Question

What is the time complexity of the following implemented algorithm?

I should notice that length of b is enough to cover the element of a as an index.

void smax(int[] a, int n){    
    int[] b = new int[n];
    for (int i=0;i<b.length;i++){
        b[i]=0;
    }
    int m=0;
    while (m<b.length) {
        int k=a[0];
        for (int i=0;i<a.length;i++) {
            if (a[i]> k && b[a[i]]!=1) {
               b[a[i]]=1;
            }
         }
         m++;
    }
    for (int i=0;i<a.length;i++){
         if (b[a[i]]!=1){
            b[a[i]]=1;
         }
    }
    for (int j=0;j<b.length;j++){
         if (b[j]==1){
            System.out.println(j);
         }
    }
}

Answer 1

Looks like homework so the best answer would not only be the final answer but also one you could learn from.

Let n = a.Lengh, m = b.Length

for (int i=0;i<b.length;i++){
     b[i]=0;
  }

makes one pass on b’s elements so it would contribute m steps.

for (int i=0;i<a.length;i++){
     if (b[a[i]]!=1){
        b[a[i]]=1;
     }
  }

makes one pass on a’s elements so it would contribute n steps.

for (int j=0;j<b.length;j++){
     if (b[j]==1){
        System.out.println(j);
     }
  }

makes one pass on b’s elements so it would contribute m steps.

so far we have 2m+n

int m=0;
  while (m<b.length) {
     int k=a[0];
     for (int i=0;i<a.length;i++) {
        if (a[i]> k && b[a[i]]!=1) {
           b[a[i]]=1;
        }
     }
     m++;
  }

for every element of b there is a pass on all a’s elements which contribute mn steps.

the sum of all steps is 2m+n+mn which in the asymptotic notation is O(mn).