Question

Can I get some help with this? please and thanks

Cost vouw 1. A simple linear Search algorithm on an unsorted array (30 points) Statements 1 int linearSearch( int [ ]a, int n, int target ) { 2 int i = 0; 3 while (i<n) { if (target == array[i]) return i; i++; 7 } 8 return -1; 9 } 2. The number of statements executed for calculating the average value of a two-dimensional array with n*n elements (30 points) Statements Cost 1 float find Avg2D( int [[ ]a, int n) { float sum = 0; int i = 0; while (i<n) { int j = 0; while (j<n) { sum += a[i]; j++; i++; if (n>0) return sum/n; else return 0.Of; 16 }

Answer 1

Solution:

(1)

Explanation:

=>An array a[] is given, an integer variable n is given and a variable target of integer type is given.

=>while(i<n) loop will be executed for the case of i<n otherwise it will not be executed.

=> if(target == array[i]) is true then only it will be returning i value otherwise not.

=>So in worst case from i = 0 to i = n-1 loop will be executed and target value can be found in the last index of array otherwise not found inside the array.

=>So in the worst case loop will run for n number of times from i = 0 to i = n-1

So the cost or time complexity = O(n)

(2)

Explanation:

=>A two-dimensional array a[][] is given and an integer variable n is given.

=>while(i<n) will be executed from i = 0 to i= n-1 in the worst case

=>while(j<n) will be executed from j = 0 to j= n-1 in the worst case and dependent on outer while loop.

=>So total number of times loops will be executed = n*n

=>total number of times loops will be executed = n^2

So the cost or time complexity = O(n^2)

I have explained each and every part with the help of statements attached to it.