I need way to sort according to the name .
According to the number of letters of the alphabet, the word starts from A to Z,
it's mean you want to count how many a in the two word and the word who have the largest number of letter a, you want to put this word first (swap)
And if their number of a is equal, you will compare the letter after it means b, and if the number of the word is equal, you will compare C, and this is what ... and he will tell you that this is the case Suppose that there are no students who are inspired by the same number of all letters in the same class
My Code contains a class contain a name type of string and main drive contain a array of objects
As I'm a C++ and Python Developer. I can't help you with the Java Code, but according to your query, I think Count Sort is the most suitable for this kind of problem because while sorting numbers it sorts all of them using their Digits.
Example
Input data: 1, 4, 1, 2, 7, 5, 2
Take a count array to store the count of each unique object.
Index: 0 1 2 3 4 5 6 7 8 9
Count: 0 2 2 0 1 1 0 1 0 0
Modify the count array such that each element at each index
stores the sum of previous counts.
Index: 0 1 2 3 4 5 6 7 8 9
Count: 0 2 4 4 5 6 6 7 7 7
The modified count array indicates the position of each object in
the output sequence.
Rotate the array clockwise for one time.
Index: 0 1 2 3 4 5 6 7 8 9
Count: 0 0 2 4 4 5 6 6 7 7
Output each object from the input sequence followed by
increasing its count by 1.
Process the input data: 1, 4, 1, 2, 7, 5, 2. Position of 1 is 0.
Put data 1 at index 0 in output. Increase count by 1 to place next data 1 at an index 1 greater than this index.
Above Example is Taken from https://www.geeksforgeeks.org/counting-sort/
Related
I'm solving a reverse 0/1 knapsack problem, i.e. I'm trying to recreate the list of weights and values of all the items using only the DP-table.
I have this table:
[0][1] [4][5][6] [12]
[0] 0 0 0 0 0 0 0 0 0 0 0 0 0
[1] 0 4 4 4 4 4 4 4 4 4 4 4 4
[2] 0 4 4 4 6 10 10 10 10 10 10 10 10
I don't understand how row [2] is possible.
[0] - it is clear that if we do not put anything in the knapsack, the answer total value 0.
[1] - in row [1] I see that [1][1]=4 and I hope that I correctly conclude that the first item has weight = 1 and value = 4. So, since we put only 1 item it is the only weight we can hope for in this row.
[2] - when we reach [2][4], we have 6, 6 > [2-1][4] and I assume that we use 2 items here, one weight = 1 and value = 4 (the old one) and weight = 4-1 and value = 6-4 = weight = 3 and value = 2, which is the new one.
Question: How is it possible to have [2][5] = 10? We can't put more than 1 item on a row, as I understand this chart. If we have two items in use here, shouldn't we have 6 for all the elements in row [2] starting from [2][4] to the end of the row?
This seems possible if you have two items, one with weight 1 and value 4 and one with weight 4, value 6.
How? When you're at index (2, 4) you have a weight capacity of 4 for the first time in the row that considers item 2 (weight 4, value 6). This lets you take the item with value 6 instead of the weight 1, value 4 item you previously took at index (2, 3), effectively building from the subproblem at index (2, 0).
Now, when you're at index (2, 5) with a weight capacity of 5, the total value of 10 is possible because you can take both items. That's the best you can do for the rest of the row.
See also How to find which elements are in the bag, using Knapsack Algorithm [and not only the bag's value]?
I would like to create code for a random number generator for predetermined sets of triplets (200 sets in total to randomize). I would like the sets of triplets to form a set of six numbers and the set of triplets to remain unique.
example triplets A = [1 2 3; 4 5 6; 7 8 9, 10 11 12, 13 14 15]; etc
I would like resulting triplet to retain their original sequence
1 2 3 + 4 5 6, 1 2 3 + 7 8 9, 1 2 3 + 10 11 12, 1 2 3 + 13 14 15
I am not a coder, so any help would be appreciated
You want to pick three triplets, keeping them in order. So your first triplet cannot be too close to the end -- there have to be at least two more triplets after it. Similarly, the second triplet you pick needs at least one unpicked triplet after it.
I assume that you have your triplets in an array or similar, numbered 0 to 199.
Pick a random number A in the range 0 to 197. That is the index of your first triplet.
Pick a second random number B in the range (A + 1) to 198. That is the index of your second triplet.
Pick a third random number C in the range (B + 1) to 199. That is the index of your third triplet.
The range of random numbers you pick from is affected by the numbers you have previously picked and the number of picks remaining.
I have the following data:
a b c d
5 9 6 0
3 1 3 2
Characters in the first row, numbers in the second row.
How do I get the character corresponding to the highest number in the second row, and how do I increase the corresponding number in the second row? (For example, here, column b has the highest number, 9, so increase that number by 10%.)
I use Dyalog version 17.1.
With:
⎕←data←3 4⍴'a' 'b' 'c' 'd' 5 9 6 0 3 1 3 2
a b c d
5 9 6 0
3 1 3 2
You can extract the second row with:
2⌷data
5 9 6 0
Now grade it descending, that is, find the indices that would sort it from highest to lowest:
⍒2⌷data
2 3 1 4
The first number is the column we're looking for:
⊃⍒2⌷data
2
Now we can use this to extract the character from the first row:
data[⊂1,⊃⍒2⌷data]
b
But we only need the column index, not the actual character. The full index of the number we want to increase is:
2,⊃⍒2⌷data
2 2
Extracting the data to see that we got the right index:
data[⊂2,⊃⍒2⌷data]
9
Now we can either create a new array with the target value increased by 10%:
1.1×#(⊂2,⊃⍒2⌷data)⊢data
a b c d
5 9.9 6 0
3 1 3 2
Or change it in-place:
data[⊂2,⊃⍒2⌷data]×←1.1
data
a b c d
5 9.9 6 0
3 1 3 2
Try it online!
In APL, matrices and vectors are used to hold data. I was wondering if there was a way to search within a matrix for a given value, and have that values index returned. For example, say I have the following 2-dimensional matrices:
VALUES ← 1 2 3 4 5 6 7 8 9 10 11... all the way up to 36
KINDS ← 0 0 0 2 0 0 0 3 0 ... filled with 0's the rest of the way to 36 length.
If I laminated these two matrices with
kinds,[.5] values
so that they are laminated one on top of the other
1 2 3 4 5 6 7 8 9 10...
0 0 0 2 0 0 0 3 0 ....
is there a functionally easy way to search for the index of the 2 value in the "second row" of the newly laminated matrix? eg. the column containing
4
2
and return that matrix index?
The value 2 also appears in row 1 of your newly laminated matrix (nlm), and as you stated, you really do not want to search the whole matrix, but only the second row. So, since you're only searching within a given row, getting the column index in that row gives you the complete answer:
row←2
⎕←col←nlm[row;]⍳2
4
nlm[;col] ⍝ values in matched column
4 2
Try it online!
I just got a question about counting the split points in a integer array, to ensure there is at least one duplicated integer on the two sides.
ex:
1 1 4 2 4 2 4 1
we can either split it into:
1 1 4 2 | 4 2 4 1
or
1 1 4 2 4 | 2 4 1
so that there is at least one '1', '2' ,and '4' are in both sides.
The integer can range from 1 to 100,000
The complexity requires O(n). How to solve this question?
Make one pass over the array and build count[i] = how many times the value i appears in the array. The problem is only solvable if count[i] >= 2 for all non-zero values. You can use this array to tell how many distinct values you have in your array.
Next, make another pass and using another array count2[i] (or you can reuse the first one), keep track of when you have visited each value at least once. Then use that position as your split point.
Example:
1 1 4 2 4 2 4 1
count = [3, 2, 0, 4] => 3 distinct values
1 1 4 2 4 2 4 1
^ => 1 distinct value so far
^ => 1 distinct value so far
^ => 2 distinct values so far
^ => 3 distinct values so far => this is your split point
There might be cases for which there is no solution, for example if the last 1 was at the beginning as well. To detect this, you can just make another pass over the rest of the array after you have decided on the split point and see if you still have all the values on that side.
You can avoid this last pass by using the count and count2 arrays to detect when you can no longer have a split point. This is left as an exercise.