![3) (18 pts.) For all algorithms give the analysis for the worst-case time bound. Read the problem statements carefully. You cannot use hashing (i) Given is an array A of size n containing integers in arbitrary order and an integer T. Describe and analyze an efficient algorithm determining whether there exist three entries A[i], A , and Ak] with (ii) Given is an array A of size n containing positive integers in arbitrary order and an integer T. Describe and analyze an efficient algorithm determining whether there exist two indices i and j, 1 < i < j , such that Ali Ali+1]+... + Aj] < T. When a solution exists, determine the one with j - i a maximum over all solutions. For example, consider A , 21, 2, 22, 3, 23, 4, 24, 5) T-10 has no solution T and J-2): T-25 has a solution consisting of 3 elements (1+21+2) 22 has one solution (i 1 there exist two indices i and j, 1 icj-n such that 0 < 시i1+샤+1] + +AJ1ST 10 131+4-2ST O(n) space can be used. Clearly state what a recursive call returns and how the combine step works. (iii) Given is an array A of size n containing integers in arbitrary order and an integer T. Determine whether For example, for A3, 5, 10, -13, 1, 4, 16] and T-2, a solution exists fori3 and j-6 with Describe and analyze a Divide-and-Conquer algorithm solving the problem in O(n log n) time. Additional Note: You will get partial credit for an O(n logn ime solution. No credit will be given for a quadratic time solution.](https://d2vlcm61l7u1fs.cloudfront.net/media%2F276%2F276efea1-426b-4464-8895-1a877bfb0dc7%2FphpHMlSiy.png)
For all algorithms give the analysis for the worst-case time bound. Read the problem statements carefully. You cannot use hashing. (i) Given is an array A of size n containing integers in arbitrary order and an integer T. Describe and analyze an efficient algorithm determining whether there exist three entries A[i], A[j], and A[k] with A[i] + A[j] + A[k] lessthanorequalto T, i
Hi Friend, Please find Algorithm for Q1.
Please repost other questions in separate post.
Please let me knwo in case of any issue in Q1.
Algorithm:
1. First, sort the array.
2. Then, loop over the array with the first pointer i.
3. Now, use a second pointer j to loop up from there and a third pointer k to simultaneously loop down from the end.
Whenever you’re in a situation where A[i]+A[j]+A[k] < X,
you know that the same holds for all j<k'<k so increment your count with k-j and increment j.
I keep the hidden invariant that A[i]+A[j]+A[k+1] >= X,
so incrementing j only makes that statement stronger.
Otherwise, decrement k. When j and k meet, increment i.
You will only increment j and decrement k, so they need O(n) amortized time to meet.
Time Complexity: O(n^2) time:
In pseudocode:
count= 0
for i = 0; i < N; i++
j = i+1
k = N-1
while j < k
if A[i] + A[j] + A[k] < X
count += k-j
j++
else
k–
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