I have items with ID 1, 3, 4, 5, 6, 7. Now I have data like following.
There is an offerId for each row. Array of Ids consist of combination of the ID in an array. Discount is the value for that offerId
offerId : Array of Ids : Discount
o1 : [1] : 45
o2 : [1 3 4] : 100
o3 : [3 5] : 55
o4 : [5] : 40
o5 : [6] : 30
o6 : [6 7] : 20
Now I have to select all the offerIds which give me best combination of Ids i.e. maximum total discount.
For example in above case : possible results can be:
[o2, o4, o5] maximum discount is 170(100 + 40 + 30).
Note. the result offerId should be such that Ids don't repeat. Example for o2,o4,o6 ids are [1,3,4], [5], [6] all are distinct.
Other combination can be :
o1, o3, 06 for which ids are [1], [3,5], [6,7] However the total is 120(45+55+20) which is less then 170 as in previous case.
I need an algorithm/code which will help me to identify combination of offerIds which will give maximum discount , considering that each offer should contain distinct Ids.
NOTE I am writing my code in go language. But solutions/Logic in any language will be helpful.
NOTE : I hope I am able to explain my requirement properly. please comment if any extra information is required. Thanks.
Here is a dynamic programming solution which, for every possible subset of IDs, finds the combination of offers for which the discount is maximum possible.
This will be pseudocode.
Let our offers be structures with fields offerNumber, setOfItems and discount.
For the purposes of implementation, we first renumerate the possible items by integers from zero to number of different possible items (say k) minus one.
After that, we can represent setOfItems by a binary number of length k.
For example, if k = 6 and setOfItems = 1011102, this set includes items 5, 3, 2 and 1 and excludes items 4 and 0, since bits 5, 3, 2 and 1 are ones and bits 4 and 0 are zeroes.
Now let f[s] be the best discount we can get using exactly set s of items.
Here, s can be any integer between 0 and 2k - 1, representing one of the 2k possible subsets.
Furthermore, let p[s] be the list of offers which together allow us to get discount f[s] for the set of items s.
The algorithm goes as follows.
initialize f[0] to zero, p[0] to empty list
initialize f[>0] to minus infinity
initialize bestF to 0, bestP to empty list
for each s from 0 to 2^k - 1:
for each o in offers:
if s & o.setOfItems == o.setOfItems: // o.setOfItems is a subset of s
if f[s] < f[s - o.setOfItems] + o.discount: // minus is set subtraction
f[s] = f[s - o.setOfItems] + o.discount
p[s] = p[s - o.setOfItems] append o.offerNumber
if bestF < f[s]:
bestF = f[s]
bestP = p[s]
After that, bestF is the best possible discount, and bestP is the list of offers which get us that discount.
The complexity is O (|offers| * 2k) where k is the total number of items.
Here is another implementation which is asymptotically the same, but might be faster in practice when most subsets are unreachable.
It is "forward" instead of "backward" dynamic programming.
initialize f[0] to zero, p[0] to empty list
initialize f[>0] to -1
initialize bestF to 0, bestP to empty list
for each s from 0 to 2^k - 1:
if f[s] >= 0: // only for reachable s
if bestF < f[s]:
bestF = f[s]
bestP = p[s]
for each o in offers:
if s & o.setOfItems == 0: // s and o.setOfItems don't intersect
if f[s + o.setOfItems] < f[s] + o.discount: // plus is set addition
f[s + o.setOfItems] = f[s] + o.discount
p[s + o.setOfItems] = p[s] append o.offerNumber
Related
Problem : If you were given a number n, you will have an array with (n-1) index. with the 1st index containing 1, 2nd index containing 2, and n-1 index containing n-1. Given those sets of numbers, How can one check when + or -, the array can be equal to n?
Example :
n = 3, Array = {1,2}
+1 +2 = 3 (True)
n = 4, Array = {1,2,3}
-1 + 2 + 3 = 4 (True)
n = 5, Array = {1,2,3,4}
No possible combination
I tried too long to think about it and still haven't come up with the right answer :(
If you ale looking for simple solvable/not solvable answer, then it seems the answer if very simple
(sum - n) % 2 != 0 // => non-solvable
Here is result of an experiment:
When n gets larger it becomes easier to subtract necessary sum and there are plenty of possible solutions.
There's array of size n. The values can be between 0 and (n-1) as the indices.
For example: array[4] = {0, 2, 1, 3}
I should say if there's any number that is repeated more than 1 time.
For example: array[5] = {3,4,1,2,4} -> return true because 4 is repeated.
This question has so many different solutions and I would like to know if this specific solution is alright (if yes, please prove, else refute).
My solution (let's look at the next example):
array: indices 0 1 2 3 4
values 3 4 1 2 0
So I suggest:
count the sum of the indices (4x5 / 2 = 10) and check that the values' sum (3+4+1+2+0) is equal to this sum. if not, there's repeated number.
in addition to the first condition, get the multiplication of the indices(except 0. so: 1x2x3x4) and check if it's equal to the values' multiplication (except 0, so: 3x4x1x2x0).
=> if in each condition, it's equal then I say that there is NO repeated number. otherwise, there IS a repeated number.
Is it correct? if yes, please prove it or show me a link. else, please refute it.
Why your algorithm is wrong?
Your solution is wrong, here is a counter example (there may be simpler ones, but I found this one quite quickly):
int arr[13] = {1, 1, 2, 3, 4, 10, 6, 7, 8, 9, 10, 11, 6};
The sum is 78, and the product is 479001600, if you take the normal array of size 13:
int arr[13] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12};
It also has a sum of 78 and a product of 479001600 so your algorithm does not work.
How to find counter examples?1
To find a counter example2 3:
Take an array from 0 to N - 1;
Pick two even numbers3 M1 > 2 and M2 > 2 between 0 and N - 1 and halve them;
Replace P1 = M1/2 - 1 by 2 * P1 and P2 = M2/2 + 1 by 2 * P2.
In the original array you have:
Product = M1 * P1 * M2 * P2
Sum = 0 + M1 + P1 + M2 + P2
= M1 + M1/2 - 1 + M2 + M2/2 + 1
= 3/2 * (M1 + M2)
In the new array you have:
Product = M1/2 * 2 * P1 + M2/2 * 2 * P2
= M1 * P1 * M2 * P2
Sum = M1/2 + 2P1 + M2/2 + 2P2
= M1/2 + 2(M1/2 - 1) + M2/2 + 2(M2/2 + 1)
= 3/2 * M1 - 2 + 3/2 * M2 + 2
= 3/2 * (M1 + M2)
So both array have the same sum and product, but one has repeated values, so your algorithm does not work.
1 This is one method of finding counter examples, there may be others (there are probably others).
2 This is not exactly the same method I used to find the first counter example - In the original method, I used only one number M and was using the fact that you can replace 0 by 1 without changing the product, but I propose a more general method here in order to avoid argument such as "But I can add a check for 0 in my algorithm.".
3 That method does not work with small array because you need to find 2 even numbers M1 > 2 and M2 > 2 such that M1/2 != M2 (and reciprocally) and M1/2 - 1 != M2/2 + 1, which (I think) is not possible for any array with a size lower than 14.
What algorithms do work?4
Algorithm 1: O(n) time and space complexity.
If you can allocate a new array of size N, then:
template <std::size_t N>
bool has_repetition (std::array<int, N> const& array) {
std::array<bool, N> rep = {0};
for (auto v: array) {
if (rep[v]) {
return true;
}
rep[v] = true;
}
return false;
}
Algorithm 2: O(nlog(n)) time complexity and O(1) space complexity, with a mutable array.
You can simply sort the array:
template <std::size_t N>
bool has_repetition (std::array<int, N> &array) {
std::sort(std::begin(array), std::end(array));
auto it = std::begin(array);
auto ne = std::next(it);
while (ne != std::end(array)) {
if (*ne == *it) {
return true;
}
++it; ++ne;
}
return false;
}
Algorithm 3: O(n^2) time complexity and O(1) space complexity, with non mutable array.
template <std::size_t N>
bool has_repetition (std::array<int, N> const& array) {
for (auto it = std::begin(array); it != std::end(array); ++it) {
for (auto jt = std::next(it); jt != std::end(array); ++jt) {
if (*it == *jt) {
return true;
}
}
}
return false;
}
4 These algorithms do work, but there may exist other ones that performs better - These are only the simplest ones I could think of given some "restrictions".
What's wrong with your method?
Your method computes some statistics of the data and compares them with those expected for a permutation (= correct answers). While a violation of any of these comparisons is conclusive (the data cannot satisfy the constraint), the inverse is not necessarily the case. You only look at two statistics, and these are too few for sufficiently large data sets. Owing to the fact that the data are integer, the smallest number of data for which your method may fail is larger than 3.
If you are searching duplicates in your array there is simple way:
int N =5;
int array[N] = {1,2,3,4,4};
for (int i = 0; i< N; i++){
for (int j =i+1; j<N; j++){
if(array[j]==array[i]){
std::cout<<"DUPLICATE FOUND\n";
return true;
}
}
}
return false;
Other simple way to find duplicates is using the std::set container for example:
std::set<int> set_int;
set_int.insert(5);
set_int.insert(5);
set_int.insert(4);
set_int.insert(4);
set_int.insert(5);
std::cout<<"\nsize "<<set_int.size();
the output will be 2, because there is 2 individual values
A more in depth explanation why your algorithm is wrong:
count the sum of the indices (4x5 / 2 = 10) and check that the values' sum (3+4+1+2+0) is equal to this sum. if not, there's repeated number.
Given any array A which has no duplicates, it is easy to create an array that meets your first requirement but now contains duplicates. Just take take two values and subtract one of them by some value v and add that value to the other one. Or take multiple values and make sure the sum of them stays the same. (As long as new values are still within the 0 .. N-1 range.) For N = 3 it is already possible to change {0,1,2} to {1,1,1}. For an array of size 3, there are 7 compositions that have correct sum, but 1 is a false positive. For an array of size 4 there are 20 out of 44 have duplicates, for an array of size 5 that's 261 out of 381, for an array of size 6 that's 3612 out of 4332, and so on. It is save to say that the number of false positives grows much faster than real positives.
in addition to the first condition, get the multiplication of the indices(except 0. so: 1x2x3x4) and check if it's equal to the values' multiplication (except 0, so: 3x4x1x2x0).
The second requirement involves the multiplication of all indices above 0. It is easy to realize this is could never be a very strong restriction either. As soon as one of the indices is not prime, the product of all indices is no longer uniquely tied to the multiplicands and a list can be constructed of different values with the same result. E.g. a pair of 2 and 6 can be replaced with 3 and 4, 2 and 9 can be replaced with 6 and 3 and so on. Obviously the number of false positives increases as the array-size gets larger and more non-prime values are used as multiplicands.
None of these requirements is really strong and the cannot compensate for the other. Since 0 is not even considered for the second restriction a false positive can be created fairly easy for arrays starting at size 5. any pair of 0 and 4 can simply be replaced with two 2's in any unique array, for example {2, 1, 2, 3, 2}
What you would need, is to have a result that is uniquely tight to the occurring values. You could tweak your second requirement to a more complex approach and skip over the non-prime values and take 0 into account. For example you could use the first prime as multiplicand (2) for 0, use 3 as multiplicand for 1, 5 as multiplicand for 2, and so on. That would work (you would not need the first requirement), but this approach would be overly complex. An simpler way to get a unique result would be to OR the i-th bit for each value (0 => 1 << 0, 1 => 1 << 1, 2 => 1 << 2, and so on. (Obviously it is faster to check wether a bit was already set by a reoccurring value, rather than wait for the final result. And this is conceptually the same as using a bool array/vector from the other examples!)
I am reading book on programming pearls.
Question: Given a sequential file that contains at most four billion
32 bit integers in random order, find a 32-bit integer that isn't in
the file (and there must be at least one missing). This problem has to
be solved if we have a few hundred bytes of main memory and several
sequential files.
Solution: To set this up as a binary search we have to define a range,
a representation for the elements within the range, and a probing
method to determine which half of a range holds the missing integer.
How do we do this?
We'll use as the range a sequence of integers known to contain atleast
one missing element, and we'll represent the range by a file
containing all the integers in it. The insight is that we can probe a
range by counting the elements above and below its midpoint: either
the upper or the lower range has atmost half elements in the total
range. Because the total range has a missing element, the smaller half
must also have a mising element. These are most ingredients of a
binary search algorithm for above problem.
Above text is copy right of Jon Bently from programming pearls book.
Some info is provided at following link
"Programming Pearls" binary search help
How do we search by passes using binary search and also not followed with the example given in above link? Please help me understand logic with just 5 integers rather than million integers to understand logic.
Why don't you re-read the answer in the post "Programming Pearls" binary search help. It explains the process on 5 integers as you ask.
The idea is that you parse each list and break it into 2 (this is where binary part comes from) separate lists based on the value in the first bit.
I.e. showing binary representation of actual numbers
Original List "": 001, 010, 110, 000, 100, 011, 101 => (broken into)
(we remove the first bit and append it to the "name" of the new list)
To form each of the bellow lists we took values starting with [0 or 1] from the list above
List "0": 01, 10, 00, 11 (is formed from subset 001, 010, 000, 011 of List "" by removing the first bit and appending it to the "name" of the new list)
List "1": 10, 00, 01 (is formed from subset 110, 100, 101 of List "" by removing the first bit and appending it to the "name" of the new list)
Now take one of the resulting lists in turn and repeat the process:
List "0" becomes your original list and you break it into
List "0***0**" and
List "0***1**" (the bold numbers are again the 1 [remaining] bit of the numbers in the list being broken)
Carry on until you end up with the empty list(s).
EDIT
Process step by step:
List "": 001, 010, 110, 000, 100, 011, 101 =>
List "0": 01, 10, 00, 11 (from subset 001, 010, 000, 011 of the List "") =>
List "00": 1, 0 (from subset 01, 00 of the List "0") =>
List "000": 0 [final result] (from subset 0 of the List "00")
List "001": 1 [final result] (from subset 1 of the List "00")
List "01": 0, 1 (from subset 10, 11 of the List "0") =>
List "010": 0 [final result] (from subset 0 of the List "01")
List "011": 1 [final result] (from subset 1 of the List "01")
List "1": 10, 00, 01 (from subset 110, 100, 101 of the List "") =>
List "10": 0, 1 (from subset 00, 01 of the List "1") =>
List "100": 0 [final result] (from subset 0 of the List "10")
List "101": 1 [final result] (from subset 1 of the List "10")
List "11": 0 (from subset 10 of the List "1") =>
List "110": 0 [final result] (from subset 0 of the List "11")
List "111": absent [final result] (from subset EMPTY of the List "11")
The positive of this method is that it will allow you to find ANY number of missing numbers in the set - i.e. if more than one is missing.
P.S. AFAIR for 1 single missing number out of the complete range there is even more elegant solution of XOR all numbers.
The idea is to solve easier problem:
Is the missing value in range [minVal, X] or (X, maxVal).
If you know this, you can move X and check again.
For example, you have 3, 4, 1, 5 (2 is missing).
You know that minVal = 1, maxVal = 5.
Range = [1, 5], X = 3, there should be 3 integers in range [1, 3] and 2 in range [4, 5]. There are only 2 in range [1, 3], so you are looking in range [1, 3]
Range = [1, 3], X = 2. There are only 1 value in range [1, 2], so you are looking in range [1, 2]
Range = [1, 2], X = 1. There are no values in range [2, 2] so it is your answer.
EDIT: Some pseudo-C++ code:
minVal = 1, maxVal = 5; //choose correct values
while(minVal < maxVal){
int X = (minVal + maxVal) / 2
int leftNumber = how much in range [minVal, X]
int rightNumber = how much in range [X + 1, maxVal]
if(leftNumber < (X - minVal + 1))maxVal = X
else minVal = X + 1
}
Here's a simple C solution which should illustrate the technique. To abstract away any tedious file I/O details, I'm assuming the existence of the following three functions:
unsigned long next_number (void) reads a number from the file and returns it. When called again, the next number in the file is returned, and so on. Behavior when the end of file is encountered is undefined.
int numbers_left (void) returns a true value if there are more numbers available to be read using next_number(), false if the end of the file has been reached.
void return_to_start (void) rewinds the reading position to the start of the file, so that the next call to next_number() returns the first number in the file.
I'm also assuming that unsigned long is at least 32 bits wide, as required for conforming ANSI C implementations; modern C programmers may prefer to use uint32_t from stdint.h instead.
Given these assumptions, here's the solution:
unsigned long count_numbers_in_range (unsigned long min, unsigned long max) {
unsigned long count = 0;
return_to_start();
while ( numbers_left() ) {
unsigned long num = next_number();
if ( num >= min && num <= max ) {
count++;
}
}
return count;
}
unsigned long find_missing_number (void) {
unsigned long min = 0, max = 0xFFFFFFFF;
while ( min < max ) {
unsigned long midpoint = min + (max - min) / 2;
unsigned long count = count_numbers_in_range( min, midpoint );
if ( count < midpoint - min + 1 ) {
max = midpoint; // at least one missing number below midpoint
} else {
min = midpoint; // no missing numbers below midpoint, must be above
}
}
return min;
}
One detail to note is that min + (max - min) / 2 is the safe way to calculate the average of min and max; it won't produce bogus results due to overflowing intermediate values like the seemingly simpler (min + max) / 2 might.
Also, even though it would be tempting to solve this problem using recursion, I chose an iterative solution instead for two reasons: first, because it (arguably) shows more clearly what's actually being done, and second, because the task was to minimize memory use, which presumably includes the stack too.
Finally, it would be easy to optimize this code, e.g. by returning as soon as count equals zero, by counting the numbers in both halves of the range in one pass and choosing the one with more missing numbers, or even by extending the binary search to n-ary search for some n > 2 to reduce the number of passes. However, to keep the example code as simple as possible, I've left such optimizations unmade. If you like, you may want to, say, try modifying the code so that it requires at most eight passes over the file instead of the current 32. (Hint: use a 16-element array.)
Actually, if we have range of integers from a to b. Sample: [a..b].
And in this range we have b-a integers. It means, that only one is missing.
And if only one is missing, we can calculate result using only single cycle.
First we can calculate sum of all integers in range [a..b], which equals:
sum = (a + b) * (b - a + 1) / 2
Then we calcualate summ of all integers in our sequence:
long sum1 = 0;
for (int i = 0; i < b - a; i++)
sum1 += arr[i];
Then we can find missing element as difference of those two sums:
long result = sum1 - sum;
when you've seen 2^31 zeros or ones in the ith digit place then your answer has a one or zero in the ith place. (Ex: 2^31 ones in 5th binary position means the answer has a zero in the 5th binary position.
First draft of c code:
uint32_t binaryHistogram[32], *list4BILLION, answer, placesChecked[32];
uint64_t limit = 4294967296;
uint32_t halfLimit = 4294967296/2;
int i, j, done
//General method to point to list since this detail is not important to the question.
list4BILLION = 0000000000h;
//Initialize array to zero. This array represents the number of 1s seen as you parse through the list
for(i=0;i<limit;i++)
{
binaryHistogram[i] = 0;
}
//Only sum up for first half of the 4 billion numbers
for(i=0;i<halfLimit;i++)
{
for(j=0;j<32;j++)
{
binaryHistogram[j] += ((*list4BILLION) >> j);
}
}
//Check each ith digit to see if all halfLimit values have been parsed
for(i=halfLimit;i<limit;i++)
{
for(j=0;j<32;j++)
{
done = 1; //Dont need to continue to the end if placesChecked are all
if(placesChecked[j] != 0) //Dont need to pass through the whole list
{
done = 0; //
binaryHistogram[j] += ((*list4BILLION) >> j);
if((binaryHistogram[j] > halfLimit)||(i - binaryHistogram[j] == halfLimit))
{
answer += (1 << j);
placesChecked[j] = 1;
}
}
}
}
I am trying to do the following in Matlab. Take two lists of numbers, possibly containing repeated elements, and subtract one set from the other set.
Ex: A=[1 1 2 4]; B=[1 2 4];
Desired result would be A-B=C=[1]
Or, another example, E=[3 3 5 5]; F=[3 3 5];
Desired result would be E-F=G=[5]
I wish I could do this using Matlab's set operations, but their function setdiff does not respect the repeated elements in the matrices. I appreciate that this is correct from a strict set theory standpoint, but would nevertheless like to tackle problems like: "I have 3 apples and 4 oranges, and you take 2 apples and 1 orange, how many of each do I have left." My range of possible values in these sets is in the thousands, so building a large matrix for tallying elements and then subtracting matrices does not seem feasible for speed reasons. I will have to do thousands of these calculations with thousands of set elements during a gui menu operation.
Example of what I would like to avoid for tackling the second example above:
E=[0 0 2 0 2]; F=[0 0 2 0 1];
G=E-F=[0 0 0 0 1];
Thanks for your help!
This can be done with the accumarray command.
A = [1 1 2 4]';
B = [1 2 4]'; % <-make these column vectors
X = accumarray(A, 1);
Y = accumarray(B, 1);
This will produce the output
X = [2 1 0 1]'
and
Y = [1 1 0 1]'
Where X(i) represents the number of incidents of the number i, in vector A, and Y(i) represents the number of incidents of number i in vector B.
Then you can just take X - Y.
One caveat: if the maximum values of A and B are different, the output from accummarray will have different lengths. If that is the case, you can just assign the output to be a subset of a vector of zeros that is the size of the larger vector.
I just want to improve on Prototoast's answer.
In order to avoid pitfalls involving non-positive numbers in A or B use hist:
A = [-10 0 1 1 2 4];
B = [1 2 4];
We need the minimum and maximum values in the union of A and B:
U = [A,B];
range_ = min(U):max(U);
So that we can use hist to give us same length vectors:
a = hist(A,range_)
b = hist(B,range_)
Now you need to subtract the histograms:
r = a-b
If you wish the set difference operator be symmetric then use:
r = abs(a-b)
The following will give you which items are in A \ B (\ here is your modified set difference):
C = range_(logical(r))
Hope this helps.
I was trying to write a program for the problem I mentioned above, the numbers (i.e the lists) can be of unequal length, I was not able to figure out a way to do this other than the most commonly thought of approach i.e
reverse list-1
reverse list-2
find the sum and store it in a new list represented by list-3
reverse the list.
The complexity of this should be of the O(n+m). Is there anyway to reduce it, or do it better?
Ideally the first thing I would do is store the numbers in reverse digit order, so 43,712 is stored as:
2 -> 1 -> 7 -> 3 -> 4
It makes arithmetic operations much easier.
Displaying a number can be done either iteratively or more simply with a recursive algorithm. Note: all this assumes singly-linked lists.
Edit: But you've since stated you have no choice in the storage format. As such, your best bet is to reverse both the lists, do the addition and then reverse the result.
You can do better without list reversal. WLOG I'll assume that both lists have equal length (prepend with 0 if necessary).
Start the addition from left to right (from most significant to least significant digit). You have three cases, depending of the sum of two digits:
= 9: keep the nine and increase a counter
< 9: write counter x nine, write sum, reset counter
9: increase last digit, write counter x zero, write sum (modulo 10), reset counter
I'll work on the following example:
2 568 794 +
1 438 204
--------- =
4 006 998
Add 2 + 1 = 3: case 3.
list = (3), counter = 0
Add 5 + 4 = 9: case 1
list = (3), counter = 1
Add 6 + 4 = 9: case 1
list = (3), counter = 2
Add 8 + 8 = 16: case 3
list = (4, 0, 0, 6), counter = 0
Add 7 + 2 = 9: case 1
list = (4, 0, 0, 6), counter = 1
Add 9 + 0 = 9: case 1
list = (4, 0, 0, 6), counter = 2
Add 4 + 4 = 8: case 2
list = (4, 0, 0, 6, 9, 9, 8), counter = 0
If you can use a doubly-linked list then you can quickly traverse to the end of each list, and then just work your way back, adding the numbers at each point and add that to a new list.
You will need to determine which list is longer, and add up based on the length of the shorter list, and then just finish summing and adding the longer list.
But, you will have some issues with the fact that the sum may go over one digit, so if that happens you will need to keep track of the overflow and add that to the next node.
I don't think of a better solution to the problem as stated. The root problem is that you have to process the list elements in the reverse order. In theory you could implement the algorithm recursively, avoiding the need for explicit reversal steps. But that requires O(max(m,n)) stack space, and would most likely be slower.
But I think that is really saying that you've chosen a poor representation. If you represent the numbers as doubly linked lists of int or arrays of int (with an explicit size), the complexity will be O(max(m,n)) with a smaller constant of proportionality.
Note: O(max(m,n)) and O(m+n) are both abuses of O notation. Strictly speaking, O is a defined in terms of a limit as a single variable goes to infinity. Looked at it this way, O(max(m,n)) and O(m+n) both reduce to O(m) or O(n). However, I understand what you are trying to say :-).
The only potential optimization, which would come at the cost of some code clarity, would be to combine the initial reversals into a single loop. You then go from O(n+m+m) to O(m+m), although the steps inside the loop are costlier.