By : user2954353
Date : November 22 2020, 01:01 AM

To fix the issue you can do Assume that we're given a set of pairs S={(x_1,y_1),...,(x_n,y_n)} of integers. What is the most efficient way of computing a maximal sequence of elements (a_1,b_1),...,(a_m,b_m) in S with the property that , Yes, it is possible to do it O(n log n). code :
Share :

LINQ: Finding maximal element (not maximal value)
By : BRusinov
Date : March 29 2020, 07:55 AM

subsequence of numbers with maximal sum, solution isn't clear
By : Tiffanymiau
Date : March 29 2020, 07:55 AM
With these it helps So as you stated, you iterate over the array in one loop and accumulate sum S. On each step you do S += a[i] and you compare your S to current maximum like max = Math.max(S, a[i]). Your question is why should we zero S when S < 0. Let's say it happened on step i. Then (S + a[i+1]) < a[i+1] (because S < 0). Since we are looking for maximum subsequent sum its better to start new subsequence from a[i+1] i.e. set S = 0.

How does finding a Longest Increasing Subsequence that ends with a particular element leads to the solution of finding L
By : David Amram
Date : March 29 2020, 07:55 AM
hope this fix your issue I have understood that to find the solution of LIS problem, we need to find a LIS for every subsequence starting from initial element of the array to the each element that ends with a particular element(the last element), but I am not able to understand how would that help in finally finding a LIS of a given unsorted array, I also understand that this leads to an optimal substructure property and then can be solved, but as mentioned, I dont see how finding LIS(j) that ends with arr[j] will help us. , Consider this sequence as an example: code :
a[] : 10 20 1 2 5 30 6 8 50 5 7
a[] : 10 20 1 2 5 30 6 8 50 5 7
LIS[] : 1 2 1 2 3 4 4 5 6 3 4
50 8 6 5 2 1
1 2 5 6 8 50

Why am I finding the longest increasing subsequence instead of the longest decreasing subsequence?
By : Juan Sarmiento O
Date : March 29 2020, 07:55 AM
I hope this helps . If you are open to any way of looking at it, Wikipedia has some pseudocode which transferred easily into Python and flipped for decreasing subsequence. code :
N = len(X)
P = np.zeros(N, dtype=np.int)
M = np.zeros(N+1, dtype=np.int)
L = 0
for i in range(0, N1):
# Binary search for the largest positive j ≤ L
# such that X[M[j]] <= X[i]
lo = 1
hi = L
while lo <= hi:
mid = (lo+hi)//2
if X[M[mid]] >= X[i]:
lo = mid+1
else:
hi = mid1
# After searching, lo is 1 greater than the
# length of the longest prefix of X[i]
newL = lo
# The predecessor of X[i] is the last index of
# the subsequence of length newL1
P[i] = M[newL1]
M[newL] = i
#print(i)
if newL > L:
# If we found a subsequence longer than any we've
# found yet, update L
L = newL
# Reconstruct the longest increasing subsequence
S = np.zeros(L, dtype=np.int)
k = M[L]
for i in range(L1, 1, 1):
S[i] = X[k]
k = P[k]
S
array([38, 20, 15, 14, 6])

Length of the longest sorted subsequence
By : Anton Emelyanov
Date : March 29 2020, 07:55 AM
wish helps you This is called the Longest Ascending Subsequence problem. You can find it using a simple dynamic programming algorithm described in the article. If all you need is the length of the longest subsequence, you can do it like this:



Related Posts :
