incorrect

(defn longest-unique-substring [str]
  (let [n (count str)
        res (atom 0)
        last-index (vec (repeat 256 -1))
        i (atom 0)]
    (doseq [j (range n)]
      (reset! i (max @i (+ (@last-index (get str j)) 1)))
      (reset! res (max @res (- j @i 1)))
      (assoc! last-index (int (get str j)) j))
    @res))

(let [str "geeksforgeeks"]
  (println "The input string is" str)
  (let [len (longest-unique-substring str)]
    (println "The length of the longest non-repeating character substring is" len)))
    
(defn -main [& args]
  (let [s "abcabcbb"]
    (println (str s " : " (length-of-longest-substring s))))
  
  (let [s "bbbbb"]
    (println (str s " : " (length-of-longest-substring s))))
  
  (let [s "bbabcdb"]
    (println (str s " : " (length-of-longest-substring s))))
  
  (when-let [s (first args)]
    (println (str s " : " (length-of-longest-substring s)))))

Now let's analyze the time and space complexity of the Clojure code:

Time Complexity Analysis:

1. The code has a single loop that iterates over each character of the input string using dotimes. The loop runs for n iterations, where n is the length of the string.
2. Inside the loop, the operations of accessing and updating elements in the m array, as well as performing comparisons and calculations, all take constant time O(1).
3. Therefore, the overall time complexity is O(n), where n is the length of the input string.

Space Complexity Analysis:

1. The code uses additional space for the m array, which has a fixed size of 256 elements. Therefore, it requires constant space O(1).
2. The variables n, max-len, and last-repeat-pos require constant space as well.
3. Therefore, the overall space complexity is O(1).

In summary, the time complexity of the length-of-longest-substring function is O(n), and the space complexity is O(1), where n is the length of the input string.