Question

I have a list of numbers like this (array)

1 2 3 4

So my goal is the check given another array if this array if a permutation of the original example, the array (3 4 1 2) and (1 2 4 3) are permutations of the original but (1 2 1 1) or (1 5 4 3) not.

Answer 1

Two possible solutions are:

(1) O(n) space & average time solution will be to create a histogram, based on a hash table, of the data – and check if the histograms are identicals. The idea is – count how many each element appears in each list, and then check to see each element appears exactly the same times in each array.

pseudo code:

map1 = new map //first histogram
map2 = new map //second histogram
for each element in arr1: //create first histogram
   if (element in map1):
         map1.put(element,map1.get(element)+1)
   else:
         map1.put(element,1)
for each element in arr2: //create second histogram
   if (element in map2):
         map2.put(element,map2.get(element)+1)
   else:
         map2.put(element,1)
for each key in map 1: //check all elements in arr1 appear in arr2
   if map1.get(key) != map2.get(key):
        return false
//make sure the sizes also match, it means that each element in arr2 appears in arr1.
return arr1.length == arr2.length 

(2) O(nlogn) time solution will be to sort both arrays, and then iterate and check if they are identical.