I got this as an interview question and was thinking on how to solve it.
Lets say a number is an array ={7,6,3} and its consequtive substring ares {7},{6},{3},{7,6},{6,3},{7,6,3}({7,3} is not valid) check if product of any two subsets are equal).
so {6,2,3} fails as {6}={2*3}
can anyone give me a nudge in right direction.
If there are n numbers in the array. So there are n * (n - 1) / 2 consequtive subsets. You can pre-process thess subsets' product, and insert into a map. Then enum the subsets to find if the product have more than twice in the map. Then you can solve this question.
Related
I am having two lists with repeating values and I wanted to take the intersection of the repeating values along with the values that have occurred only once in any one of the lists.
I am just a beginner and would love to hear simple suggestions!
Method 1
l1=[1,2,3,4]
l2=[1,2,5]
intersection=[value for value in l1 if value in l2]
for x in l1+l2:
if x not in intersection:
intersection.append(x)
print(intersection)
Method 2
print(list(set(l1+l2)))
I'm trying to do dynamic programming backtracking of maximum sum of non adjacent elements to construct the optimal solution to get the max sum.
Background:
Say if input list is [1,2,3,4,5]
The memoization should be [1,2,4,6,9]
And my maximum sum is 9, right?
My solution:
I find the first occurence of the max sum in memo (as we may not choose the last item) [this is O(N)]
Then I find the previous item chosen by using this formula:
max_sum -= a_list[index]
As in this example, 9 - 5 = 4, which 4 is on index 2, we can say that the previous item chosen is "3" which is also on the index 2 in the input list.
I find the first occurence of 4 which is on index 2 (I find the first occurrence because of the same concept in step 1 as we may have not chosen that item in some cases where there are multiple same amounts together) [Also O(N) but...]
The issue:
The third step of my solution is done in a while loop, let's say the non adjacent constraint is 1, the max amount we have to backtrack when the length of list is 5 is 3 times, approx N//2 times.
But the 3rd step, uses Python's index function to find the first occurence of the previous_sum [which is O(N)] memo.index(that_previous_sum)
So the total time complexity is about O(N//2 * N)
Which is O(N^2) !!!
Am I correct on the time complexity? Or am I wrong? Is there a more efficient way to backtrack the memoization list?
P.S. Sorry for the formatting if I done it wrong, thanks!
Solved:
I looped from behind checking if the item in front is same or not
If it's same, means it's not first occurrence. If not, it's first occurrence.
Tada! No Python's index function to find from the first index! We find it now from the back
So the total time complexity is about O(N//2 * N)
Now O(N//2 + 1), which is O(N).
I'm finding it difficult to understand why/how the worst and average case for searching for a key in an array/list using binary search is O(log(n)).
log(1,000,000) is only 6. log(1,000,000,000) is only 9 - I get that, but I don't understand the explanation. If one did not test it, how do we know that the avg/worst case is actually log(n)?
I hope you guys understand what I'm trying to say. If not, please let me know and I'll try to explain it differently.
Worst case
Every time the binary search code makes a decision, it eliminates half of the remaining elements from consideration. So you're dividing the number of elements by 2 with each decision.
How many times can you divide by 2 before you are down to only a single element? If n is the starting number of elements and x is the number of times you divide by 2, we can write this as:
n / (2 * 2 * 2 * ... * 2) = 1 [the '2' is repeated x times]
or, equivalently,
n / 2^x = 1
or, equivalently,
n = 2^x
So log base 2 of n gives you x, which is the number of decisions being made.
Finally, you might ask, if I used log base 2, why is it also OK to write it as log base 10, as you have done? The base does not matter because the difference is only a constant factor which is "ignored" by Big O notation.
Average case
I see that you also asked about the average case. Consider:
There is only one element in the array that can be found on the first try.
There are only two elements that can be found on the second try. (Because after the first try, we chose either the right half or the left half.)
There are only four elements that can be found on the third try.
You can see the pattern: 1, 2, 4, 8, ... , n/2. To express the same pattern going in the other direction:
Half the elements take the maximum number of decisions to find.
A quarter of the elements take one fewer decision to find.
etc.
Since half of the elements take the maximum amount of time, it doesn't matter how much less time the other elements take. We could assume that all elements take the maximum amount of time, and even if half of them actually take 0 time, our assumption would not be more than double whatever the true average is. We can ignore "double" since it is a constant factor. So the average case is the same as the worst case, as far as Big O notation is concerned.
For binary search, the array should be arranged in ascending or descending order.
In each step, the algorithm compares the search key value with the key value of the middle element of the array.
If the keys match, then a matching element has been found and its index, or position, is returned.
Otherwise, if the search key is less than the middle element's key, then the algorithm repeats its action on the sub-array to the left of the middle element.
Or, if the search key is greater,then the algorithm repeats its action on the sub-array to the right.
If the remaining array to be searched is empty, then the key cannot be found in the array and a special "not found" indication is returned.
So, a binary search is a dichotomic divide and conquer search algorithm. Thereby it takes logarithmic time for performing the search operation as the elements are reduced by half in each of the iteration.
For sorted lists which we can do a binary search, each "decision" made by the binary search compares your key to the middle element, if greater it takes the right half of the list, if less it will take the left half of the list (if it's a match it will return the element at that position) you effectively reduce your list by half for every decision yielding O(logn).
Binary search however, only works for sorted lists. For un-sorted lists you can do a straight search starting with the first element yielding a complexity of O(n).
O(logn) < O(n)
Although it entirely depends on how many searches you'll be doing, your inputs, etc what your best approach would be.
For Binary search the prerequisite is a sorted array as input.
• As the list is sorted:
• Certainly we don't have to check every word in the dictionary to look up a word.
• A basic strategy is to repeatedly halve our search range until we find the value.
• For example, look for 5 in the list of 9 #s below.v = 1 1 3 5 8 10 18 33 42
• We would first start in the middle: 8
• Since 5<8, we know we can look at just the first half: 1 1 3 5
• Looking at the middle # again, narrow down to 3 5
• Then we stop when we're down to one #: 5
How many comparison is needed: 4 =log(base 2)(9-1)=O(log(base2)n)
int binary_search (vector<int> v, int val) {
int from = 0;
int to = v.size()-1;
int mid;
while (from <= to) {
mid = (from+to)/2;
if (val == v[mid])
return mid;
else if (val > v[mid])
from = mid+1;
else
to = mid-1;
}
return -1;
}
A string is given to you and it contains characters consisting of only 3 characters. Say, x y z.
There will be million queries given to you.
Query format: x z i j
Now in this we need to find all possible different substrings which begins with x and ends in z. i and j denotes the lower and upper bound of the range where the substring must lie. It should not cross this.
My Logic:-
Read the string. Have 3 arrays which will store the count of x y z respectively, for i=0 till strlen
Store the indexes of each characters separately in 3 more arrays. xlocation[], ylocation[], zlocation[]
Now, accordingly to the query, (a b i j) find all the indices of b within the range i and j.
Calculate the answer, for each index of b and sum it to get the result.
Is it possible to pre-process this string before the query? So, like that it takes O(1) time to answer the query.
As the others suggested, you can do this with a divide and conquer algorithm.
Optimal substructure:
If we are given a left half of the string and a right half and we know how many substrings there are in the left half and how many there are in the right half then we can add the two numbers together. We will be undercounting by all the strings that begin in the left and end in the right. This is simply the number of x's in the left substring multiplied by the number of z's in the right substring.
Therefore we can use a recursive algorithm.
This would be a problem however if we tried to solve for everything single i and j combination as the bottom level subproblems would be solved many many times.
You should look into implementing this with a dynamic programming algorithm keeping track of substrings in range i,j, x's in range i,j, and z's in range i,j.
Would it be reasonable to systematically try all possible placements in a word search?
Grids commonly have dimensions of 15*15 (15 cells wide, 15 cells tall) and contain about 15 words to be placed, each of which can be placed in 8 possible directions. So in general it seems like you can calculate all possible placements by the following:
width*height*8_directions_to_place_word*number of words
So for such a grid it seems like we only need to try 15*15*8*15 = 27,000 which doesn't seem that bad at all. I am expecting some huge number so either the grid size and number of words is really small or there is something fishy with my math.
Formally speaking, assuming that x is number of rows and y is number of columns you should sum all the probabilities of every possible direction for every possible word.
Inputs are: x, y, l (average length of a word), n (total words)
so you have
horizontally a word can start from 0 to x-l and going right or from l to x going left for each row: 2x(x-l)
same approach is used for vertical words: they can go from 0 to y-l going down or from l to y going up. So it's 2y(y-l)
for diagonal words you shoul consider all possible start positions x*y and subtract l^2 since a rect of the field can't be used. As before you multiply by 4 since you have got 4 possible directions: 4*(x*y - l^2).
Then you multiply the whole result for the number of words included:
total = n*(2*x*(x-l)+2*y*(y-l)+4*(x*y-l^2)