Difference between revisions of "TADM2E 4.10"

Revision as of 14:27, 23 July 2020

Here the main thing to notice is that we need a O($ n^{k-1}logn $) solution.
For various values of k,
k Solution Time Complexity
1 O($ n^0logn $)
2 O($ n^1logn $)
3 O($ n^2logn $)
4 O($ n^3logn $)

for k = 2 onwards
1. sort the array ( nlogn )
2. use k-1 loops for iterating over array, where ith loop starts from i-1 + 1 element of array, and then use binary search
eg:
for k = 3
for( i = 0; i < n; i++ )
for( j = i+1; j < n; j++ )

1. now use binary search to find ( T - a[i] - a[j] ) from j+1 element to end of array
   

Another recursive solution

First, we note that an O($ n \lg n $) solution for $ k = 2 $ exists , which is already within the required bounds. Now, for $ k \ge 3 $, we can do something like this:

sort S ascending;
CheckSumOfK(S, k, T); // defined below

// S must be aorted array of integers
function CheckSumOfK(S, k, T)
    if k <= 2:
        Use any method in the solution of ex 4.8 and return the result. This can be done in O(n) time because S is sorted.

    Initialize array A of n - 1 elements;
    for i from 0 to n - 1:
        k = 0
        // Collect in A all numbers in S but S[i]. Note that A will still be sorted.
        for j from 0 to n - 1:
            if i != j:
                A[k] = S[j];
                k = k + 1
        // If S[i] is included in the k integers that sum up to T, then there must
        // exactly (k - 1) integers in the rest of S (i.e., A) that sum to (T - S[i]).
        // We can find that out by calling ourselves recursively.
        if CheckSumOfK(A, k - 1, T - S[i]):
            return True
    return False

Complexity

For each item in S, there is $ O(n) $ work done in assembling the array A, except at the $ {k - 1}^{th} $ recursive call, which completes in $ O(n \lg n) $ time. So, for each number in S, we have $ O(kn) + O(n \lg n) $ work, and since $ k <= n $, each iteration of the outer loop takes $ O(n^2) + O(n \lg n) = O(n^2) $ work. Now since the outer loop goes on for at most $ n $ iterations, we have a total runtime of $ O(n^3) $.

A trick can be used to lower this runtime to $ O(n^2 \lg n) $ by having the routines in this solution take an index to ignore when iterating in their inner loops. With this, we save the $ O(n) $ construction of A for every item, and then each iteration of the outer loop becomes $ O(n \lg n) $ (How? $ O(k) = O(n) $ constant time recursive calls plus $ O(n \lg n) $ time spent in the $ {k - 1}^{th} $ calls), giving a total runtime of $ O(n^2 \lg n) $ for $ n $ elements in S.

Note that this cost includes the initial $ O(n \lg n) $ cost for sorting S. Also, this algorithm is always going to be $ O(n^{k-1} \lg n) $ for all $ k >= 2 $. The exercise just states an upper bound.