I want number of ways to divide an array of possitive integers such that maximum value of left part of array is greater than or equal to the maximum value of right part of the array.
For example,
6 4 1 2 1 can be divide into:
[[6,4,1,2,1]] [[6][4,1,2,1]] [[6,4][1,2,1]] [[6,4,1][2,1]] [[6,4,1,2][1]] [[6][4,1][2,1]] [[6][4][1,2,1]] [[6][4][1,2][1]] [[6][4,1,2][1]] [[6,4,1][2][1]] [[6][4,1][2][1]] [[6,4][1,2][1]]
which are total 12 ways of partitioning.
I tried a recursive approach but it fails beacause of termination due to exceed of time limit. Also this approach is not giving correct output always.
In this another approach, I took the array ,sort it in decreasing order and then for each element I checked weather it lies on right of the original array, and if does then added it's partitions to it's previous numbers too.
I want an approach to solve this, any implementation or pseudocode or just an idea to do this would be appreciable.
I designed a simple recursive algorithm. I will try to explain on your example;
First, check if [6] is a possible/valid part of a partition.
It is a valid partition because maximum element of ([6]) is bigger than remaining part's ([4,1,2,1]) maximum value.
Since it is a valid partition, we can use recursive part of the algorithm.
concatenate([6],algorithm([4,1,2,1]))
now the partitions
[[6][4,1,2,1]], [[6][4,1][2,1]], [[6][4,1][2,1]] [[6][4][1,2,1]] [[6][4][1,2][1]] [[6][4,1,2][1]]
are in our current solution set.
Check if [6,4] is a possible/valid part of a partition.
Continue like this until reaching [6,4,1,2,1].
Related
given an array of size n, n<=10^5 what is efficient approach to count number of sub arrays whose product is even ?
i am using naive approach with (On^3) time complexity ?
please suggest some efficient approach?
Be careful: from your explanation I have the impression that you are taking all sub-arrays, calculate the product and check if it is even.
However there's one very important mathematical rule: when you have a series of natural numbers, as soon as there's one even number, the product will be even.
So, I'd advise you to program following algorithm:
Search in your array for an even number.
Count the amount of sub-arrays, containing that even number.
Search in your array for the next even number.
Count the amount of sub-arrays, containing that next even number, but not containing the previous even number.
Continue until you've processed all even numbers in your array.
I'm working on a problem where I have an entire table from a database in memory at all times, with a low range and high range of 9-digit numbers. I'm given a 9-digit number that I need to use to lookup the rest of the columns in the table based on whether that number falls in the range. For example, if the range was 100,000,000 to 125,000,000 and I was given a number 117,123,456, then I would know that I'm in the 100-125 mil range, and whatever vector of data that points to is what I will be using.
Now the best I can think of for lookup time is log(n) run time. This is OK, at best, but still pretty slow. The table has at least 100,000 entries and I will need to look up values in this table tens-of-thousands, if not hundred-thousands of times, per execution of this application (10+ times/day).
So I was wondering if it was possible to use an unordered_set instead, writing my own Hash function that ALWAYS returns the same hash-value for every number in range. Using the same example above, 100,000,000 through 125,000,000 will always return, for example, a hash value of AB12CD. Then when I use the lookup value of 117,123,456, I will get that same AB12CD hash and have a lookup time of O(1).
Is this possible, and if so, any ideas how?
Thanks in advance.
Yes. Assuming that you can number your intervals in order, you could fit a polynomial to your cutoff values, and receive an index value from the polynomial. For instance, with cutoffs of 100,000,000, 125,000,000, 250,000,000, and 327,000,000, you could use points (100, 0), (125, 1), (250, 2), and (327, 3), restricting the first derivative to [0, 1]. Assuming that you have decently-behaved intervals, you'll be able to fit this with an (N+2)th-degree polynomial for N cutoffs.
Have a table of desired hash values; use floor[polynomial(i)] for the index into the table.
Can you write such a hash function? Yes. Will evaluating it be slower than a search? Well there's the catch...
I would personally solve this problem as follows. I'd have a sorted vector of all values. And then I'd have a jump table of indexes into that vector based on the value of n >> 8.
So now your logic is that you look in the jump table to figure out where you are jumping to and how many values you should consider. (Just look at where you land versus the next index to see the size of the range.) If the whole range goes to the same vector, you're done. If there are only a few entries, do a linear search to find where you belong. If they are a lot of entries, do a binary search. Experiment with your data to find when binary search beats a linear search.
A vague memory suggests that the tradeoff is around 100 or so because predicting a branch wrong is expensive. But that is a vague memory from many years ago, so run the experiment for yourself.
for example if
a[3]={1,2,3}
1,2,3,1+2,2+3,1+2+3
so my program should print
1
2
3
3
5
8
I know there exits a formula for calculating the total sum.
But what i want is efficient method to calculate individual sums.
A seg tree advisable?
Assuming that by subarray you mean what some authors conventionally call subarray being a contiguous slice of an array, you are looking for Kadane's algorithm.
It works by incrementally finding the biggest subarray. At any given point of the search, the maximum subarray on that index is either the empty array (sum == zero) or consists of one more element than the maximum subarray that ended at the previous position. You keep track of what is the best you've ever found so you can compare subarrays with the best so far and return the actual best solution.
It may also be extended to multiple dimensions.
I've been working on the exercise and stumbled upon a problem.
Given an array of integers, determine whether or not it can be partitioned into two arrays, each of which is in increasing order. For instance 3,1,5,2,4 can, but 4,8,1,5,3 can't.
The problem is here. I couldn't understand why 1st array can but the 2nd one can't.
There is a hint given:
If we successfully partitioned an initial segment of the array, one of the parts must contain the maximum element seen so far. It is obviously in our best interest that the largest element of the other part be as small as possible. So, given the next element, if it's the maximum to this point add it to the "maximum containing part". If not there is no choice but to add it to the other part if possible (e.g: if it is larger than the largest element of this part, but it is not the current maximum). If this procedure fails then no partition is possible, and if it succeeds then we have demonstrated a partition.
The most important part is to understand the logic of this partitioning.
Thank you in advance.
Let's use the given algorithm on {3,1,5,2,4}.
First number is 3. Our partition is {3},{}.
Next comes 1. We can't add that to {3}, so we add it to the other: {3},{1}.
Next comes 5. We will add it to {3}, so as to save the {1} for smaller numbers: {3,5},{1}.
Next comes 2. we must add it to {1}: {3,5},{1,2}. (Now we see why it was good not to add 5 to {1}.)
Next comes 4: again, we have no choice: {3,5},{1,2,4}.
Given a integer array like
int numbers[8]={1, 3, 5, 7, 8, 6, 4, 2};
The half side in the front array are odd numbers, and the rest (the equal amount number)
are even. The odd numbers are in an ascending order and even part are in a descending order. After the sorting, the order of the numbers can't be changed.
How can I sort them alternatively with time complexity less than O(n^2) and space complexity O(1)?
For this example, the result would be: {1,8,3,6,5,4,7,2};
I can't use external array storage but temporary variables are acceptable.
I have tried to use two pointers(oddPtr, evenPtr) to point odd and even numbers separately, and move evenPtrto insert the even values to the middles of odd numbers.(Like insertion sort)
But it takes O(n^2).
UPDATED
As per Dukeling's comment I realized that the solution I propose in fact is not linear,but linearithmic and even worse - you can't control if it takes extra memory or not. On my second thought I realized you know a lot about the array you to implement a more specific, but probably easier solution.
I will make an assumption that all values in the array are positive. I need this so that I can use negative values as kind of 'already processed' flag. My idea is the following - iterate over the array from left to right. For each element if it is already processed(i.e. its value is negative) simply continue with the next one. Otherwise you will have a constant formula where is the position where this element should be:
If the value is odd and its index is i it should move to i*2
If the value is even and its index is i it should move to (i - n/2)*2 + 1
Store this value into a temporary and make the value at the current index of the array 0. Now until the position where the value we 'have at hand' is not zero, swap it with the value staying at the position we should place it according to the formula above. Also when you place the value at hand negate it to 'mark it as processed'. Now we have a new value 'at hand' and again we calculate where it should go according to the formula above. We continue moving values until the value we 'have at hand' should go to the position with 0. With a little thought you can prove that you will never have a negative('processed') value at hand and that eventually you will end up at the empty spot of the array.
After you process all the values iterate once over the array to negate all values and you will have the array you need. The complexity of the algorithm I describe is linear- each value will be no more than once 'at hand' and you will iterate over it no more than once.