You may also refer to the course notes.
Big-O example problems
First problem
In the following method, let the problem size N be rowsCols, which is the number of rows and columns in the square two-dimensional array matrix.
public static boolean isUpperTriangular(double[][] matrix, int rowsCols) {
for (int j = 1; j < rowsCols; j++) {
for (int i = 0; i < j; i++) {
if (matrix[j][i] != 0.0) {
return false;
}
}
}
return true;
}As a function of the problem size N, what is the big-O upper bound on the worst-case running time of this method?
Also: would our answer be different if N was the number of elements in matrix?
Second problem
Assume that, in the following problem, a and b are both square two-dimensional arrays (same number of rows and columns). Let the problem size N be the number of rows and columns.
public static double[][] matrixMult(double[][] a, double[][] b) {
if (a[0].length != b.length) {
throw new IllegalArgumentException();
}
int resultRows = a.length;
int resultCols = b[0].length;
double[][] result = new double[resultRows][resultCols];
for (int j = 0; j < resultRows; j++) {
for (int i = 0; i < resultCols; i++) {
// compute dot product of row j in a and col i in b
double sum = 0;
for (int k = 0; k < b.length; k++) {
sum += a[j][k] * b[k][i];
}
result[j][i] = sum;
}
}
return result;
}As a function of the problem size N, what is the big-O upper bound on the worst-case running time of this method?
Again: would the answer be different if N was the number of elements in the two arrays?
Third problem
Assume that, in the following algorithm, the problem size N is the number of elements in the array.
// Return the index of the element of the array which is equal to
// searchVal. Returns -1 if the array does not contain
// any element equal to searchVal.
// Important: the array must be in sorted order for
// this algorithm to work.
public static<E extends Comparable<E>>
int binarySearch(E[] arr, E searchVal) {
int min = 0, max = arr.length;
while (min < max) {
int mid = min + (max-min)/2;
int cmp = arr[mid].compareTo(searchVal);
if (cmp == 0) {
return mid;
} else if (cmp < 0) {
max = mid;
} else {
min = mid + 1;
}
}
return -1;
}As a function of the problem size N, what is the big-O upper bound on the worst-case running time of this method?
