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:
- the array is of lenght N
- each integer in the array goes from 0 to N-1 (an array of size 5 would look something like this: [0,1,3,4,2]
- the algorithm shall work also for arrays that could be composed of 100,000 cells
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:
- N = 5, arr = {2,0,4,3,1] -> total_swaps = -1 (cannot be solved)
- N = 6, arr = {2,3,0,5,4,1} -> total_swaps = 3 (can be solved)
- N = 200, arr = {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,115,87,88,89,90,91,92,93,94,95,96,97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,112,113,114,86,116,117,118,119,120,121,122,123,124,125,126,127,128,129,130,131,132,133,134,135,136,137,138,139,140,141,142,143,144,145,146,147,148,149,150,151,152,153,154,155,156,157,158,159,160,161,162,163,164,165,166,167,168,169,170,171,172,173,174,175,176,177,178,179,180,181,182,183,184,185,186,187,188,189,190,191,192,193,194,195,196,197,198,199}, total_swaps = -1 (cannot be solved)
8
The question isn’t asking you to do the sort as specified, it is asking you to compute how it would perform.
You’ve correctly identified that the problem is asking you to examine the properties of a modified bubble sort. Bubble sort is very inefficient (O(n^2) in the average case). However, there are algorithms you can use to calculate the number of swaps required by bubble sort in O(n). So you’d calculate these values for each half of the array and return the sum. You’ll find a description of how to count the number of swaps here: /a/65119332/529630 . You can leverage the fact if a < b && a < c then b < c to count the number of swaps in one loop over the array.
But that only works for when the array IS sortable. You need to figure out a way to check if the array is sortable by “bubble sort i+2”. But no one said you had to use bubble sort to figure this out. Instead you can use a more efficient sort. If you know the range of possible values that an item can take in the array, then you could even use integer sort (which is O(n)). And in these sorts of exercises, they are usually set up so you can use integer sort. So you’d use integer sort to calculate the two sorted halves of the array, merge them, and them compare the result with another array that was sorted altogether. If they’re equal then you know “bubble sort i+2” can sort the array.
It’s a lot of extra loops and code, but because both parts of calculation are O(n), the whole calculation remains O(n), rather than O(n^2) had you actually implemented “bubble sort i+2”.
2
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;
};