You are given the post order traversal, P, of a binary search tree on the n element, 1,2,....,n. You have to determine the unique binary search tree that has P as its post order traversal. What is the time complexity of the most efficient algorithm for doing this?

The minimum number of comparisons required to determine if an integer appears more than n/2 times in a sorted array of n integers is

Consider the following C functions:

int f1(int n){
 if(n == 0 || n == 1){
    return n;
 }
 return (2 * f1(n - 1) + 3 * f1(n - 2));
}
int f2(int n){
 int i;
 int X[N], Y[N], Z[N];
 X[0] = Y[0] = Z[0] = 0;
 X[1] = 1; Y[1] = 2; Z[1] = 3;
 for(i = 2; i <= n; i++){
  X[i] = Y[i - 1] + Z[i - 2];
  Y[i] = 2 * X[i];
  Z[i] = 3 * X[i];
 }
 return X[n];
}

The returning time of f1(n) and f2(n) are

Consider the following C functions:

int f1(int n){
 if(n == 0 || n == 1){
    return n;
 }
 return (2 * f1(n - 1) + 3 * f1(n - 2));
}
int f2(int n){
 int i;
 int X[N], Y[N], Z[N];
 X[0] = Y[0] = Z[0] = 0;
 X[1] = 1; Y[1] = 2; Z[1] = 3;
 for(i = 2; i <= n; i++){
  X[i] = Y[i - 1] + Z[i - 2];
  Y[i] = 2 * X[i];
  Z[i] = 3 * X[i];
 }
 return X[n];
}

f1(8) and f2(8) return the values