**Question:** Find all the characters, which are present in the character array given. The order of character should remain intact.**Example:**i/p: acdaacbdac

o/p: acdb

**Answer:**The naïve solution will be to have a result array and check for every character in input string in that. If the character is already present in the result, go to next character. This will be a O(n^2) solution. Print the result array, as it is to save the order. Lets see if we can find some better solution.

Lets have an array ‘result’ of 255 characters (to cover all characters). Initialize it to zero. Now for every element in input, set result[input[i]] to 1. So now ‘result’ array will have 1 at indexes, which are present in input and ‘0’ otherwise. Now in 2nd scan of input array, if result[input[i]] is 1, output it, make it ‘0’ else go on.

**Algorithm:**

1 n <- length(input) - 12 for i <- 1 to n3 ....do result[input[i]] = 14 for i ← 1 to n5 ....if result[input[i]]6 ........do output = output + result[input[i]]7 ........do result[input[i]] = 08 return result

**Code:**I have written the code in ruby. Also I have used Hash in pace of result array as I didn’t had much time. I have also checked if hash element already present or not.

def unique(string)

len = string.length

h = Hash.new

for i in 0..len-1

if not h[string[i]]

h[string[i]] = 1

end

end

for i in 0..len-1

if h[string[i]] == 1

p string[i]

h.delete(string[i])

end

end

end

**Complexity:**Since we are just traversing the array twice. Complexity of this solution is O(n).

**PS:**One can use 1 bit per character to represent presence in the input array. But since we have just 255 length array, it won't help much in reducing space complexity.

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ManishI think there is no need for second traversal as well:

for i in 0..len-1

if not h[string[i]]

h[string[i]] = 1

p string[i]end

end

--Now delete the complete hash in one go--

Akash@Manish: Absolutely correct :)