I am trying to find out total number of ways to reach a target in subset sum problem. Below is my approach.
Let DP[i, j] be 1 if sum of 'j' elements sum up to 'i' else it is 0 where 'a' be the input. So,
DP[i, j] = DP[i, j-1] + DP[i - a[j], j-1]
For input [10, 13, 15, 18, 20, 15] and target = 30
; we are looking for DP[30, 6] as the answer.
I am able to get it working with recursion (http://ideone.com/0sHhDL) but i need to do it with DP.
Once you have a recursive function written, all we need to add to make it efficient is to cache the return values. The modification is: Before making a recursive call, check that the computation with those arguments was not already done.
int NMAX = 100;
int cache[NMAX][NMAX];
int answer;
// Returns true if there is a subset of set[] with sun equal to given sum
int isSubsetSum(int set[], int n, int sum)
{
// Base Cases
if (sum == 0) {
answer++;
return 0; //return false here as we are after all the sums
}
if (n == 0 && sum != 0)
return 0;
if(cache[n][sum] == -1) {
// If last element is greater than sum, then ignore it
if (set[n-1] > sum)
return isSubsetSum(set, n-1, sum);
/* else, check if sum can be obtained by any of the following
(a) including the last element
(b) excluding the last element */
cache[n][sum] = isSubsetSum(set, n-1, sum) || isSubsetSum(set, n-1, sum-set[n-1]);
}
return cache[n][sum];
}
// Driver program to test above function
int main()
{
int set[] = {3, 34, 4, 12, 5, 2};
int sum = 9;
int n = sizeof(set)/sizeof(set[0]);
for(int i=0; i<NMAX; i++) for(int j=0; j<NMAX; j++)
cache[i][j] = -1;
isSubsetSum(set, n, sum);
printf("total %d\n", answer);
return 0;
}
In the code, the line
cache[n][sum] = isSubsetSum(set, n-1, sum) || isSubsetSum(set, n-1, sum-set[n-1]);
is equivalent to the recursive formula
DP[i, j] = DP[i, j-1] + DP[i - a[j], j-1]
The difference is that one is top-bottom, the other is bottom-top.
Related
This question already has an answer here:
subset sum find all subsets that add up to a number
(1 answer)
Closed 2 years ago.
I have 2 input
array: {3,6,9,0,2,1,3} // positive number and can repeat also
Sum = 9
Need to find a combination(order not mandatory) of array element which has total to Sum(here for example it's 9).
Output expected :
{3,6}
{9}
{6,3}
{3,2,1,3}
I am not able to solve it. So, please don't ask for my solution. Please help by solving in java.
This problem can be solved by printing all the subsets with given sum.
Have a look at the following implementation:
// A Java program to count all subsets with given sum.
import java.util.ArrayList;
public class SubSet_sum_problem
{
// dp[i][j] is going to store true if sum j is
// possible with array elements from 0 to i.
static boolean[][] dp;
static void display(ArrayList<Integer> v)
{
System.out.println(v);
}
// A recursive function to print all subsets with the
// help of dp[][]. Vector p[] stores current subset.
static void printSubsetsRec(int arr[], int i, int sum,
ArrayList<Integer> p)
{
// If we reached end and sum is non-zero. We print
// p[] only if arr[0] is equal to sun OR dp[0][sum]
// is true.
if (i == 0 && sum != 0 && dp[0][sum])
{
p.add(arr[i]);
display(p);
p.clear();
return;
}
// If sum becomes 0
if (i == 0 && sum == 0)
{
display(p);
p.clear();
return;
}
// If given sum can be achieved after ignoring
// current element.
if (dp[i-1][sum])
{
// Create a new vector to store path
ArrayList<Integer> b = new ArrayList<>();
b.addAll(p);
printSubsetsRec(arr, i-1, sum, b);
}
// If given sum can be achieved after considering
// current element.
if (sum >= arr[i] && dp[i-1][sum-arr[i]])
{
p.add(arr[i]);
printSubsetsRec(arr, i-1, sum-arr[i], p);
}
}
// Prints all subsets of arr[0..n-1] with sum 0.
static void printAllSubsets(int arr[], int n, int sum)
{
if (n == 0 || sum < 0)
return;
// Sum 0 can always be achieved with 0 elements
dp = new boolean[n][sum + 1];
for (int i=0; i<n; ++i)
{
dp[i][0] = true;
}
// Sum arr[0] can be achieved with single element
if (arr[0] <= sum)
dp[0][arr[0]] = true;
// Fill rest of the entries in dp[][]
for (int i = 1; i < n; ++i)
for (int j = 0; j < sum + 1; ++j)
dp[i][j] = (arr[i] <= j) ? (dp[i-1][j] ||
dp[i-1][j-arr[i]])
: dp[i - 1][j];
if (dp[n-1][sum] == false)
{
System.out.println("There are no subsets with" +
" sum "+ sum);
return;
}
// Now recursively traverse dp[][] to find all
// paths from dp[n-1][sum]
ArrayList<Integer> p = new ArrayList<>();
printSubsetsRec(arr, n-1, sum, p);
}
//Driver Program to test above functions
public static void main(String args[])
{
int arr[] = {3, 6, 9, 0, 2, 1, 3};
int n = arr.length;
int sum = 9;
printAllSubsets(arr, n, sum);
}
}
Output:
[6, 3]
[9]
[0, 6, 3]
[0, 9]
[1, 2, 6]
[1, 2, 0, 6]
[3, 6]
[3, 0, 6]
[3, 1, 2, 3]
[3, 1, 2, 0, 3]
I'm going through an exercise to partition a set into K subsets with equal sum.
Let's say
Input : arr = [2, 1, 4, 5, 6], K = 3
Output : Yes
we can divide above array into 3 parts with equal
sum as [[2, 4], [1, 5], [6]]
I found a solution here,
http://www.geeksforgeeks.org/partition-set-k-subsets-equal-sum/
// C++ program to check whether an array can be
// partitioned into K subsets of equal sum
#include <bits/stdc++.h>
using namespace std;
// Recursive Utility method to check K equal sum
// subsetition of array
/**
array - given input array
subsetSum array - sum to store each subset of the array
taken - boolean array to check whether element
is taken into sum partition or not
K - number of partitions needed
N - total number of element in array
curIdx - current subsetSum index
limitIdx - lastIdx from where array element should
be taken */
bool isKPartitionPossibleRec(int arr[], int subsetSum[], bool taken[],
int subset, int K, int N, int curIdx, int limitIdx)
{
if (subsetSum[curIdx] == subset)
{
/* current index (K - 2) represents (K - 1) subsets of equal
sum last partition will already remain with sum 'subset'*/
if (curIdx == K - 2)
return true;
// recursive call for next subsetition
return isKPartitionPossibleRec(arr, subsetSum, taken, subset,
K, N, curIdx + 1, N - 1);
}
// start from limitIdx and include elements into current partition
for (int i = limitIdx; i >= 0; i--)
{
// if already taken, continue
if (taken[i])
continue;
int tmp = subsetSum[curIdx] + arr[i];
// if temp is less than subset then only include the element
// and call recursively
if (tmp <= subset)
{
// mark the element and include into current partition sum
taken[i] = true;
subsetSum[curIdx] += arr[i];
bool nxt = isKPartitionPossibleRec(arr, subsetSum, taken,
subset, K, N, curIdx, i - 1);
// after recursive call unmark the element and remove from
// subsetition sum
taken[i] = false;
subsetSum[curIdx] -= arr[i];
if (nxt)
return true;
}
}
return false;
}
// Method returns true if arr can be partitioned into K subsets
// with equal sum
bool isKPartitionPossible(int arr[], int N, int K)
{
// If K is 1, then complete array will be our answer
if (K == 1)
return true;
// If total number of partitions are more than N, then
// division is not possible
if (N < K)
return false;
// if array sum is not divisible by K then we can't divide
// array into K partitions
int sum = 0;
for (int i = 0; i < N; i++)
sum += arr[i];
if (sum % K != 0)
return false;
// the sum of each subset should be subset (= sum / K)
int subset = sum / K;
int subsetSum[K];
bool taken[N];
// Initialize sum of each subset from 0
for (int i = 0; i < K; i++)
subsetSum[i] = 0;
// mark all elements as not taken
for (int i = 0; i < N; i++)
taken[i] = false;
// initialize first subsubset sum as last element of
// array and mark that as taken
subsetSum[0] = arr[N - 1];
taken[N - 1] = true;
// call recursive method to check K-substitution condition
return isKPartitionPossibleRec(arr, subsetSum, taken,
subset, K, N, 0, N - 1);
}
// Driver code to test above methods
int main()
{
int arr[] = {2, 1, 4, 5, 3, 3};
int N = sizeof(arr) / sizeof(arr[0]);
int K = 3;
if (isKPartitionPossible(arr, N, K))
cout << "Partitions into equal sum is possible.\n";
else
cout << "Partitions into equal sum is not possible.\n";
}
This works in all scenarios.
Question
Let's say if I pass arr = [4, 4, 1, 3, 2, 3, 2, 1] and k = 4,the algorithm tried to solve it by adding 1+2+2 and then, 3+3 or 3+1 and so on. It doesn't gets the partition and finally solves it to [[4,1], [4,1], [3,2], [3,2]]. I am not sure how does this algorithm finds the alternative? I'm not able to follow up with the recursion.
What are the ways to solve it? Is the backtracking the only way?
Thanks!
Given an array A and a sum, I want to find out if there exists a subsequence of length K such that the sum of all elements in the subsequence equals the given sum.
Code:
for i in(1,N):
for len in (i-1,0):
for sum in (0,Sum of all element)
Possible[len+1][sum] |= Possible[len][sum-A[i]]
Time complexity O(N^2.Sum). Is there any way to improve the time complexity to O(N.Sum)
My function shifts a window of k adjacent array items across the array A and keeps the sum up-to-data until it matches of the search fails.
int getSubSequenceStart(int A[], size_t len, int sum, size_t k)
{
int sumK = 0;
assert(len > 0);
assert(k <= len);
// compute sum for first k items
for (int i = 0; i < k; i++)
{
sumK += A[i];
}
// shift k-window upto end of A
for (int j = k; j < len; j++)
{
if (sumK == sum)
{
return j - k;
}
sumK += A[j] - A[j - k];
}
return -1;
}
Complexity is linear with the length of array A.
Update for the non-contiguous general subarray case:
To find a possibly non-contiguous subarray, you could transform your problem into a subset sum problem by subtracting sum/k from every element of A and looking for a subset with sum zero. The complexity of the subset sum problem is known to be exponential. Therefore, you cannot hope for a linear algorithm, unless your array A has special properties.
Edit:
This could actually be solved without the queue in linear time (negative numbers allowed).
C# code:
bool SubsequenceExists(int[] a, int k, int sum)
{
int currentSum = 0;
if (a.Length < k) return false;
for (int i = 0; i < a.Length; i++)
{
if (i < k)
{
currentSum += a[i];
continue;
}
if (currentSum == sum) return true;
currentSum += a[i] - a[i-k];
}
return false;
}
Original answer:
Assuming you can use a queue of length K something like that should do the job in linear time.
C# code:
bool SubsequenceExists(int[] a, int k, int sum)
{
int currentSum = 0;
var queue = new Queue<int>();
for (int i = 0; i < a.Length; i++)
{
if (i < k)
{
queue.Enqueue(a[i]);
currentSum += a[i];
continue;
}
if (currentSum == sum) return true;
currentSum -= queue.Dequeue();
queue.Enqueue(a[i]);
currentSum += a[i];
}
return false;
}
The logic behind that is pretty much straightforward:
We populate a queue with first K elements while also storing its sum somewhere.
If the resulting sum is not equal to sum then we dequeue an element from the queue and add the next one from A (while updating the sum).
We repeat step 2 until we either reach the end of sequence or find the matching subsequence.
Ta-daa!
Let is_subset_sum(int set[], int n, int sum) be the function to find whether there is a subset of set[] with sum equal to sum. n is the number of elements in set[].
The is_subset_sum problem can be divided into two subproblems
Include the last element, recur for n = n-1, sum = sum – set[n-1]
Exclude the last element, recur for n = n-1.
If any of the above subproblems return true, then return true.
Following is the recursive formula for is_subset_sum() problem.
is_subset_sum(set, n, sum) = is_subset_sum(set, n-1, sum) || is_subset_sum(set, n-1, sum-set[n-1])
Base Cases:
is_subset_sum(set, n, sum) = false, if sum > 0 and n == 0
is_subset_sum(set, n, sum) = true, if sum == 0
We can solve the problem in Pseudo-polynomial time using Dynamic programming. We create a boolean 2D table subset[][] and fill it in a bottom-up manner. The value of subset[i][j] will be true if there is a subset of set[0..j-1] with sum equal to i., otherwise false. Finally, we return subset[sum][n]
The time complexity of the solution is O(sum*n).
Implementation in C
// A Dynamic Programming solution for subset sum problem
#include <stdio.h>
// Returns true if there is a subset of set[] with sun equal to given sum
bool is_subset_sum(int set[], int n, int sum) {
// The value of subset[i][j] will be true if there is a
// subset of set[0..j-1] with sum equal to i
bool subset[sum+1][n+1];
// If sum is 0, then answer is true
for (int i = 0; i <= n; i++)
subset[0][i] = true;
// If sum is not 0 and set is empty, then answer is false
for (int i = 1; i <= sum; i++)
subset[i][0] = false;
// Fill the subset table in botton up manner
for (int i = 1; i <= sum; i++) {
for (int j = 1; j <= n; j++) {
subset[i][j] = subset[i][j-1];
if (i >= set[j-1])
subset[i][j] = subset[i][j] || subset[i - set[j-1]][j-1];
}
}
/* // uncomment this code to print table
for (int i = 0; i <= sum; i++) {
for (int j = 0; j <= n; j++)
printf ("%4d", subset[i][j]);
printf("\n");
} */
return subset[sum][n];
}
// Driver program to test above function
int main() {
int set[] = {3, 34, 4, 12, 5, 2};
int sum = 9;
int n = sizeof(set)/sizeof(set[0]);
if (is_subset_sum(set, n, sum) == true)
printf("Found a subset with given sum");
else
printf("No subset with given sum");
return 0;
}
There's an array A containing (positive and negative) integers. Find a (contiguous) subarray whose elements' absolute sum is minimal, e.g.:
A = [2, -4, 6, -3, 9]
|(−4) + 6 + (−3)| = 1 <- minimal absolute sum
I've started by implementing a brute-force algorithm which was O(N^2) or O(N^3), though it produced correct results. But the task specifies:
complexity:
- expected worst-case time complexity is O(N*log(N))
- expected worst-case space complexity is O(N)
After some searching I thought that maybe Kadane's algorithm can be modified to fit this problem but I failed to do it.
My question is - is Kadane's algorithm the right way to go? If not, could you point me in the right direction (or name an algorithm that could help me here)? I don't want a ready-made code, I just need help in finding the right algorithm.
If you compute the partial sums
such as
2, 2 +(-4), 2 + (-4) + 6, 2 + (-4) + 6 + (-3)...
Then the sum of any contiguous subarray is the difference of two of the partial sums. So to find the contiguous subarray whose absolute value is minimal, I suggest that you sort the partial sums and then find the two values which are closest together, and use the positions of these two partial sums in the original sequence to find the start and end of the sub-array with smallest absolute value.
The expensive bit here is the sort, so I think this runs in time O(n * log(n)).
This is C++ implementation of Saksow's algorithm.
int solution(vector<int> &A) {
vector<int> P;
int min = 20000 ;
int dif = 0 ;
P.resize(A.size()+1);
P[0] = 0;
for(int i = 1 ; i < P.size(); i ++)
{
P[i] = P[i-1]+A[i-1];
}
sort(P.begin(),P.end());
for(int i = 1 ; i < P.size(); i++)
{
dif = P[i]-P[i-1];
if(dif<min)
{
min = dif;
}
}
return min;
}
I was doing this test on Codility and I found mcdowella answer quite helpful, but not enough I have to say: so here is a 2015 answer guys!
We need to build the prefix sums of array A (called P here) like: P[0] = 0, P[1] = P[0] + A[0], P[2] = P[1] + A[1], ..., P[N] = P[N-1] + A[N-1]
The "min abs sum" of A will be the minimum absolute difference between 2 elements in P. So we just have to .sort() P and loop through it taking every time 2 successive elements. This way we have O(N + Nlog(N) + N) which equals to O(Nlog(N)).
That's it!
The answer is yes, Kadane's algorithm is definitely the way to go for solving your problem.
http://en.wikipedia.org/wiki/Maximum_subarray_problem
Source - I've closely worked with a PhD student who's entire PhD thesis was devoted to the maximum subarray problem.
def min_abs_subarray(a):
s = [a[0]]
for e in a[1:]:
s.append(s[-1] + e)
s = sorted(s)
min = abs(s[0])
t = s[0]
for x in s[1:]:
cur = abs(x)
min = cur if cur < min else min
cur = abs(t-x)
min = cur if cur < min else min
t = x
return min
You can run Kadane's algorithmtwice(or do it in one go) to find minimum and maximum sum where finding minimum works in same way as maximum with reversed signs and then calculate new maximum by comparing their absolute value.
Source-Someone's(dont remember who) comment in this site.
Here is an Iterative solution in python. It's 100% correct.
def solution(A):
memo = []
if not len(A):
return 0
for ind, val in enumerate(A):
if ind == 0:
memo.append([val, -1*val])
else:
newElem = []
for i in memo[ind - 1]:
newElem.append(i+val)
newElem.append(i-val)
memo.append(newElem)
return min(abs(n) for n in memo.pop())
Short Sweet and work like a charm. JavaScript / NodeJs solution
function solution(A, i=0, sum =0 ) {
//Edge case if Array is empty
if(A.length == 0) return 0;
// Base case. For last Array element , add and substart from sum
// and find min of their absolute value
if(A.length -1 === i){
return Math.min( Math.abs(sum + A[i]), Math.abs(sum - A[i])) ;
}
// Absolute value by adding the elem with the sum.
// And recusrively move to next elem
let plus = Math.abs(solution(A, i+1, sum+A[i]));
// Absolute value by substracting the elem from the sum
let minus = Math.abs(solution(A, i+1, sum-A[i]));
return Math.min(plus, minus);
}
console.log(solution([-100, 3, 2, 4]))
Here is a C solution based on Kadane's algorithm.
Hopefully its helpful.
#include <stdio.h>
int min(int a, int b)
{
return (a >= b)? b: a;
}
int min_slice(int A[], int N) {
if (N==0 || N>1000000)
return 0;
int minTillHere = A[0];
int minSoFar = A[0];
int i;
for(i = 1; i < N; i++){
minTillHere = min(A[i], minTillHere + A[i]);
minSoFar = min(minSoFar, minTillHere);
}
return minSoFar;
}
int main(){
int A[]={3, 2, -6, 4, 0}, N = 5;
//int A[]={3, 2, 6, 4, 0}, N = 5;
//int A[]={-4, -8, -3, -2, -4, -10}, N = 6;
printf("Minimum slice = %d \n", min_slice(A,N));
return 0;
}
public static int solution(int[] A) {
int minTillHere = A[0];
int absMinTillHere = A[0];
int minSoFar = A[0];
int i;
for(i = 1; i < A.length; i++){
absMinTillHere = Math.min(Math.abs(A[i]),Math.abs(minTillHere + A[i]));
minTillHere = Math.min(A[i], minTillHere + A[i]);
minSoFar = Math.min(Math.abs(minSoFar), absMinTillHere);
}
return minSoFar;
}
int main()
{
int n; cin >> n;
vector<int>a(n);
for(int i = 0; i < n; i++) cin >> a[i];
long long local_min = 0, global_min = LLONG_MAX;
for(int i = 0; i < n; i++)
{
if(abs(local_min + a[i]) > abs(a[i]))
{
local_min = a[i];
}
else local_min += a[i];
global_min = min(global_min, abs(local_min));
}
cout << global_min << endl;
}
Suppose I have an array of integers int a[] = {0, 1, ... N-1}, where N is the size of a. Now I need to generate all permutations of a s that a[i] != i for all 0 <= i < N. How would you do that?
Here's some C++ implementing an algorithm based on a bijective proof of the recurrence
!n = (n-1) * (!(n-1) + !(n-2)),
where !n is the number of derangements of n items.
#include <algorithm>
#include <ctime>
#include <iostream>
#include <vector>
static const int N = 12;
static int count;
template<class RAI>
void derange(RAI p, RAI a, RAI b, int n) {
if (n < 2) {
if (n == 0) {
for (int i = 0; i < N; ++i) p[b[i]] = a[i];
if (false) {
for (int i = 0; i < N; ++i) std::cout << ' ' << p[i];
std::cout << '\n';
} else {
++count;
}
}
return;
}
for (int i = 0; i < n - 1; ++i) {
std::swap(a[i], a[n - 1]);
derange(p, a, b, n - 1);
std::swap(a[i], a[n - 1]);
int j = b[i];
b[i] = b[n - 2];
b[n - 2] = b[n - 1];
b[n - 1] = j;
std::swap(a[i], a[n - 2]);
derange(p, a, b, n - 2);
std::swap(a[i], a[n - 2]);
j = b[n - 1];
b[n - 1] = b[n - 2];
b[n - 2] = b[i];
b[i] = j;
}
}
int main() {
std::vector<int> p(N);
clock_t begin = clock();
std::vector<int> a(N);
std::vector<int> b(N);
for (int i = 0; i < N; ++i) a[i] = b[i] = i;
derange(p.begin(), a.begin(), b.begin(), N);
std::cout << count << " permutations in " << clock() - begin << " clocks for derange()\n";
count = 0;
begin = clock();
for (int i = 0; i < N; ++i) p[i] = i;
while (std::next_permutation(p.begin(), p.end())) {
for (int i = 0; i < N; ++i) {
if (p[i] == i) goto bad;
}
++count;
bad:
;
}
std::cout << count << " permutations in " << clock() - begin << " clocks for next_permutation()\n";
}
On my machine, I get
176214841 permutations in 13741305 clocks for derange()
176214841 permutations in 14106430 clocks for next_permutation()
which IMHO is a wash. Probably there are improvements to be made on both sides (e.g., reimplement next_permutation with the derangement test that scans only the elements that changed); that's left as an exercise to the reader.
If you have access to C++ STL, use next_permutation, and do an additional check of a[i] != i in a do-while loop.
If you want to avoid the filter approach that others have suggested (generate the permutations in lexicographic order and skip those with fixed points), then you should generate them based on cycle notation rather than one-line notation (discussion of notation).
The cycle-type of a permutation of n is a partition of n, that is a weakly decreasing sequence of positive integers that sums to n. The condition that a permutation has no fixed points is equivalent to its cycle-type having no 1s. For example, if n=5, then the possible cycle-types are
5
4,1
3,2
3,1,1
2,2,1
2,1,1,1
1,1,1,1,1
Of those, only 5 and 3,2 are valid for this problem since all others contain a 1. Therefore the strategy is to generate partitions with smallest part at least 2, then for each such partition, generate all permutations with that cycle-type.
The permutations you are looking for are called derangements. As others have observed, uniformly randomly distributed derangements can be generated by generating uniformly randomly distributed permutations and then rejecting permutations that have fixed points (where a[i] == i). The rejection method runs in time e*n + o(n) where e is Euler's constant 2.71828... . An alternative algorithm similar to #Per's runs in time 2*n + O(log^2 n). However, the fastest algorithm I've been able to find, an early rejection algorithm, runs in time (e-1)*(n-1). Instead of waiting for the permutation to be generated and then rejecting it (or not), the permutation is tested for fixed points while it is being constructed, allowing for rejection at the earliest possible moment. Here's my implementation of the early rejection method for derangements in Java.
public static int[] randomDerangement(int n)
throws IllegalArgumentException {
if (n<2)
throw new IllegalArgumentException("argument must be >= 2 but was " + n);
int[] result = new int[n];
boolean found = false;
while (!found) {
for (int i=0; i<n; i++) result[i] = i;
boolean fixed = false;
for (int i=n-1; i>=0; i--) {
int j = rand.nextInt(i+1);
if (i == result[j]) {
fixed = true;
break;
}
else {
int temp = result[i];
result[i] = result[j];
result[j] = temp;
}
}
if (!fixed) found = true;
}
return result;
}
For an alternative approach, see my post at Shuffle list, ensuring that no item remains in same position.
Just a hunch: I think lexicographic permutation might be possible to modify to solve this.
Re-arrange the array 1,2,3,4,5,6,... by swapping pairs of odd and even elements into 2,1,4,3,6,5,... to construct the permutation with lowest lexicographic order. Then use the standard algorithm, with the additional constraint that you cannot swap element i into position i.
If the array has an odd number of elements, you will have to make another swap at the end to ensure that element N-1 is not in position N-1.
Here's a small recursive approach in python:
def perm(array,permutation = [], i = 1):
if len(array) > 0 :
for element in array:
if element != i:
newarray = list(array)
newarray.remove(element)
newpermutation = list(permutation)
newpermutation.append(element)
perm(newarray,newpermutation,i+1)
else:
print permutation
Running perm(range(1,5)) will give the following output:
[2, 1, 4, 3]
[2, 3, 4, 1]
[2, 4, 1, 3]
[3, 1, 4, 2]
[3, 4, 1, 2]
[3, 4, 2, 1]
[4, 1, 2, 3]
[4, 3, 1, 2]
[4, 3, 2, 1]