Pancake sorting

Given an an unsorted array, sort the given array. You are allowed to do only following operation on array.

flip(arr, i): Reverse array from 0 to i 

Unlike a traditional sorting algorithm, which attempts to sort with the fewest comparisons possible, the goal is to sort the sequence in as few reversals as possible.

The idea is to do something similar to Selection Sort. We one by one place maximum element at the end and reduce the size of current array by one.

Following are the detailed steps. Let given array be arr[] and size of array be n.
1) Start from current size equal to n and reduce current size by one while it’s greater than 1. Let the current size be curr_size. Do following for every curr_size
……a) Find index of the maximum element in arr[0..curr_szie-1]. Let the index be ‘mi’
……b) Call flip(arr, mi)
……c) Call flip(arr, curr_size-1)

See following video for visualization of the above algorithm.

[iframe width=”340″ height=”252″ src=”http://www.youtube.com/embed/kk-_DDgoXfk” frameborder=”0″]

/* A C++ program for Pancake Sorting */ #include <stdlib.h> #include <stdio.h> /* Reverses arr[0..i] */ void flip(int arr[], int i) {     int temp, start = 0;     while (start < i)     {         temp = arr[start];         arr[start] = arr[i];         arr[i] = temp;         start++;         i--;     } } /* Returns index of the maximum element in arr[0..n-1] */ int findMax(int arr[], int n) {    int mi, i;    for (mi = 0, i = 0; i < n; ++i)        if (arr[i] > arr[mi])               mi = i;    return mi; } // The main function that sorts given array using flip operations int pancakeSort(int *arr, int n) {     // Start from the complete array and one by one reduce current size by one     for (int curr_size = n; curr_size > 1; --curr_size)     {         // Find index of the maximum element in arr[0..curr_size-1]         int mi = findMax(arr, curr_size);         // Move the maximum element to end of current array if it's not         // already at the end         if (mi != curr_size-1)         {             // To move at the end, first move maximum number to beginning             flip(arr, mi);             // Now move the maximum number to end by reversing current array             flip(arr, curr_size-1);         }     } } /* A utility function to print an array of size n */ void printArray(int arr[], int n) {     for (int i = 0; i < n; ++i)         printf("%d ", arr[i]); } // Driver program to test above function int main() {     int arr[] = {23, 10, 20, 11, 12, 6, 7};     int n = sizeof(arr)/sizeof(arr[0]);     pancakeSort(arr, n);     puts("Sorted Array ");     printArray(arr, n);     return 0; }

Output:

Sorted Array
6 7 10 11 12 20 23

Total O(n) flip operations are performed in above code. The overall time complexity is O(n^2).