jump | ✌️ | Command Line Interface library
kandi X-RAY | jump Summary
kandi X-RAY | jump Summary
Jump integrates with your shell and learns about your navigational habits by keeping track of the directories you visit. It gives you the most visited directory for the shortest search term you type.
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Currently covering the most popular Java, JavaScript and Python libraries. See a Sample of jump
jump Key Features
jump Examples and Code Snippets
def jump_search(arr: list, x: int) -> int:
"""
Pure Python implementation of the jump search algorithm.
Examples:
>>> jump_search([0, 1, 2, 3, 4, 5], 3)
3
>>> jump_search([-5, -2, -1], -1)
2
>&g
def _add_jump_node(self, ast_node, guards):
"""Grows the graph by adding a jump node.
Jump nodes are added to the current leaf set, and the leaf set becomes
empty. If the jump node is the last in a cond section, then it may be added
def min_jumps_2(nums):
n = len(nums)
if n <= 1:
return 0
jumps = 0
max_jump = 0
new_max_jump = 0
for i in range(n):
if max_jump < i:
max_jump = new_max_jump
jumps += 1
t
Community Discussions
Trending Discussions on jump
QUESTION
I was checking "minimum number of jumps to reach the end" problem in GeekforGeeks https://www.geeksforgeeks.org/minimum-number-of-jumps-to-reach-end-of-a-given-array/ . I got confused about the time complexity mentioned there which is O(n^n).
...ANSWER
Answered 2021-Jun-15 at 07:34I can see the recurse relation as T(n) = T(n-1) + T(n-2) + T(n-3) + T(n-4) + ... + T(0)
, since the loop is from l to h (ignore the if
condition for now). So for an interval [l,h]
every value in that interval will be called in the worst case that is minJumps(l+1, h), minJumps(l+2, h) ... minJumps(h, h)
and it can be noticed that the above recurse relation holds here.
Now, solving the relation, we can write it as T(n) = T(n-1) + T(n-1)
as T(n-1) = T(n-2) + T(n-3) + T(n-4) + ... + T(0)
. Hence T(n) = 2 * T(n-1)
which boils down to O(2^n)
.
The time complexity of the mentioned algorithm should be O(2^n)
.
QUESTION
I have a concave hull (not convex) that I have the points for eg: A,B,C,D,E
. I've gotten the pairs of points that make up the outer edges. [A,B],[A,E],[C,D],[B,C],[E,D]
. (This is a very simplified version)
I want to get the connected points in order (CW or CCW doesn't matter) so I can use them as a contour.
But the pairs are not ordered, you can see A goes to B, then A goes to E, etc. The only solution I had was searching for each point and its next pair sequentially in a loop
Is there a way to solve this using numpy only in a vectorized manner so that its fast for a large array of edges? I know shapely exists but I have trouble installing it and I'd prefer no external dependancies
this is my code:
...ANSWER
Answered 2021-Jun-15 at 08:27You can do this efficiently with a dictionary:
QUESTION
The problem is the following: I got a png file : example.png
that I filter using chan vese of
skimage.segmentation.chan_vese
- It's return a png file in black and white.
i detect segments around my new png file with
cv2.ximgproc.createFastLineDetector()
- it's return a list a segment
But the list of segments represent disjoint segments.
I use two naive methods to polygonize this list of segment:
-It's seems that cv2.ximgproc.createFastLineDetector()
create a almost continuous list so I just join by creating new segments:
ANSWER
Answered 2021-Jun-15 at 06:36So I use another library to solve this problem: OpenCV-python
We got have also the detection of segments( which are not disjoint) but with a hierarchy with the function findContours
. The hierarchy is useful since the function detects different polygons. This implies no problems of connections we could have with the other method like explain in the post
QUESTION
I am new to MIPS assembly. I am trying to convert a java code to MIPS code but I can't figure it out that how can I load and store double values in MIPS. I get this error "address not aligned on doubleword boundary 0x10010001". Here is the java code:
...ANSWER
Answered 2021-Jun-14 at 17:49Use syscall function code 3 to print doubles (not function code 2 — that's for single float).
Use mov.d
to copy one register to another, e.g. into $f12
for sycalls (then no need to load 0.0 into $f0
).
(Also, use l.d
instead of ldc1
.)
The loop:
loop isn't a loop (there is no backward branch for it).
As @Peter says, your scaling/offsets are not properly accounting for the size of double, which is 8. Scaling needs to multiply by 8 (shift by 3, not 2), and offsets need to be multiples of 8.
Your first loop, when its done, should not EXIT the program but rather continue with the next statement (i.e. the printing loop).
For this statement numbers[i+2] = numbers[i+1] + numbers[i];
there are two loads and one store as array references, whereas the assembly code is doing 3 loads and no store.
You use $s5
, but never put anything in it.
The comparison of numbers[i+1] < 150
has to be done in (double) floating point, whereas the assembly code is attempting this in integer registers.
Floating point comparison is done using c.eq.d
, c.lt.d
, or c.le.d
. Like with integer compares, constants need to be in registers before the compare instruction. 150.0 would best come directly from memory (as a double constant) before the loop and remain there for the loop duration (or you can move integer 150 into a floating point register and convert that to double (cvt.w.d
)).
Conditional branches on floating point compare conditions use either bc1t
or bc1f
(depending on the sense you want).
The C-like pseudo code probably doesn't run properly. The exit condition for the first loop is suspect — it will probably run off the end of the array (depending on its initial data values).
Translating non-working C code into assembly is a recipe for frustration. Suggest getting it working in C first; run it to make sure it works.
QUESTION
we are using a STM32H743VIT6 on a custom board with a JLink debugger. Out of the blue the processor jumps in a state where it isn't possible to flash the ECU anymore. The board is running but nether JMem nor our IDE (uVision) are able to access or detect the controller. Has anyone else encountered this behaviour so far? Google wasn't helpful either.
...ANSWER
Answered 2021-May-30 at 19:21It is almost impossible to archive unless you enable RDP (which is very hard to archive if it was not the intention of the programmer).
You probably have screw-up the board design. You should have pull-up resistors on the debug lines and NRST connected to the programmer.
If you do not have NRST available simple solder the wire to the NRST, and when the programming probe connects to the uC, connect it to the GND.
If the NRST line is connected to the programmer you need to select nn the configuration "Connect under Reset"
QUESTION
Taking the following C code
...ANSWER
Answered 2021-Jun-14 at 11:23If you read the assembler code from the top you will see that it reaches .L3
, plus it also jumps to it with jne .L3
, which is your for
loop in C.
QUESTION
I have a div with contenteditable="true"
and resize: both
attributes which has centered text via flexbox
ANSWER
Answered 2021-Jun-11 at 13:28With JavaScript you can insert a space ('
') when you detect the contenteditable is empty. This pushes the caret position to the center.
QUESTION
I am new to unity and trying to make the player jump on start. I have the following code
...ANSWER
Answered 2021-Jun-14 at 06:34You need an instance of class Rigidbody
QUESTION
I have recovered an old 6502 emulator I did years ago to implement some new features. During testing I discovered something wrong, surely due to an error in my implementation.
I have to loop through a 16 bit subtraction until the result is negative: quite simple, no? Here is an example:
ANSWER
Answered 2021-May-25 at 12:22loop through a 16 bit subtraction until the result is negative
"Branch" to Label if result is >0,
Do you see that these descriptions contradict each other?
The 1st one continues on 0, the 2nd one stops on 0.
Only you can decide which one is correct!
From a comment:
Do ... Loop While GE 0This code is part of a Bin to Ascii conversion, made by power of ten subtraction. The bin value could be >$8000, so it is 'negative' but this does not matter. In the first iteration I sub 10000 each cycle until the result is 'below 0', then I restore the previous value and continue with the remainder. The problem is how to detect the 'below 0' condition as said in the post
Next example subtracts 10000 ($2710) from the unsigned word stored at zero page address $90. The low byte is at $90, the high byte is at $91 (little endian).
QUESTION
I try to finish a login function on my web application; however, when I enter the correct password and username already registered in my database, it always returns 404.
I want to use sessions to identify each unique user. And what I also want to know how to jump to a new webpage after login in successfully.
Here is my code in app.js:
...ANSWER
Answered 2021-Jun-14 at 03:52Edit page2.html
in both app.js post and html form action to page2
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