# 4.9

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

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

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

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

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