I have an array of integers that I need to sort. However, the only operation allowed is to swap an element at index i with the element at index i+2. This means traditional sorting algorithms like quicksort or mergesort won’t work directly because they rely on adjacent swaps.
the constraints are these:
How can I fully sort the array given the i and i+2 swap constraint? Are there any existing algorithms or strategies that I can adapt to handle this specific problem? Also, how can I count the exact number of swaps required to sort the array completely?
Here's what I came up with so far
#include <iostream>
#include <algorithm>
using namespace std;
// Function to perform bubble sort on a static array and count swaps
int bubble_sort_with_swap_count(int arr[], int n) {
int swap_count = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n - i - 1; j++) {
if (arr[j] > arr[j + 1]) {
swap(arr[j], arr[j + 1]);
swap_count++;
}
}
}
return swap_count;
}
// Function to sort the array with the given constraints and count steps
void sort_special_with_steps(int arr[], int n, int &total_swaps) {
// Separate even and odd indexed elements
int even_elements[n / 2 + 1], odd_elements[n / 2 + 1];
int even_idx = 0, odd_idx = 0;
for (int i = 0; i < n; i += 2) {
even_elements[even_idx++] = arr[i];
}
for (int i = 1; i < n; i += 2) {
odd_elements[odd_idx++] = arr[i];
}
// Sort both subsequences and count swaps
int even_swaps = bubble_sort_with_swap_count(even_elements, even_idx);
int odd_swaps = bubble_sort_with_swap_count(odd_elements, odd_idx);
// Merge the sorted subsequences back into the original array
even_idx = 0;
odd_idx = 0;
for (int i = 0; i < n; i++) {
if (i % 2 == 0) {
arr[i] = even_elements[even_idx++];
} else {
arr[i] = odd_elements[odd_idx++];
}
}
total_swaps = even_swaps + odd_swaps;
}
int main() {
int arr[] = {4, 1, 3, 2, 5};
int n = sizeof(arr) / sizeof(arr[0]);
int total_swaps = 0;
sort_special_with_steps(arr, n, total_swaps);
cout << "Sorted array: ";
for (int i = 0; i < n; i++) {
cout << arr[i] << " ";
}
cout << endl;
cout << "Number of steps: " << total_swaps << endl;
return 0;
}
this is not nearly fast enough, I saw online that there's some sort of tree structure that can be used but I never got it to work.
here are some test cases with each their expected outputs:
in the end, I managed to solve it thanks to your suggestions, here's my final O(N log N) solution:
#include <vector>
#include <algorithm>
using namespace std;
// Function to merge two halves and count inversions
int mergeAndCount(std::vector<int> &arr, int left, int mid, int right)
{
std::vector<int> leftSub(arr.begin() + left, arr.begin() + mid + 1);
std::vector<int> rightSub(arr.begin() + mid + 1, arr.begin() + right + 1);
int i = 0, j = 0, k = left, swaps = 0;
while (i < leftSub.size() && j < rightSub.size())
{
if (leftSub[i] <= rightSub[j])
{
arr[k++] = leftSub[i++];
}
else
{
arr[k++] = rightSub[j++];
swaps += leftSub.size() - i;
}
}
while (i < leftSub.size())
{
arr[k++] = leftSub[i++];
}
while (j < rightSub.size())
{
arr[k++] = rightSub[j++];
}
return swaps;
}
// Function to implement merge sort and count inversions
int mergeSortAndCount(std::vector<int> &arr, int left, int right)
{
int swaps = 0;
if (left < right)
{
int mid = left + (right - left) / 2;
swaps += mergeSortAndCount(arr, left, mid);
swaps += mergeSortAndCount(arr, mid + 1, right);
swaps += mergeAndCount(arr, left, mid, right);
}
return swaps;
}
// Function to count inversions
int countInversions(std::vector<int> &arr)
{
return mergeSortAndCount(arr, 0, arr.size() - 1);
}
long long flip_sort(int N, int V[])
{
long long swaps = 0;
vector<int> V_even;
vector<int> V_odd;
for (int i = 0; i < N; i++)
{
if (i % 2 == 0)
if (V[i] % 2 != 0)
{
sort(V, V + N);
return -1;
}
else
{
V_even.push_back(V[i]);
}
else if (V[i] % 2 == 0)
{
sort(V, V + N);
return -1;
}
else
{
V_odd.push_back(V[i]);
}
}
sort(V, V + N);
swaps = countInversions(V_even) + countInversions(V_odd);
return swaps;
};