/**

* Title: Pr#3 Review Solutions

* Course: CS100J

* @author Ivan Lysiuk

*/

//problem 1

static int[] rightShift(int[] array)

{

int[] temp = new int[array.length];

int clk;

for (clk= 0; clk < array.length-1; clk++)

temp[clk+1] = array[clk];

temp[0] = array[clk];

return temp;

}

//problem 2

static int[] delete(int[] a, int elem)

{

/*2 steps:

-count # of "non- elem" in init array

-create an array of appropriate size

-copy "non-elem" into the newly created array

*/

int numNonElem=0;

for (int clk=0; clk < a.length; clk++)

if (a[clk]!=elem)

numNonElem++;

int[] b = new int[numNonElem];

//additional counter to move on the new array

int added =0;

for(int clk=0; clk<a.length; clk++)

{

if (a[clk]!=elem)

{

b[added] = a[clk];

added++;

}

}

return b;

}

// problem 3

static boolean isPalindrome(char [] a)

{

int numChecks = a.length/2;

for (int clk=0; clk < numChecks; clk++)

if (a[clk]!=a[(a.length-1)-clk])

return false;

return true;

}

//problem 4

static int[][] copy(int[][] matx)

{

int[][] temp = new int[matx.length][];

for (int inx1=0; inx1<matx.length; inx1++)

{

temp[inx1] = new int[matx[inx1].length];

for (int inx2=0; inx2<matx[inx1].length;inx2++)

temp[inx1][inx2] = matx[inx1][inx2];

}

return temp;

}

//problem 5

static int[][] rightShift(int[][] matrix)

{

/*

the picture says: column major!

column-major <= => the first index counts columns

*/

int[][] temp = new int[matrix.length][];

for (int col=0; col <matrix.length-1; col++)

{

temp[col+1] = new int[matrix[col].length];

for (int cnt=0; cnt <matrix[col].length;cnt++)

{

temp[col+1][cnt] = matrix[col][cnt];

}

}

for (int cnt=0; cnt <matrix[0].length;cnt++)

temp[0][cnt] = matrix[matrix.length-1][cnt];

return temp;

}

//problem 6

static double aver(int[][] matx)

{

int sum =0;

int numElems=0;

//visit ALL elements

//sum them and find their number

for (int inx1=0; inx1 < matx.length; inx1++)

for (int inx2=0; inx2 < matx[inx2].length; inx2++)

{

sum += matx[inx1][inx2];

numElems++;

}

return (double)sum/numElems;

}