Might be a quite stupid question and I'm not sure if it belongs here or to math.
My problem:
I have several elements of type X which have a boolean attribute Y.
To calculate the percentage of elements where Y is true, I count all X where Y is true and divide it by the number of elements.
But I don't want to iterate all the time above all elements to update that percentage-value.
My idea was:
If I had 33% for 3 elements, and am adding a fourth one where Y is true:
(0.33 * 3 + 1) / 4 = 0.4975
Obviously that does not work well because of the 0.33.
Is there any way for getting an accurate solution without iteration or saving the number of items where Y is true?
Keep a count of the total number of elements and of the "true" ones. Global vars, object member variables, whatever. I assume that sometime back when the program is starting, you have zero elements. Every time an element is added, removed, or its boolean attribute changes, increment or decrement those counts as appropriate. You'll never have to iterate over the list (except maybe for testing) but at the cost of every change to the list having to include fiddling with those variables.
Your idea doesn't work because 0.33 does not equal 1/3. It's an approximation. If you take the exact value, you get the right answer:
(1/3 * 3 + 1) / 4 = (1 + 1) / 4 = 1/2
My question is, if you can store the value of 33% without iterating, why not just store the values of 1 and 3 and calculate them? That is, just keep a running total of the number of true values and number of objects. Increment when you get new ones. Calculate on demand. It's not necessary to iterate every time is way.
Related
I have a 5 lines below with each line representing a,b,c in ax+by+c = 0
1 0 0
1 2 3
3 4 5
30 40 0
30 40 50
I want to find the largest set of non colinear parallel lines from these lines. The result in the above case will be:-
set of 2 lines
3 4 5
30 40 0
The brute force approach would be to go through all the possible combinations which would be O(n*(n+1)/2) and update the largest possible set after each iteration.
Is there any way to find the set size faster?
A solution would be to transform the coordinates into (angle, distance from origin). Finding the largest set will then be O(nlogn).
Assuming that (a,b) are never both 0, find the distance to origin d using d = c/|(a,b)|. Then, find the angle θ using θ = atan2(b,a). You then have a list of coordinates that looks like this:
[[θ0,d0],
[θ1,d1],
...
]
Sort this list using θ as the key.
Remove all elements that you consider colinear given a threshold. Simply parse the list to check if some pairs of consecutive elements have approximately the same value. Do not forget to test the last element with the first element to account for 360° = 0°. If you encounter a colinear pair, remove one of the element.
Using a minimum and maximum index starting at 0, increase the maximum index until the difference between the first angle (at min index) and the last angle (at max index) pass the angle tolerance that you can accept as being parallel (do not forget to consider that 359.999° is close to 0°). If the size of the set (max index - min index) is bigger than your current best set, note it as the current best. Then, increase the minimum index by one and continue increasing the maximum index until the angle difference test fail again. Continue to do so until the minimum index reaches the end of the list and do not forget to make the maximum index wrap to 0 to consider the cases close to 0° and 360°.
To make it easier to find the elements in the user provided list, you can add the original index to the transformed list, e.g., [[θ0,d0,0],[θ1,d1,1],...].
Some implementation details to consider to avoid making it accidentally O(n^2): Removing an element from an contiguous array is O(n), so instead of removing colinear elements every time you encounter one, note the index in a separate list and recreate the array in a second pass. If you instead use a linked list, the min/max index should be replaced by iterators to avoid the O(n) random access using an index to access an element.
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 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]
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;
}
I participated in code jam, I successfully solved small input of The Repeater Challenge but can't seem to figure out approach for multiple strings.
Can any one give the algorithm used for multiple strings. For 2 strings ( small input ) I am comparing strings character by character and doing operations to make them equal. However this approach would time out for large input.
Can some one explain their algorithm they used. I can see solutions of other users but can't figure out what have they done.
I can tell you my solution which worked fine for both small and large inputs.
First, we have to see if there is a solution, you do that by bringing all strings to their "simplest" form. If any of them does not match, there there is no solution.
e.g.
aaabbbc => abc
abbbbbcc => abc
abbcca => abca
If only the first two were given, then a solution would be possible. As soon as the third is thrown into the mix, then it's impossible. The algorithm to do the "simplification" is to parse the string and eliminate any double character you see. As soon as a string does not equal the simplified form of the batch, bail out.
As for actual solution to the problem, i simply converted the strings to a [letter, repeat] format. So for example
qwerty => 1q,1w,1e,1r,1t,1y
qqqwweeertttyy => 3q,2w,3e,1r,3t,2y
(mind you the outputs are internal structures, not actual strings)
Imagine now you have 100 strings, you have already passed the test that there is a solution and you have all strings into the [letter, repeat] representation. Now go through every letter and find the least 'difference' of repetitions you have to do, to reach the same number. So for example
1a, 1a, 1a => 0 diff
1a, 2a, 2a => 1 diff
1a, 3a, 10a => 9 diff (to bring everything to 3)
the way to do this (i'm pretty sure there is a more efficient way) is to go from the min number to the max number and calculate the sum of all diffs. You are not guaranteed that the number will be one of the numbers in the set. For the last example, you would calculate the diff to bring everything to 1 (0,2,9 =11) then for 2 (1,1,8 =10), the for 3 (2,0,7 =9) and so on up to 10 and choose the min again. Strings are limited to 1000 characters so this is an easy calculation. On my moderate laptop, the results were instant.
Repeat the same for every letter of the strings and sum everything up and that is your solution.
This answer gives an example to explain why finding the median number of repeats produces the lowest cost.
Suppose we have values:
1 20 30 40 100
And we are trying to find the value which has shortest total distance to all these values.
We might guess the best answer is 50, with cost |50-1|+|50-20|+|50-30|+|50-40|+|50-100| = 159.
Split this into two sums, left and right, where left is the cost of all numbers to the left of our target, and right is the cost of all numbers to the right.
left = |50-1|+|50-20|+|50-30|+|50-40| = 50-1+50-20+50-30+50-40 = 109
right = |50-100| = 100-50 = 50
cost = left + right = 159
Now consider changing the value by x. Providing x is small enough such that the same numbers are on the left, then the values will change to:
left(x) = |50+x-1|+|50+x-20|+|50+x-30|+|50+x-40| = 109 + 4x
right(x) = |50+x-100| = 50 - x
cost(x) = left(x)+right(x) = 159+3x
So if we set x=-1 we will decrease our cost by 3, therefore the best answer is not 50.
The amount our cost will change if we move is given by difference between the number to our left (4) and the number to our right (1).
Therefore, as long as these are different we can always decrease our cost by moving towards the median.
Therefore the median gives the lowest cost.
If there are an even number of points, such as 1,100 then all numbers between the two middle points will give identical costs, so any of these values can be chosen.
Since Thanasis already explained the solution, I'm providing here my source code in Ruby. It's really short (only 400B) and following his algorithm exactly.
def solve(strs)
form = strs.first.squeeze
strs.map { |str|
return 'Fegla Won' if form != str.squeeze
str.chars.chunk { |c| c }.map { |arr|
arr.last.size
}
}.transpose.map { |row|
Range.new(*row.minmax).map { |n|
row.map { |r|
(r - n).abs
}.reduce :+
}.min
}.reduce :+
end
gets.to_i.times { |i|
result = solve gets.to_i.times.map { gets.chomp }
puts "Case ##{i+1}: #{result}"
}
It uses a method squeeze on strings, which removes all the duplicate characters. This way, you just compare every squeezed line to the reference (variable form). If there's an inconsistency, you just return that Fegla Won.
Next you use a chunk method on char array, which collects all consecutive characters. This way you can count them easily.