Question

Individually or in pairs, please write pseudo-code to solve the following problems and state which Big-θ category the algorithm belongs to (you don’t have to prove it).

1. An array, A[0…n-2], contains n-1 integers from 0 to n-1 in increasing order – one integer is missing in the range. Design the most efficient algorithm you can to find the missing integer.
2. You have an unsorted sequence of college applications in an array, A[0…n-1]. Each application A contains a name A.name, an SAT score A.sat and a Boolean value indicating whether the student applied early decision, A.early. Design the most efficient algorithm you can to rearrange the array so that all the early decision applicants appear in the first i cells of the array A[0…i-1] and all the remaining applicants appear in the last j cells of the array A[i … n-1].

Answer 1

I am writing the pseudo - codes individually .

1. we know that the formula for sum of first N positive numbers is (N * (N+1)) / 2 . So , compute the value of sum of first n-1 numbers and place the value in x . Now, compute the sum of all the A[0....n-2] numbers by traversing the array and place the value in y.

The difference between x and y is the missing number .

The time complexity is O(n) , since we need to traverse whole array for finding the sum value .

PSEUDO - CODE :

int x , y ;

x = ( (n-1) * (( n-1) +1) ) / 2 ;

int sum =0;

for(int i=0;i<=n-2;i++)

sum += A[i] ;

y = sum ;

printf("the missing value is : %d ", x-y);

2. Let i and j are two variables which are used for traversing the array A. we initialize i and j to first two indices  in A.

we have conditions to check ;-

Here , i and j corresponds to the index values

a. if i == late and j == early then swap the values in i and j and increment both the values of i and j

b. if i == late and j == late then increment j

c. if i == early and j == late then increment j

d.if i == early and j == early then increment the values of both i and j .

I keep the boolean value 1 for early in A.early and boolean value 0 for late in A.early  

PSEUDO - CODE :

int i = 0 , j=1 ;

for ( ; j < n ; )

{

if (A[i].early == 0 && A[j].early == 1 )

{

swap ((A[i] , A[j]) ;

i++,j++ ;

}

if (A[i].early == 0 && A[j].early == 0 ) j++;

if (A[i].early == 1 && A[j].early == 0 ) j++;

if (A[i].early == 1 && A[j].early == 1 ) i++ , j++ ;

}