Question

What is the running time of the following method in terms of n and k: public static int foo(int[ ] a, int[ ] b) { int n = a.length; int k = b.length; int counter = 0; for (int i=0; i=0; j--) { for (int x=i; x>=0; x--) { if (a[i]*b[j] == a[x]) { counter++; } } for (int x=j; x

Answer 1

The running time of the provided method, which includes a nested loop structure, is
`O(n^2 * k + n * k^2)`, which simplifies to
`O(n^2 * k)` if n is larger than or equal to k, or
`O(n * k^2)` if k is larger than n.

Step-by-step explanation:

The question asks to determine the running time of a given method in terms of n and k, where n is the length of array a, and k is the length of array b. The method consists of a nested loop structure where the outer loop runs for i from 0 to n-1 and for each iteration of i, an inner loop runs for j from k-1 to 0. Inside this inner loop, there are two additional loops. The first additional loop runs a variable x back over the range from i to 0, while the second additional loop runs another variable x from j to 0. The condition inside checks if the product of a[i] and b[j] is equal to a[x], incrementing a counter if so.

The time complexity of this method can be analyzed by looking at the nesting levels of the loops. The outermost loop and inner loop provide a base of n*k iterations. The two additional loops are dependent on the values of i and j respectively, so they run at most n times for the first and k times for the second. The overall running time is, therefore,
`O(n^2 * k + n * k^2)`, which simplifies to
`O(n^2 * k)` if we assume n is larger than or equal to k, or
`O(n * k^2)` if k is larger than n.