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How many unique arrays of m elements exist such that they contain numbers in the range [1,n] and there exists atleast one subsequence {1,2,3,4....n}?
Constraints: m > n
I thought of combinations approach. But there will be repetitions.
In my approach, I first lay out all the numbers from 1 to n.
For example, if m=n+1, answer is n^2. (n spots available, each number in range [1,n])
Now, I think there might be a DP relation for further calculation, but I am not being able to figure it out.
Here's an example for n=3 and m=5. The green squares are the subsequence. The subsequence consists of the first 1 in the array, the first 2 that's after the first 1, etc. Squares that aren't part of the subsequence can either take n values if they are after the end of the subsequence, or n-1 values otherwise.
So the answer to this example is 1*9 + 3*6 + 6*4 = 51, which is easily verified by brute force. The coefficients 1,3,6 appear to be related to Pascal's triangle. The rest is left to the reader.
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Given array A[] and B[], to find how frequently each possible sums A[i] + B[j] appears we can use FFT
For example
A = [1,2,3], B = [2,4].
The sum 3 can be obtained in 1 way, sum 4 : 1 way, sum 5 : 2 ways, sum 6 : 1 way, sum 7 : 1 way
The way we can do this is to construct two polynomials, P and Q with their power corresponding to the element of the array. And apply the regular FFT.
So is there an efficient way of backtracking the numbers that forms the above. To elaborate, we know 3 can be formed in 1 way. But how do we know which two numbers form it?
One way to do it would be the classic two sum algorithm, i.e given an array and a target sum. Find the pairs that create the sum. Which is O(n)
Given we can have N different targets, the resulting algorithm is O(n^2). But I want to keep it under O(nlogn).
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Given a list of 2n-1 numbers: all between 1 to n, all but one occur twice. Determine the number that occurs only once. Multiple ways preferred.
I think the problem is at fault, how can you determine which number without knowing the list of numbers?
[O(1) space, O(n) time]: Just take the XOR of all the numbers. Since all the numbers occur two times except one, XOR of those numbers will be zero and the single occurring number will be the result.
[O(1) space, O(n) time]: As said by user3386109 in comments, we can sum all the given numbers and compare that to the sum of numbers in the range [1, n] which will be n*(n+1) (since all numbers are supposed to occur twice). The difference of the two numbers is the answer.
[O(n) space, O(n) time]: Create an array of size n and keep the count of all the elements in the array at their corresponding positions. At the end, traverse the array, and find the number whose count is only 1.
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I have an array of elements A1,A2,...,An.
The probability of a user searching for each element are P1,P2,...,Pn.
If the elements are rearranged, will the average case of the algorithm change?
Edit : I have posted the question, which appeared in my exam.
The expected number of comparisons is sum_{i=1...n}(i * p_i).
Re-ordering the elements in descending order reduces the expectation. That's intuitive since by looking at the most probable choices first will reduce, on average, the number of elements looked at before you find a particular choice.
As an example, suppose there's three items k1, k2, k3 with match probabilities 10%, 30% and 60%.
Then in the order k1, k2, k3, the expected number of comparisons is 1*0.1 + 2*0.3 + 3*0.6 = 2.5
With the most likely first: k3, k2, k1, the expected number of comparisons is 1*0.6 + 2*0.3 + 3*0.1 = 1.5
No, because it takes O(1) time to access an element in array and it does not depend on a position of this element in array. So arr[0] and arr[10000] should take the same amount of time.
If you will have something like a linked list or a binary tree, then it makes sense to put elements that are accessed with higher probability closer to the beginning.
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Wondering if anyone have worked on similar problem before? And my question is, why {8, 6} are peaks? I think 8 is peak, but since 6 is smaller than 8, it should not be a peak? Thanks.
In an array of integers, a "peak" is an element which is greater than or equal to the adjacent integers and a "valley" us an element which is less than or equal to the adjacent integers. For example, in the array {5,8,6,2,3,4,6} {8,6} are peaks and {5,2} are valleys. Given an array of integers, sort the array into an alternating sequence of peaks and valleys.
Example,
Input: {5,3,1,2,3}
Output: {5,1,3,2,3}
thanks in advance,
Lin
The 6 referred to is the 2nd 6 at the end of the sequence. This fits the description well (if not made very clear) and is backed up by 5 being a valley.
An alternating sequence of peaks and valleys is a sequence such that either all of the elements in odd-numbered positions are peaks and those in even-numbered positions are valleys, or vice-versa. The output sequence in the example demonstrates the elements of the input sequence sorted in such a way.
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Imagine you are given a matrix of positive integer numbers (maximum 25*15, value of number does not exceed 3000000). When you do column sums and pick the smallest and the largest one, the difference between them must be the smallest possible.
You can swap numbers in every row (permute rows), not in column, how many times you want.
How would you solve this task?
I'm not asking for your code but your ideas.
Thanks in advance
I would make an attempt to solve the problem using Simulated Annealing. Here is a sketch of the plan:
Let the distance to optimize the difference between the max and min column sums.
Set the goal to be 0 (i.e., try to reach as close as possible to a matrix with no difference between sums)
Initialize the problem by calculating the array of sums of all columns to their current value.
Let a neighbor of the current matrix be the matrix that results from swapping two entries in the same row of the matrix.
Represent neighbors by their row index and two swapping column indexes.
When accepting a neighbor, do not compute all sums again. Just adjust the array of sums in the columns that have been swapped and by the difference of the swap (which you can deduce from the swapped row index)
Step 6 is essential for the sake of performance (large matrices).
The bad news is that this problem without the limits is NP-hard, and exact dynamic programming at scale seems out of the question. I think that my first approach would be large-neighborhood local search: repeatedly choose a random submatrix (rows and columns) small enough to be amenable to brute force and choose the optimal permutations while leaving the rest of the matrix undisturbed.