I am working on my thesis and I was wondering the following:
I have a set N with index i.
Then, based on the elements of set N, I want to create a set that contains certain pairs of (i, i'), where i and i' are both from set N.
How can I do this?
Thank you very much in advance.
I would say, just add another index i2 to set N. This way you'll have a set N with index i and i2. Now you can define for example MyPairs(i,i2) where i and i2 are elements of N.
Related
given a set of points D and some number K I want to find all numbers that are in D such that the distance between K and any found number is less or equal to integer N?
Example:
suppose we have D={5,9,0,6,7} and K=8 and N=1 then the result should be {9,7}
I was thinking to use k-d tree or VP tree but both as I understand (correct me if I am wrong please) find nearest neighbors and do not care about N in my example.
To summarize all the comment:
Solve this problem as brute force will take O(n) time as iterate on each element in D and check if its distance from k is less then n.
You have big data set but a lot of queries it is better to do pre-processing on D (with O(nlogn) and the you can get the answer in O(logn) -> by sorting D as pre-processes (in O(nlogn) as dimple sort of array.
Now, on given query search for k - notice binary search will stop if the number missing but he do stop at the closest value. From that index start spread to both side of D and for each check if still in n range. Notice the spreading in allow as it is include of O(|output|).
In your example: sorted D yield: [0,5,6,7,9]. Try finding k=8 will give false but index 3 or 4 (depended on the implementation). Let say is return index 3. for 3 till last index check if arr[i] - k < n if so print - if bigger stop. For the other side check k - arr[i] < n - if so print and if bigger stop -> this will give you 7,9
Hope that helps!
Compare two numbers in the form of string lexicographically multiple times.
What I tried
the question is straight forward, to compare string after change but I am getting Time limit exceeded error because of multiple queries.
I was searching internet and came across segment tree to solve range queries. But alas I am not able to visualise, how it can help here.
Any hint is appreciated.
Segment tree seems overkill for this problem. When will B be lexicographically larger than A? When at index i, A[i] = 0, B [i] = 1 and A[0:i] = B [0:i]. Iterate over both strings at same time and keep in a set all indexes where they are different.
For each query at index i, update B to 1. Then check if B[i] = A[i]. IF they are equal, erase i from the indexes set. Otherwise, add it to the set. If there is no index left in the set, A and B are now equal => answer YES.
If there is at least 1 element, get the lowest index. If this index is j, that means A[0:j] = B[0:j] but A[j] != B[j]. So either A is 0 and B is 1 or A is 1 and B is 0. Depending on that, answer YES or NO.
This has complexity of O(Q log N), being Q the amount of queries
I've been struggling with something in excel which is quite easy to do individual cases of using an array, but I want to do in a single cell.
Effectively, in row C I have the multipliers I need, lets call them i_k, for j from 1 to n. The equation I want to calculate in mathematical notation is;
Sigma(from j = 0 to n) (Pi(from k = j to n) (i_k))
But I'm not quite sure how best to go about this. Effectively it should be;
(i_1)^n + (i_2)^(n-1) + (i_3)^(n-2) + ...
In the end. Any help?
I dont know if this is an elegant solution but I think this might solve it for you.
The key is to make a table so that your formulas continue.
The screenshot I have taken explains the formula I have used with its explanation I have used on top of it.
-- The first column will have the intended values of I
-- The second column's first row will have the value of N
-- The third column will have a formula which says:
=IF(ISNUMBER([#[Values of N]]),[#[Values of N]],C2-1)
This will give you the decreasing value of N
Now just multiply N with I on the 4th column using a simple multiplication and add the final result:
I came across a solution to the Coin Change problem here : Coin Change. Here I was able to understand the first recursive method, the second method which uses DP with a 2D array. But am not able to understand the logic behind the third solution.
As far as I have thought, the last method works for problems in which the sequence of coins used in coin change is considered. Am I correct? Can anyone please explain me if I am wrong.
Well I figured it out myself!
This can be easily proved using induction. Let table[k] denote the ways change can be given for a total of k. Now the algorithm consists of two loops, one which is controlled by i and iterates through the array containing all the different coins and the other is the j controlled loop which for a given i, updates all the values of elements in array table. Now consider for a fixed i we have calculated the number of ways change can be given for all values from 1 to n and these values are stored in table from table[1] to table[n]. When the i controlled loop iterates for i+1, the value in table[j] for an arbitrary j is incremented by table[j-S[i + 1]] which is nothing but the ways we can create j using at least one coin with value S[i + 1] (the array which stores coin values). Thus the total value in table[j] equals the number of ways we can create a change with coins of value S[1]....S[i] (this was already stored before) and the value table[j-S[i + 1]]. This is same as the optimal substructure of the problem used in the recursive algorithm.
int arr[size];
memset(arr,0,sizeof(size));
int n;
cin>>n;
int sum;
cin>>sum;
int a[size];
fi(i,n)
cin>>a[i];
arr[0]=1;
fi(i,n)
for(int j=arr[i]; j<=n; j++)
a[j]+=a[j-arr[i]];
cout<<arr[n];
The array arr is initialised as 0 so as to show that the number of ways a sum of ican be represented is zero(that is not initialised). However, the number of ways in which a sum of 0 can be represented is 1 (zero way).
Further, we take each coin and start initialising each position in the array starting from the coin denomination.
a[j]+=a[j-arr[i]] means that we are basically incrementing the possible ways to represent the sum jby the previous number of ways, required (j-arr[i]).
In the end, we output the a[n]
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.