'Move all odd positioned element to left half and even positioned to right half in-place
Given an array with positive and negative integers, move all the odd indexed elements to the left and even indexed elements to the right.
The difficult part of the problem is to do it in-place while maintaining the order.
e.g.
7, 5, 6, 3, 8, 4, 2, 1
The output should be:
5, 3, 4, 1, 7, 6, 8, 2
If the order didn't matter, we could have been used partition() algorithm of quick sort.
How to do it in O( N )?
Solution 1:[1]
I tried to implement as Evgeny Kluev said, and here is the result:
#pragma once
#include <iterator>
#include <algorithm>
#include <type_traits>
#include <limits>
#include <deque>
#include <utility>
#include <cassert>
template< typename Iterator >
struct perfect_shuffle_permutation
{
static_assert(std::is_same< typename std::iterator_traits< Iterator >::iterator_category, std::random_access_iterator_tag >::value,
"!");
using difference_type = typename std::iterator_traits< Iterator >::difference_type;
using value_type = typename std::iterator_traits< Iterator >::value_type;
perfect_shuffle_permutation()
{
for (difference_type power3_ = 1; power3_ < std::numeric_limits< difference_type >::max() / 3; power3_ *= 3) {
powers3_.emplace_back(power3_ + 1);
}
powers3_.emplace_back(std::numeric_limits< difference_type >::max());
}
void
forward(Iterator _begin, Iterator _end) const
{
return forward(_begin, std::distance(_begin, _end));
}
void
backward(Iterator _begin, Iterator _end) const
{
return backward(_begin, std::distance(_begin, _end));
}
void
forward(Iterator _begin, difference_type const _size) const
{
assert(0 < _size);
assert(_size % 2 == 0);
difference_type const left_size_ = *(std::upper_bound(powers3_.cbegin(), powers3_.cend(), _size) - 1);
cycle_leader_forward(_begin, left_size_);
difference_type const rest_ = _size - left_size_;
if (rest_ != 0) {
Iterator middle_ = _begin + left_size_;
forward(middle_, rest_);
std::rotate(_begin + left_size_ / 2, middle_, middle_ + rest_ / 2);
}
}
void
backward(Iterator _begin, difference_type const _size) const
{
assert(0 < _size);
assert(_size % 2 == 0);
difference_type const left_size_ = *(std::upper_bound(powers3_.cbegin(), powers3_.cend(), _size) - 1);
std::rotate(_begin + left_size_ / 2, _begin + _size / 2, _begin + (_size + left_size_) / 2);
cycle_leader_backward(_begin, left_size_);
difference_type const rest_ = _size - left_size_;
if (rest_ != 0) {
Iterator middle_ = _begin + left_size_;
backward(middle_, rest_);
}
}
private :
void
cycle_leader_forward(Iterator _begin, difference_type const _size) const
{
for (difference_type leader_ = 1; leader_ != _size - 1; leader_ *= 3) {
permutation_forward permutation_(leader_, _size);
Iterator current_ = _begin + leader_;
value_type first_ = std::move(*current_);
while (++permutation_) {
assert(permutation_ < _size);
Iterator next_ = _begin + permutation_;
*current_ = std::move(*next_);
current_ = next_;
}
*current_ = std::move(first_);
}
}
void
cycle_leader_backward(Iterator _begin, difference_type const _size) const
{
for (difference_type leader_ = 1; leader_ != _size - 1; leader_ *= 3) {
permutation_backward permutation_(leader_, _size);
Iterator current_ = _begin + leader_;
value_type first_ = std::move(*current_);
while (++permutation_) {
assert(permutation_ < _size);
Iterator next_ = _begin + permutation_;
*current_ = std::move(*next_);
current_ = next_;
}
*current_ = std::move(first_);
}
}
struct permutation_forward
{
permutation_forward(difference_type const _leader, difference_type const _size)
: leader_(_leader)
, current_(_leader)
, half_size_(_size / 2)
{ ; }
bool
operator ++ ()
{
if (current_ < half_size_) {
current_ += current_;
} else {
current_ = 1 + (current_ - half_size_) * 2;
}
return (current_ != leader_);
}
operator difference_type () const
{
return current_;
}
private :
difference_type const leader_;
difference_type current_;
difference_type const half_size_;
};
struct permutation_backward
{
permutation_backward(difference_type const _leader, difference_type const _size)
: leader_(_leader)
, current_(_leader)
, half_size_(_size / 2)
{ ; }
bool
operator ++ ()
{
if ((current_ % 2) == 0) {
current_ /= 2;
} else {
current_ = (current_ - 1) / 2 + half_size_;
}
return (current_ != leader_);
}
operator difference_type () const
{
return current_;
}
private :
difference_type const leader_;
difference_type current_;
difference_type const half_size_;
};
std::deque< difference_type > powers3_;
};
Solution 2:[2]
I modified the code here to get this algorithm:
void PartitionIndexParity(T arr[], size_t n)
{
using std::swap;
for (size_t shift = 0, k; shift != n; shift += k)
{
k = (size_t)pow(3, ceil(log(n - shift) / log(3)) - 1) + 1;
for (size_t i = 1; i < k; i *= 3) // cycle-leader algorithm
{
size_t j = i;
do { swap(arr[(j = j / 2 + (j % 2) * (k / 2)) + shift], arr[i + shift]); } while (j != i);
}
for (size_t b = shift / 2, m = shift, e = shift + (k - k / 2), i = m; shift != 0 && k != 0; ) // or just use std::rotate(arr, b, m, e)
{
swap(arr[b++], arr[i++]);
if (b == m && i == e) { break; }
if (b == m) { m = i; }
else if (i == e) { i = m; }
}
}
}
Solution 3:[3]
Here's a Java implementation of Peiyush Jain's algorithm:
import java.util.Arrays;
public class InShuffle {
public static void main(String[] args) {
Integer[] nums = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 };
inShuffle(nums);
System.out.println(Arrays.toString(nums));
}
public static <T> void inShuffle(T[] array) {
if (array == null) {
return;
}
inShuffle(array, 0, array.length - 1);
}
private static <T> void inShuffle(T[] array, int startIndex, int endIndex) {
while (endIndex - startIndex + 1 > 1) {
int size = endIndex - startIndex + 1;
int n = size / 2;
int k = (int)Math.floor(Math.log(size + 1) / Math.log(3));
int m = (int)(Math.pow(3, k) - 1) / 2;
rotateRight(array, startIndex + m, startIndex + n + m - 1, m);
for (int i = 0, cycleStartIndex = 0; i < k; ++i, cycleStartIndex = cycleStartIndex * 3 + 2) {
permuteCycle(array, startIndex, cycleStartIndex, 2 * m + 1);
}
endIndex = startIndex + 2 * n - 1;
startIndex = startIndex + 2 * m;
}
}
private static <T> void rotateRight(T[] array, int startIndex, int endIndex, int amount) {
reverse(array, startIndex, endIndex - amount);
reverse(array, endIndex - amount + 1, endIndex);
reverse(array, startIndex, endIndex);
}
private static <T> void reverse(T[] array, int startIndex, int endIndex) {
for (int leftIndex = startIndex, rightIndex = endIndex; leftIndex < rightIndex; ++leftIndex, --rightIndex) {
swap(array, leftIndex, rightIndex);
}
}
private static <T> void swap(T[] array, int index1, int index2) {
T temp = array[index1];
array[index1] = array[index2];
array[index2] = temp;
}
private static <T> void permuteCycle(T[] array, int offset, int startIndex, int mod) {
for (int i = ((2 * startIndex + 2) % mod) - 1; i != startIndex; i = ((2 * i + 2) % mod) - 1) {
swap(array, offset + i, offset + startIndex);
}
}
}
And doing an out-shuffle is as simple as:
public static <T> void outShuffle(T[] array) {
if (array == null) {
return;
}
inShuffle(array, 1, array.length - 1);
}
Solution 4:[4]
public class OddToLeftEvenToRight {
private static void doIt(String input){
char[] inp = input.toCharArray();
int len = inp.length;
for(int j=1; j< len; j++)
{
for(int i=j; i<len-j; i+=2)
{
swap(inp, i, i+1);
}
}
System.out.print(inp);
}
private static void swap(char[] inp, int i, int j) {
char tmp = inp[i];
inp[i]= inp[j];
inp[j]=tmp;
}
public static void main(String[] args)
{
doIt("a1b");
}
}
This program does it in O(n^2).
Solution 5:[5]
There is a simple in-place algorithm that does exactly N swaps described here: https://stackoverflow.com/a/55112294/10396
Define a source index function to find the value that belongs at a given location in the desired result.
index(loc, N):
mid = N-N/2
return (loc*2)+1 if loc < mid else (loc-mid)*2
then walk from beginning to the end, swapping every item with the one at the calculated index, but if that index is earlier than the current location, recalculate to see where it was swapped to.
N = len(A)
for i in range(N):
source = index(i)
while source < i: source = index(source,N)
swap(A[i],A[source])
The number of calls to index is exponential, but for large arrays on real hardware, the time cost is minimal compared to the cost of the swaps.
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | |
| Solution 2 | user541686 |
| Solution 3 | John Kurlak |
| Solution 4 | johnnyRose |
| Solution 5 | AShelly |