java - Does break statement make my code faster? -

i supposed write code if array has duplicates. runtime wasn't important. think below code have o(n²) because used nested for-loop. know there better , faster codes 1 have written, my question if break statement made inside if-statement make code (a bit) faster? should make faster because program knows "hey, found duplicate , can stop searching more". once heard fellow student code better / more stable if avoid statements return or break. bad didn't care enough ask why. maybe can tell me if true?

and if right , these statements "hurt" code, there better workarounds?

public class findduplicate{     public static void main(string[] args){         int[] a={1,2,3,4,5,6,7,8,4};         boolean bool=false;         for(int i=0; i<a.length; i++){             for(int j=0; j<a.length; j++){                 if(a[i]==a[j] && i!=j){                     bool=true;                     break;                 }             }         }         if(bool==true){             system.out.print("duplicate found");         }else{             system.out.print("no duplicate found");         }     } } 

my question if break statement made inside if-statement make code (a bit) faster?

not in every case, however, in cases, make code faster considering don't have keep iterating when you've found you're after.

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the algorithm below contains 2 nested loops. outer loop iterates on array’s n items, takes o(n) steps. each trip through outer loop, inner loop iterates on n items in array, takes o(n) steps. because 1 loop nested inside other, combined performance o(n × n) = o(n2).

for(int = 0; < a.length; i++){     for(int j=0; j < a.length; j++){        if(a[i] == a[j] && != j){            bool = true;            break;        }     } } 

we can make algorithm bit faster not going j = 0 @ each iteration of outer loop.

for(int = 0; < a.length; i++){   for(int j = i+1; j < a.length; j++){      if(a[i] == a[j]){          bool = true;          break;      }   } } 

note in case don't need check && != j because never equal.

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i once heard fellow student code better / more stable if avoid statements return or break

the jvm specification not state either existence or absence of performance loss when using break. put simply, there no proof whatsoever using break or return makes code unstable(not know of anyway). scenario in "oh not practise" when overuse word break. however, in many cases break possibility accomplish task faster, example current solution. basically, why carry on iterating when you've found after?. consider return not being "a bad practise" because similar break, why carry on executing code when don't need to, surely makes code faster.

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can make find duplicate algorithm faster?

sure can, consider set interface in java doesn't allow duplicates , it's based upon hash table data structure insertion take o(1) time in average case. using hashset, general purpose set implementation, can find duplicates in o(n) time. since hashset allows unique elements, add() method fail , return false when try add duplicates.

solution:

public static boolean hasduplicate(int[] array) {       set<integer> dupes = new hashset<integer>();       (integer : array) {           if (!dupes.add(i)) {              return true; // have found duplicate           }       }       return false; // no duplicate }