TimSort – Data Structures and Algorithms Tutorials

Last Updated : 20 Nov, 2023

Tim Sort is a hybrid sorting algorithm derived from merge sort and insertion sort. It was designed to perform well on many kinds of real-world data. Tim Sort is the default sorting algorithm used by Python’s sorted() and list.sort() functions.

Tim Sort Algorithms

The main idea behind Tim Sort is to exploit the existing order in the data to minimize the number of comparisons and swaps. It achieves this by dividing the array into small subarrays called runs, which are already sorted, and then merging these runs using a modified merge sort algorithm.

How does Tim Sort work?

Let’s consider the following array as an example: arr[] = {4, 2, 8, 6, 1, 5, 9, 3, 7}.

Step 1: Define the size of the run

• Minimum run size: 32 (we’ll ignore this step since our array is small)

Step 2: Divide the array into runs

• In this step, we’ll use insertion sort to sort the small subsequences (runs) within the array.
• The initial array: [4, 2, 8, 6, 1, 5, 9, 3, 7]
• No initial runs are present, so we’ll create runs using insertion sort.
• Sorted runs: [2, 4], [6, 8], [1, 5, 9], [3, 7]
• Updated array: [2, 4, 6, 8, 1, 5, 9, 3, 7]

Step 3: Merge the runs

• In this step, we’ll merge the sorted runs using a modified merge sort algorithm.
• Merge the runs until the entire array is sorted.
• Merged runs: [2, 4, 6, 8], [1, 3, 5, 7, 9]
• Updated array: [2, 4, 6, 8, 1, 3, 5, 7, 9]

Step 4: Adjust the run size

• After each merge operation, we double the size of the run until it exceeds the length of the array.
• The run size doubles: 32, 64, 128 (we’ll ignore this step since our array is small)

Step 5: Continue merging

• Repeat the merging process until the entire array is sorted.
• Final merged run: [1, 2, 3, 4, 5, 6, 7, 8, 9]

The final sorted array is [1, 2, 3, 4, 5, 6, 7, 8, 9].

Below is the implementation for the TimSort:

C++

 // C++ program to perform TimSort. #include 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)

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



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

Complexity Analysis:

Complexity Comparison with Merge and Quick Sort:

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