What is the Big Oh of these two implementations of the same algorithm? -

i writing algorithm return true if string contains no duplicate characters, , false otherwise. have 2 different implementations:

the first uses hashmap:

public static boolean hasuniquecharsfast(string s) {     char[] input = s.tochararray();     map<character, character> map = new hashmap<character, character>();     (char c: input) {         if (map.get(c) != null) {             return false;         } else {             map.put(c, c);         }     }     return true; } 

the second in-place , compares each character every other character:

public static boolean hasuniquechars(string s) {         char[] input = s.tochararray();         //compare string (i.e. ignore chars @ same index)         (int i=0;i<input.length;i++) {             (int j=0;j<input.length;j++) {                 if (input[i] == input[j] && != j ) {                     return false;                 }             }         }         return true;     } 

is correct big oh of first implementation o(n), , big oh of second implementation o(n^2)?

i guess trade-offs first 1 uses additional storage space, whereas second implementation doesn't use additional storage (i.e. in-place)?

is correct big oh of first implementation o(n), , big oh of second implementation o(n^2)?

yes. hash operations considered have constant cost.

i guess trade-offs first 1 uses additional storage space, whereas second implementation doesn't use additional storage (i.e. in-place)?

yes. should note constant value of first higher. large strings first algorithm crush second.

note code can optimized twice fast.

first algorithm (assuming put returns previous value, in java):

public static boolean hasuniquecharsfast(string s) {     char[] input = s.tochararray();     map<character, character> map = new hashmap<character, character>();     (char c: input) {         if (map.put(c, c) != null) {             return false;         }     }     return true; } 

second algorithm:

public static boolean hasuniquechars(string s) {     char[] input = s.tochararray();     //compare string (i.e. ignore chars @ same index)     (int i=0;i<input.length;i++) {         (int j=i+1;j<input.length;j++) {  // changed starting index             if (input[i] == input[j]) {       // removed == j check                 return false;             }         }     }     return true; }