# Generating permutations of NSArray elements

Let’s say I have an NSArray of NSNumbers like this: 1, 2, 3

Then the set of all possible permutations would look something like this:

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• 1, 2, 3

1, 3, 2

2, 1, 3

2, 3, 1

3, 1, 2

3, 2, 1

What’s a good way to do this in objective-c?

### 5 Solutions Collect From Internet About “Generating permutations of NSArray elements”

I have used the code from Wevah’s answer above and discovered some problems with it so here my changes to get it to work properly:

NSArray+Permutation.h

``````@interface NSArray(Permutation)

- (NSArray *)allPermutations;

@end
``````

NSArray+Permutation.m

``````#import "NSArray+Permutation.h"

#define MAX_PERMUTATION_COUNT   20000

NSInteger *pc_next_permutation(NSInteger *perm, const NSInteger size);
NSInteger *pc_next_permutation(NSInteger *perm, const NSInteger size)
{
// slide down the array looking for where we're smaller than the next guy
NSInteger pos1;
for (pos1 = size - 1; perm[pos1] >= perm[pos1 + 1] && pos1 > -1; --pos1);

// if this doesn't occur, we've finished our permutations
// the array is reversed: (1, 2, 3, 4) => (4, 3, 2, 1)
if (pos1 == -1)
return NULL;

assert(pos1 >= 0 && pos1 <= size);

NSInteger pos2;
// slide down the array looking for a bigger number than what we found before
for (pos2 = size; perm[pos2] <= perm[pos1] && pos2 > 0; --pos2);

assert(pos2 >= 0 && pos2 <= size);

// swap them
NSInteger tmp = perm[pos1]; perm[pos1] = perm[pos2]; perm[pos2] = tmp;

// now reverse the elements in between by swapping the ends
for (++pos1, pos2 = size; pos1 < pos2; ++pos1, --pos2) {
assert(pos1 >= 0 && pos1 <= size);
assert(pos2 >= 0 && pos2 <= size);

tmp = perm[pos1]; perm[pos1] = perm[pos2]; perm[pos2] = tmp;
}

return perm;
}

@implementation NSArray(Permutation)

- (NSArray *)allPermutations
{
NSInteger size = [self count];
NSInteger *perm = malloc(size * sizeof(NSInteger));

for (NSInteger idx = 0; idx < size; ++idx)
perm[idx] = idx;

NSInteger permutationCount = 0;

--size;

NSMutableArray *perms = [NSMutableArray array];

do {
NSMutableArray *newPerm = [NSMutableArray array];

for (NSInteger i = 0; i <= size; ++i)

} while ((perm = pc_next_permutation(perm, size)) && ++permutationCount < MAX_PERMUTATION_COUNT);
free(perm);

return perms;
}

@end
``````

There might be a better way to do this (in-place, or something), but this seems to work:

``````@interface NSArray (PermutationAdditions)

- (NSArray *)allPermutations;

@end
``````

Implemetation:

``````@implementation NSArray (PermutationAdditions)

NSInteger *pc_next_permutation(NSInteger *p, const NSInteger size) {
// slide down the array looking for where we're smaller than the next guy
NSInteger i;
for (i = size - 1; p[i] >= p[i + 1]; --i) { }

// if this doesn't occur, we've finished our permutations
// the array is reversed: (1, 2, 3, 4) => (4, 3, 2, 1)
if (i == -1)
return NULL;

NSInteger j;
// slide down the array looking for a bigger number than what we found before
for (j = size; p[j] <= p[i]; --j) { }

// swap them
NSInteger tmp = p[i]; p[i] = p[j]; p[j] = tmp;

// now reverse the elements in between by swapping the ends
for (++i, j = size; i < j; ++i, --j) {
tmp = p[i]; p[i] = p[j]; p[j] = tmp;
}

return p;
}

- (NSArray *)allPermutations {
NSInteger size = [self count];
NSInteger *perm = malloc(size * sizeof(NSInteger));

for (NSInteger idx = 0; idx < size; ++idx)
perm[idx] = idx;

NSInteger j = 0;

--size;

NSMutableArray *perms = [NSMutableArray array];

do {
NSMutableArray *newPerm = [NSMutableArray array];

for (NSInteger i = 0; i <= size; ++i)

} while ((perm = pc_next_permutation(perm, size)) && ++j);

return perms;
}

@end
``````

To use it, simply `#import` the header file, and call `[yourArray allPermutations]`; the method will return an array containing arrays for each permutation.

[Code adapted from PHP code here.]

You can change `NSString *characters` to `id something` and use it for any type of object

``````NSArray *array = [NSArray arrayWithObjects:@"a",@"b",@"c", nil];

NSMutableArray *permutations = nil;

int i = 0;
for (i = 0; i < array.count ; i++){

if (!permutations){
permutations = [NSMutableArray array];
for (NSString *character in array){
}

} else {

//make copy of permutations array and clean og array
NSMutableArray *aCopy = [[permutations copy] autorelease];
[permutations removeAllObjects];

for (NSString *character in array){

//loop through the copy
for (NSArray *oldArray in aCopy){

//check if old string contains looping char..
if ([oldArray containsObject:character] == NO){

//update array
NSMutableArray *newArray = [NSMutableArray arrayWithArray:oldArray];

}

}
}
}

}
NSLog(@"permutations = \n %@",permutations);
``````

I recently stumbled upon the same problem, and wrote a recursive solution that I believe is more readable. It relies on the core principal that is noted here. I’m including it here in case it will help someone:

``````- (NSMutableArray <NSArray *> *)permutationOfArray:(NSMutableArray *)array withPrefixArray:(NSMutableArray *)prefixArray withPermutations:(NSMutableArray <NSArray *> *)permutations {
NSUInteger numArrayElements = [array count];
if (numArrayElements == 0) {
return permutations;
}

for (NSUInteger i = 0; i < numArrayElements; i++) {
// Remove object and add it to the prefix array.
// Note: we must create new copies of the arrays as we
// are running using recursive call and these objects
// are called by reference.
NSMutableArray *newArray = [NSMutableArray arrayWithArray:array];
[newArray removeObjectAtIndex:i];
NSMutableArray *newPrefixArray = [NSMutableArray arrayWithArray:prefixArray];

[self permutationOfArray:newArray withPrefixArray:newPrefixArray withPermutations:permutations];
}

return permutations;
}

- (NSArray <NSArray *> *)allPermutationsOfArray:(NSArray *)array {
return [self permutationOfArray:[NSMutableArray arrayWithArray:array] withPrefixArray:[NSMutableArray array] withPermutations:[NSMutableArray array]];
}
``````

Sample call:

``````NSArray <NSArray *> *permutations = [self allPermutationsOfArray:@[@1, @2, @3]];
for (NSArray *singlePermutation in permutations) {
NSLog(@"%@", [singlePermutation componentsJoinedByString:@", "]);
}
``````

Result:

1, 2, 3

1, 3, 2

2, 1, 3

2, 3, 1

3, 1, 2

3, 2, 1

-log prints for clarity

-work with any object

-more explanations

-cover edge cases where error can result

``````[self allPermutationsOfArray:@[@1,@2,@3,@4]]; // usage

...

-(void)allPermutationsOfArray:(NSArray*)array
{
NSMutableArray *permutations = [NSMutableArray new];

for (int i = 0; i < array.count; i++) { // for each item in the array
NSLog(@"iteration %d", i);
if (permutations.count == 0) { // first time only

NSLog(@"creating initial list");
for (id item in array) { // create a 2d array starting with each of the individual items
NSMutableArray* partialList = [NSMutableArray arrayWithObject:item];
[permutations addObject:partialList]; // where array = [1,2,3] permutations = [ [1] [2] [3] ] as a starting point for all options
}

} else { // second and remainder of the loops

NSMutableArray *permutationsCopy = [permutations mutableCopy]; // copy the original array of permutations
[permutations removeAllObjects]; // remove all from original array

for (id item in array) { // for each item in the original list
NSLog(@"adding item %@ where it doesnt exist", item);

for (NSMutableArray *partialList in permutationsCopy) { // loop through the arrays in the copy

if ([partialList containsObject:item] == false) { // add an item to the partial list if its not already

// update a copy of the array
NSMutableArray *newArray = [NSMutableArray arrayWithArray:partialList];

// add to the final list of permutations
}
}
}
}
[self printArrayInLine:permutations];
}
NSLog(@"%lu permutations",(unsigned long)permutations.count);
}

-(void)printArrayInLine:(NSArray*)twoDimensionArray
{
NSString* line = @"\n";
for (NSArray* array in twoDimensionArray) {
line = [line stringByAppendingString:@"["];
for (id item in array) {
line = [line stringByAppendingString:[NSString stringWithFormat:@"%@,",item]];
}
line = [line stringByAppendingString:@"]\n"];
}
NSLog(@"%@", line);
}
``````

log output for input @[@1,@2,@3]

``````iteration 0
creating initial list
[1,]
[2,]
[3,]
iteration 1
adding item 1 where it doesnt exist and creates a new list
adding item 2 where it doesnt exist and creates a new list
adding item 3 where it doesnt exist and creates a new list
[2,1,]
[3,1,]
[1,2,]
[3,2,]
[1,3,]
[2,3,]
iteration 2
adding item 1 where it doesnt exist and creates a new list
adding item 2 where it doesnt exist and creates a new list
adding item 3 where it doesnt exist and creates a new list
[3,2,1,]
[2,3,1,]
[3,1,2,]
[1,3,2,]
[2,1,3,]
[1,2,3,]
6 permutations
``````