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Understanding return [0,size-1][nums[0]<nums[size-1]] in Python


Understanding return [0,size-1][nums[0]<nums[size-1]] in Python

By : Juan Pablo Peveri
Date : October 24 2020, 06:10 PM
will be helpful for those in need That last row is an obscure way of writing an if then else expression.
[0, size-1] creates a list of two elements. nums[0] < nums[size-1] returns either True or False when used as a list index, this True/False is implicitly converted into 1 or 0. and by that, either size-1 or 0 is picked from the list.
code :
return size - 1 if nums[0] < nums[size - 1] else 0


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return trailing set of nums from string python

return trailing set of nums from string python


By : Jak
Date : March 29 2020, 07:55 AM
this one helps. How can I return the trailing set of numbers from a list of strings? , You can use regex:
code :
>>> import re
>>> r = re.compile(r'\d+$')
>>> [int(m.group()) for m in (r.search(item) for item in mylist) if m]
[2, 20, 244, 288888]
push and popping ruby arrays nums << (nums.pop) + (nums.pop)

push and popping ruby arrays nums << (nums.pop) + (nums.pop)


By : Rendy Widi Lestari
Date : March 29 2020, 07:55 AM
it helps some times pop cuts off the last element form an array and returns it.
<< is the same as push - adds one element to the end of the array.
code :
[1,2,3] --> pop=3,pop=2,add=(2+3),push=5      --> [1,5]
[1,5]   --> pop=5,pop=1,multiply=(1*5),push=5 --> [5]
How to return the most right positive nums array cut?

How to return the most right positive nums array cut?


By : Jesus Federico
Date : March 29 2020, 07:55 AM
I hope this helps . My task: The input is a one-dimensional array. It is necessary to find the range of maximum width, the elements of which are positive (greater than 0). As a response, there should be an array of 2 elements, where: - Element number 0 - the index element of the left border of the segment; - Element №1 - the index of the element of the right border of the segment. , I see two issues:
code :
if (currentSequenceLength >= sequenceLength) {
    sequenceLength = currentSequenceLength;
    result[0] = firstIndex;
    result[1] = lastIndex;
}
public static int[] lookFor(int[] array) {
    int[] result = new int[2];
    int firstIndex = 0;
    int lastIndex = 0;
    int sequenceLength = 0;
    int currentSequenceLength = 0;

    for (int i = 0; i < array.length; i++) {
        if (array[i] > 0) {
            if ( currentSequenceLength == 0 ) {
                firstIndex = i;
            }

            currentSequenceLength += 1;
            lastIndex = i;
        } else {
            if (currentSequenceLength >= sequenceLength) {
                sequenceLength = currentSequenceLength;
                result[0] = firstIndex;
                result[1] = lastIndex;
            }
            currentSequenceLength = 0;
        }
    }
    if (currentSequenceLength >= sequenceLength) {
        sequenceLength = currentSequenceLength;
        result[0] = firstIndex;
        result[1] = lastIndex;
    }


    if (sequenceLength == 0) {
        return new int[0];
    }

    return result;
}
`nums[nums.length - 1]; ` is an range or single element?

`nums[nums.length - 1]; ` is an range or single element?


By : Tom
Date : March 29 2020, 07:55 AM
this one helps. It is a single element, because num.lengh - 1 is an int and just gives you the last accessible index of the array.
If you check wether the length of the array is > 0, then you can safely use the length of the array to determine the last accessible index.
code :
for (int j = 0; j < nums.length; j++)
for (int j = 0; j <= nums.length - 1; j++)
for (int j = 0; j <= nums.length; j++) 
Difference between nums[:] = nums[::-1] and nums = nums[::-1]

Difference between nums[:] = nums[::-1] and nums = nums[::-1]


By : adarsh sharma
Date : October 07 2020, 04:00 PM
help you fix your problem If you know about C and pointers, you can picture nums as a pointer to a list.
Consider this case:
code :
def reverse(nums):
    nums = nums[::-1]

my_list = [1, 2, 3]
reverse(my_list)
def reverse(nums):
   nums[:] = nums[::-1]

my_list = [1, 2, 3]
reverse(my_list)
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