## Find the highest subsequence

print longest increasing subsequence

maximum sum increasing subsequence

maximum sum subsequence of length k

longest increasing subsequence python

longest increasing subsequence pseudocode

number of longest increasing subsequence

maximum sum increasing subsequence hackerrank

**Closed**. This question needs details or clarity. It is not currently accepting answers.

Using `itertools.groupby`

, you can group the lists by the first element, then sort each list by the length, and take the last(=longest) sublist. The latter two operations are equivalent to taking the `max`

, as suggested by HeapOverflow.

from itertools import groupby inp = [[1], [1, 1], [1, 1, 1, 1, 1], [2, 2], [3], [3, 3]] [max(seq, key=len) for _,seq in\ groupby(sorted(inp), key=lambda x: x[0])] #[[1, 1, 1, 1, 1], [2, 2], [3, 3]]

**Maximum Sum Subsequence of length k,** If you are unknown of how to calculate the longest increasing subsequence then see the implementation going to the link. Naive Approach: In the� Longest subsequence such that every element in the subsequence is formed by multiplying previous element with a prime; Maximum subsequence sum such that no three are consecutive; Minimum sum subsequence such that at least one of every four consecutive elements is picked; Largest increasing subsequence of consecutive integers

In case order matters (note the changed example data):

>>> Input = [[1], [3], [2, 2], [1, 1, 1, 1, 1], [2], [1, 1], [3, 3]] >>> d = {} >>> for l in Input: x = l[0] if x not in d: d[x] = l elif len(l) > len(d[x]): del d[x] d[x] = l >>> list(d.values()) [[2, 2], [1, 1, 1, 1, 1], [3, 3]]

A variation suggested by @kaya3:

>>> for l in Input: x = l[0] if len(l) > len(d.get(x, [])): d.pop(x, None) d[x] = l

**Longest Consecutive Subsequence,** Update max with a maximum of count and max. filter_none find the maximum length Length of the Longest contiguous subsequence is 4. The Longest Increasing Subsequence (LIS) problem is to find the length of the longest subsequence of a given sequence such that all elements of the subsequence are sorted in increasing order. For example, the length of LIS for {10, 22, 9, 33, 21, 50, 41, 60, 80} is 6 and LIS is {10, 22, 33, 50, 60, 80}.

Using a dict to keep track of each value's longest list:

>>> list({l[0]: l for l in Input}.values()) [[1, 1, 1, 1, 1], [2, 2], [3, 3]]

If they're *not* already sorted like your example suggests, then just sort them first, either simply with `Input.sort()`

or more efficiently with `Input.sort(key=len)`

.

**Maximum Sum Increasing Subsequence,** Write a program to find the sum of maximum sum subsequence of the given array such that the integers in the subsequence are sorted in� In previous post, we have discussed about Maximum Sum Increasing Subsequence problem. However, the post only covered code related to finding maximum sum of increasing subsequence, but not to the construction of subsequence. In this post, we will discuss how to construct Maximum Sum Increasing Subsequence itself. Let arr[0..n-1] be the input array.

Assuming your input data always has subsequences with identical numbers, you could use a `collections.defaultdict()`

to group by the first value of each subsequence, then output the longest subsequence with `max()`

and `key=len`

.

from collections import defaultdict lst = [[1], [1, 1], [1, 1, 1, 1, 1], [2, 2], [3], [3, 3]] d = defaultdict(list) for x in lst: d[x[0]].append(x) print([max(x, key=len) for x in d.values()]) # [[1, 1, 1, 1, 1], [2, 2], [3, 3]]

Note that dictionaries in Python 3.7+ are ordered by insertion. If you are using a lower Python version, then simply sort the resulting dictionary by the key: `[max(v, key=len) for _, v in sorted(d.items())]`

.

**Find the subsequence with largest sum of elements in an array ,** If you want the largest sum of sequential numbers then something like this might work: $cur = $max = 0; foreach ($seq as $n) { $cur += $n; if ($cur < 0) $cur = 0;� Write a program to find the sum of maximum sum subsequence of the given array such that the intgers in the subsequence are sorted in increasing order. The longest increasing subsequence is [2, 3, 7, 101], therefore the length is 4.

**Solving the maximum subsequence sum and related problems ,** Given a sequence of n numbers, the task of finding the contiguous subsequence, with maximum sum over all subsequences of the given� Maximum sum increasing subsequence Given an array, this function will find the sum of maximum subsequence of the given array, that is the integers in the subsequence are in sorted order. Note: A single number is supposed to be an increasing subsequence of size 1.

**Maximum subarray problem,** In computer science, the maximum sum subarray problem is the task of finding a contiguous 2012. www.algorithmist.com � alexeigor.wikidot.com � greatest subsequential sum problem on Rosetta Code � geeksforgeeks page on Kadane's � Write an efficient program to find the sum of contiguous subarray within a one-dimensional array of numbers which has the largest sum. Simple idea of the Kadane’s algorithm is to look for all positive contiguous segments of the array (max_ending_here is used for this). And keep track of maximum

**Increasing Subsequence with Maximum Sum,** Find a subsequence of a given sequence such that subsequence sum is as high as possible and subsequence's elements are in sorted order, from lowest to� Given an array of integers, find the maximum sum of subsequence of given array where subsequence contains no adjacent elements. The problem is similar to 0/1 Knapsack problem where for every item, we have two choices - to include that element in the solution or to exclude that element from solution.

##### Comments

- Okay, cool. You've told us what you want. What have you tried? How did it fail to do what you want? Please read How to Ask.
- As Chris said, please provide some code regarding your attempts. The problem seems fairly straightforward so if you show us what you tried and where you fail, we might probably be able to help you with the necessary adjustments :)
- @CeliusStingher I put my last attempt, but I am new using Python and, i didn't post beacuse i tought that it was useless since i think that's the wrong way
`r`

doesn't match your original input, and`r[j][0]`

would raise a`TypeError`

(which maybe is the reason for the question).- What's up with that
`r=[1,2,2,2,2,1,1,1,1]`

in your attempt? Doesn't look anything like the input. - Could be simplified to avoid the two branches:
`if len(l) > len(d.get(x, ())):`

. Also, no need to`del d[x]`

before putting a new value there. - @kaya3 Have you tried that? As expected, it gives me the wrong order.
- Oh, you want the output in a specific order. I took
*"in case order matters"*just to mean that the input might not have lists with the same values adjacent to each other. In that case,`d.pop(x, None)`

removes it if present, avoiding the need to branch. - @kaya3 Not sure how exactly you mean it, but I think that would also give me a wrong order.
- @kaya3 Ah, together with
`get`

. Yeah, ok. Hmm. I'm actually not sure which one I like better...