    {
        document.write(arr[i] + " ");
    }
    document.write("<br>");
}
 
// Driver code
let arr = [ -2, 7, 15, -14, 0, 15, 0, 7,
            -7, -4, -13, 5, 8, -14, 12 ];
let n = arr.length;
document.write("Given Array is<br>");
printArray(arr, n);
timSort(arr, n);
 
document.write("After Sorting Array is<br>");
printArray(arr, n);
 
// This code is contributed by rag2127
 
</script>

Output:

Given Array is
-2, 7, 15, -14, 0, 15, 0, 7, -7, -4, -13, 5, 8, -14, 12
After Sorting Array is
-14  -14  -13  -7  -4  -2  0  0  5  7  7  8  12  15  15

References :
https://svn.python.org/projects/python/trunk/Objects/listsort.txt
https://en.wikipedia.org/wiki/Timsort#Minimum_size_.28minrun.29
This article is contributed by Aditya Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes