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It would definitely help to see what you are doing first, but I'll try explaining what I'm doing. What helped me a lot with problems like this is to create multiple variables for top, bottom, left, and right. This allows the code to be much more readable and to have fewer weird index offsets.

My code is below. With this in mind, I encourage you to do a similar problem, such as spiral matrix 1.

public void rotate(int[][] matrix) { if (matrix == null || matrix.length == 0) { return; } int top = 0; int bot = matrix.length - 1; int left = 0; int right = matrix[0].length - 1; while (top < bot && left < right) { if (top != 0) {

} for (int i = 0; left + i < right; ++i) {

int oldTopLeft = matrix[top][left + i]; int oldTopRight = matrix[top + i][right]; int oldBotRight = matrix[bot][right - i]; int oldBotLeft = matrix[bot - i][left]; if (top != 0) { System.out.println(oldTopLeft + ", " + oldTopRight + ", " + oldBotRight + ", " + oldBotLeft); } matrix[top][left + i] = oldBotLeft; matrix[top + i][right] = oldTopLeft; matrix[bot][right - i] = oldTopRight; matrix[bot - i][left] = oldBotRight; } ++top; --right; --bot; ++left; } }

doesn't work, but you probably guessed that already.

ideone.com is a great scratchpad for code, btw. That and codepad.org are great lazy scratchpads when I don't have access to my home computer with all the compilers on it.

Probably not the right answer: the clock's times will match the entire day long. Anticlockwise clocks show the exact same time of day as clockwise clocks, the hands just move in the opposite direction.

I suspect you meant to ask how many times a day will the positions of the clocks' hands match exactly.

your approach works. just follow through with multiple passes.

in first pass, take out the digits with unique chars.

in second pass, numbers depleted in the first pass can be ignored for "uniqueness"

here's the python code:

class Solution: def originalDigits(self, s: str) -> str: counter = Counter(s) unique = { 'z': ('zero', 0), 'w': ('two', 2), 'u': ('four', 4), 'x': ('six', 6), 'g': ('eight', 8), } unique2 = { 'o': ('one', 1), 't': ('three', 3), 'f': ('five', 5), 'v': ('seven', 7), } unique3 = { 'n': ('nine', 9), } result = [0] * 10 self.f(unique, counter, result) self.f(unique2, counter, result) self.f(unique3, counter, result) return ''.join([str(i) * count for i, count in enumerate(result)])

def f(self, unique, counter, result): for chars, num in unique.values(): min_count = 100_001 for ch in chars: min_count = min(min_count, counter[ch]) for ch in chars: counter[ch] -= min_count if min_count != 100_001: result[num] += min_count

if anyone wants to solve it in LC: https://leetcode.com/problems/reconstruct-original-digits-from-english

I don't know the exact definition but I do know ElasticSearch has different queries with different performance. For example a terms query is pretty fast where as a keyWord or String based search (find a substring) is slower.

So depending on the query the time complexity may vary.

That's a pretty amazing countBits function but I'm not familiar enough with the language to understand everything that is happening here. I've only worked with it in school and never for bit operations; I didn't even know you could do that with a high-level language like Java until I saw this and looked it up!

I couldn't find anything on "<<=", what does this symbol do?

edit: nevermind, found it: http://www.tutorialspoint.com/java/java_basic_operators.htm

There is. You need to think in terms of invariants. Here is the article where I learned this approach (like a decade ago 😅): https://www.topcoder.com/community/competitive-programming/tutorials/binary-search

Do you mean having a merge kick off a test? You can integrate workflows into your GitHub.

You can look at continuous integration solutions such as Travis or Circle.

https://travis-ci.org

https://circleci.com

Alternatively you can repost this question in r/devops. They'll have lots if answers there.