/** * @file * @brief Counting Inversions using [Merge Sort](https://en.wikipedia.org/wiki/Merge_sort) * * @details * Program to count the number of inversions in an array * using merge-sort. * * The count of inversions help to determine how close the array * is to being sorted in ASCENDING order. * * two elements a[i] and a[j] form an inversion if a[i] > a[j] and i < j * * Time Complexity --> O(n.log n) * Space Complexity --> O(n) ; additional array temp[1..n] * ### Algorithm * 1. The idea is similar to merge sort, divide the array into two equal or almost * equal halves in each step until the base case is reached. * 2. Create a function merge that counts the number of inversions when two halves of * the array are merged, create two indices i and j, i is the index for first half * and j is an index of the second half. if `a[i]` is greater than `a[j]`, then there are (mid – i) * inversions, Because left and right subarrays are sorted, so all the remaining elements * in left-subarray (a[i+1], a[i+2] … a[mid]) will be greater than a[j]. * 3. Create a recursive function to divide the array into halves and find the answer by summing * the number of inversions is the first half, number of inversion in the second half and * the number of inversions by merging the two. * 4. The base case of recursion is when there is only one element in the given half. * 5. Print the answer * * @author [Rakshit Raj](https://github.com/rakshitraj) */ #include #include #include /** * @namespace sorting * @brief Sorting algorithms */ namespace sorting { /** * @namespace inversion * @brief Functions for counting inversions using Merge Sort algorithm */ namespace inversion { // Functions used ---> // int mergeSort(int* arr, int* temp, int left, int right); // int merge(int* arr, int* temp, int left, int mid, int right); // int countInversion(int* arr, const int size); // void show(int* arr, const int size); /** * Function to merge two sub-arrays. merge() function is called * from mergeSort() to merge the array after it split for sorting * by the mergeSort() funtion. * * In this case the merge fuction will also count and return * inversions detected when merging the sub arrays. * * @param arr input array, data-menber of vector * @param temp stores the resultant merged array * @param left lower bound of arr[] and left-sub-array * @param mid midpoint, upper bound of left sub-array, * (mid+1) gives the lower bound of right-sub-array * @param right upper bound of arr[] and right-sub-array * @returns number of inversions found in merge step */ template int merge(T* arr, T* temp, int left, int mid, int right) { int i = left; /* i --> index of left sub-array */ int j = mid + 1; /* j --> index for right sub-array */ int k = left; /* k --> index for resultant array temp */ int inv_count = 0; // inversion count while ((i <= mid) && (j <= right)) { if (arr[i] <= arr[j]) { temp[k++] = arr[i++]; } else { temp[k++] = arr[j++]; inv_count += (mid - i + 1); // tricky; may vary depending on selection of sub-array } } // Add remaining elements from the larger subarray to the end of temp while (i <= mid) { temp[k++] = arr[i++]; } while (j <= right) { temp[k++] = arr[j++]; } // Copy temp[] to arr[] for (k = left; k <= right; k++) { arr[k] = temp[k]; } return inv_count; } /** * * The mergeSort() function implements Merge Sort, a * Divide and conquer algorithm, it divides the input * array into two halves and calls itself for each * sub-array and then calls the merge() function to * merge the two halves. * * @param arr - array to be sorted * @param temp - merged resultant array * @param left - lower bound of array * @param right - upper bound of array * @returns number of inversions in array */ template int mergeSort(T* arr, T* temp, int left, int right) { int mid = 0, inv_count = 0; if (right > left) { // midpoint to split the array mid = (right + left) / 2; // Add inversions in left and right sub-arrays inv_count += mergeSort(arr, temp, left, mid); // left sub-array inv_count += mergeSort(arr, temp, mid + 1, right); // inversions in the merge step inv_count += merge(arr, temp, left, mid, right); } return inv_count; } /** * Funtion countInversion() returns the number of inversion * present in the input array. Inversions are an estimate of * how close or far off the array is to being sorted. * * Number of inversions in a sorted array is 0. * Number of inversion in an array[1...n] sorted in * non-ascending order is n(n-1)/2, since each pair of elements * contitute an inversion. * * @param arr - array, data member of std::vector, input for counting * inversions * @param array_size - number of elementa in the array * @returns number of inversions in input array, sorts the array */ template int countInversion(T* arr, const int size) { std::vector temp; temp.reserve(size); temp.assign(size, 0); return mergeSort(arr, temp.data(), 0, size - 1); } /** * UTILITY function to print array. * @param arr[] array to print * @param array_size size of input array arr[] * @returns void * */ template void show(T* arr, const int array_size) { std::cout << "Printing array: \n"; for (int i = 0; i < array_size; i++) { std::cout << " " << arr[i]; } std::cout << "\n"; } } // namespace inversion } // namespace sorting /** * @brief Test implementations * @returns void */ static void test() { // Test 1 std::vector arr1 = { 100, 99, 98, 97, 96, 95, 94, 93, 92, 91, 90, 89, 88, 87, 86, 85, 84, 83, 82, 81, 80, 79, 78, 77, 76, 75, 74, 73, 72, 71, 70, 69, 68, 67, 66, 65, 64, 63, 62, 61, 60, 59, 58, 57, 56, 55, 54, 53, 52, 51, 50, 49, 48, 47, 46, 45, 44, 43, 42, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1}; int size1 = arr1.size(); int inv_count1 = 4950; int result1 = sorting::inversion::countInversion(arr1.data(), size1); assert(inv_count1 == result1); // Test 2 std::vector arr2 = {22, 66, 75, 23, 11, 87, 2, 44, 98, 43}; int size2 = arr2.size(); int inv_count2 = 20; int result2 = sorting::inversion::countInversion(arr2.data(), size2); assert(inv_count2 == result2); // Test 3 std::vector arr3 = {33.1, 45.2, 65.4, 76.5, 1.0, 2.9, 5.4, 7.7, 88.9, 12.4}; int size3 = arr3.size(); int inv_count3 = 21; int result3 = sorting::inversion::countInversion(arr3.data(), size3); assert(inv_count3 == result3); // Test 4 std::vector arr4 = {'a', 'b', 'c', 'd', 'e'}; int size4 = arr4.size(); int inv_count4 = 0; int result4 = sorting::inversion::countInversion(arr4.data(), size4); assert(inv_count4 == result4); } // /** // * Program Body contains all main funtionality // * @returns void // */ // void body() { // // Input your own sequence // int size, input; // std::cout << "Enter number of elements:"; // std::cin >> size; // std::vector arr; // arr.reserve(size); // std::cout << "Enter elements -->\n"; // for (int i=1; i<=size; i++) { // std::cout << "Element "<< i <<" :"; // std::cin >> input; // arr.push_back(input); // } // if (size != arr.size()) { // size = arr.size(); // } // std::cout << "\n"; // sorting::inversion::show(arr.data(), size); // std::cout << "\n"; // // Counting inversions // std::cout << "\nThe number of inversions: "<< // sorting::inversion::countInversion(arr.data(), size) << "\n"; // // Output sorted array // std::cout << "\nSorted array --> \n"; // sorting::inversion::show(arr.data(), size); // } /** * @brief Main function * @returns 0 on exit */ int main() { // Run test implementations test(); // // Main Program // body(); return 0; }