I know this was talked over a lot here, but I am struggling with this problem.
We have a set of numbers, e.g [3, 1, 1, 2, 2, 1], and we need to break it into two subsets, so the each sum is equal or difference is minimal.
I've seen wikipedia entry, this page (problem 7) and a blog entry.
But every algorithm listed is giving only YES/NO result and I really don't understand how to use them to print out two subsets (e.g S1 = {5, 4} and S2 = {5, 3, 3}). What am I missing here?
The pseudo-polynomial algorithm is designed to provide an answer to the decision problem, not the optimization problem. However, note that the last row in the table of booleans in the example indicates that the current set is capable of summing up to N/2.
In the last row, take the first column where the boolean value is true. You can then check what the actual value of the set in the given column is. If the sets summed value is N/2 you have found the first set of the partition. Otherwise you have to check which set is capable of being the difference to N/2. You can use the same approach as above, this time for the difference d.
This will be O(2^N). No Dynamic Programming used here. You can print result1, result2 and difference after execution of the function. I hope this helps.
vector<int> p1,p2;
vector<int> result1,result2;
vector<int> array={12,323,432,4,55,223,45,67,332,78,334,23,5,98,34,67,4,3,86,99,78,1};
void partition(unsigned int i,long &diffsofar, long sum1,long sum2)
{
if(i==array.size())
{
long diff= abs(sum1 - sum2);
if(diffsofar > diff)
{
result1 = p1;
result2 = p2;
diffsofar = diff;
}
return;
}
p1.push_back(array[i]);
partition(i+1,diffsofar,sum1+array[i],sum2);
p1.pop_back();
p2.push_back(array[i]);
partition(i+1,diffsofar,sum1,sum2+array[i]);
p2.pop_back();
return;
}
I faced this same problem recently, and I posted a question about it (here: Variant of Knapsack). The difference in my case is that the resulting subsets must be the same size (if the original set has an even number of elements). In order to assure that, I added a few lines to #Sandesh Kobal answer;
void partition(unsigned int i,long &diffsofar, long sum1,long sum2)
{
int maxsize = (array.size()+1)/2;
if(p1.size()>maxsize)
return;
if(p2.size()>maxsize)
return;
if(i==array.size())
{
...
Also, after both calls to partition, I added if(diffsofar==0) return;. If we already found an optimal solution, it makes no sense to keep searching...
All the articles I've seen take a dynamic programming approach. Do we really need one?
Suppose the array given is arr
Use the following algorithm :
Sort the array in descending order
Create two empty arrays, a = [] and b = []
sum_a = sum_b = 0
for x in arr:
if sum_a > sum_b:
b.append(x)
sum_b += x
else:
a.append(x)
sum_a += x
The absolute difference between sum_a and sum_b would be the minimum possible difference between the two subsets.
Consider arr = [3,1,1,2,2,1]
Sorting the array : arr = [3,2,2,1,1,1]
a = [], b = []
a = [3], b = []
a = [3], b = [2]
a = [3], b = [2,2]
a = [3,1], b = [2,2]
a = [3,1,1], b = [2,2]
a = [3,1,1], b = [2,2,1]
sa = 5, sb = 5
Minimum difference : 5 - 5 = 0
Related
I was asked in an interview today below question. I gave O(nlgn) solution but I was asked to give O(n) solution. I could not come up with O(n) solution. Can you help?
An input array is given like [1,2,4] then every element of it is doubled and
appended into the array. So the array now looks like [1,2,4,2,4,8]. How
this array is randomly shuffled. One possible random arrangement is
[4,8,2,1,2,4]. Now we are given this random shuffled array and we want to
get original array [1,2,4] in O(n) time.
The original array can be returned in any order. How can I do it?
Here's an O(N) Java solution that could be improved by first making sure that the array is of the proper form. For example it shouldn't accept [0] as an input:
import java.util.*;
class Solution {
public static int[] findOriginalArray(int[] changed) {
if (changed.length % 2 != 0)
return new int[] {};
// set Map size to optimal value to avoid rehashes
Map<Integer,Integer> count = new HashMap<>(changed.length*100/75);
int[] original = new int[changed.length/2];
int pos = 0;
// count frequency for each number
for (int n : changed) {
count.put(n, count.getOrDefault(n,0)+1);
}
// now decide which go into the answer
for (int n : changed) {
int smallest = n;
for (int m=n; m > 0 && count.getOrDefault(m,0) > 0; m = m/2) {
//System.out.println(m);
smallest = m;
if (m % 2 != 0) break;
}
// trickle up from smallest to largest while count > 0
for (int m=smallest, mm = 2*m; count.getOrDefault(mm,0) > 0; m = mm, mm=2*mm){
int ct = count.getOrDefault(mm,0);
while (count.get(m) > 0 && ct > 0) {
//System.out.println("adding "+m);
original[pos++] = m;
count.put(mm, ct -1);
count.put(m, count.get(m) - 1);
ct = count.getOrDefault(mm,0);
}
}
}
// check for incorrect format
if (count.values().stream().anyMatch(x -> x > 0)) {
return new int[] {};
}
return original;
}
public static void main(String[] args) {
int[] changed = {1,2,4,2,4,8};
System.out.println(Arrays.toString(changed));
System.out.println(Arrays.toString(findOriginalArray(changed)));
}
}
But I've tried to keep it simple.
The output is NOT guaranteed to be sorted. If you want it sorted it's going to cost O(NlogN) inevitably unless you use a Radix sort or something similar (which would make it O(NlogE) where E is the max value of the numbers you're sorting and logE the number of bits needed).
Runtime
This may not look that it is O(N) but you can see that it is because for every loop it will only find the lowest number in the chain ONCE, then trickle up the chain ONCE. Or said another way, in every iteration it will do O(X) iterations to process X elements. What will remain is O(N-X) elements. Therefore, even though there are for's inside for's it is still O(N).
An example execution can be seen with [64,32,16,8,4,2].
If this where not O(N) if you print out each value that it traverses to find the smallest you'd expect to see the values appear over and over again (for example N*(N+1)/2 times).
But instead you see them only once:
finding smallest 64
finding smallest 32
finding smallest 16
finding smallest 8
finding smallest 4
finding smallest 2
adding 2
adding 8
adding 32
If you're familiar with the Heapify algorithm you'll recognize the approach here.
def findOriginalArray(self, changed: List[int]) -> List[int]:
size = len(changed)
ans = []
left_elements = size//2
#IF SIZE IS ODD THEN RETURN [] NO SOLN. IS POSSIBLE
if(size%2 !=0):
return ans
#FREQUENCY DICTIONARY given array [0,0,2,1] my map will be: {0:2,2:1,1:1}
d = {}
for i in changed:
if(i in d):
d[i]+=1
else:
d[i] = 1
# CHECK THE EDGE CASE OF 0
if(0 in d):
count = d[0]
half = count//2
if((count % 2 != 0) or (half > left_elements)):
return ans
left_elements -= half
ans = [0 for i in range(half)]
#CHECK REST OF THE CASES : considering the values will be 10^5
for i in range(1,50001):
if(i in d):
if(d[i] > 0):
count = d[i]
if(count > left_elements):
ans = []
break
left_elements -= d[i]
for j in range(count):
ans.append(i)
if(2*i in d):
if(d[2*i] < count):
ans = []
break
else:
d[2*i] -= count
else:
ans = []
break
return ans
I have a simple idea which might not be the best, but I could not think of a case where it would not work. Having the array A with the doubled elements and randomly shuffled, keep a helper map. Process each element of the array and, each time you find a new element, add it to the map with the value 0. When an element is processed, increment map[i] and decrement map[2*i]. Next you iterate over the map and print the elements that have a value greater than zero.
A simple example, say that the vector is:
[1, 2, 3]
And the doubled/shuffled version is:
A = [3, 2, 1, 4, 2, 6]
When processing 3, first add the keys 3 and 6 to the map with value zero. Increment map[3] and decrement map[6]. This way, map[3] = 1 and map[6] = -1. Then for the next element map[2] = 1 and map[4] = -1 and so forth. The final state of the map in this example would be map[1] = 1, map[2] = 1, map[3] = 1, map[4] = -1, map[6] = 0, map[8] = -1, map[12] = -1.
Then you just process the keys of the map and, for each key with a value greater than zero, add it to the output. There are certainly more efficient solutions, but this one is O(n).
In C++, you can try this.
With time is O(N + KlogK) where N is the length of input, and K is the number of unique elements in input.
class Solution {
public:
vector<int> findOriginalArray(vector<int>& input) {
if (input.size() % 2) return {};
unordered_map<int, int> m;
for (int n : input) m[n]++;
vector<int> nums;
for (auto [n, cnt] : m) nums.push_back(n);
sort(begin(nums), end(nums));
vector<int> out;
for (int n : nums) {
if (m[2 * n] < m[n]) return {};
for (int i = 0; i < m[n]; ++i, --m[2 * n]) out.push_back(n);
}
return out;
}
};
Not so clear about the space complexity required in the question, so this is my top-of-the-mind attempt to this question if this requires O(n) time complexity.
If the length of the input array is not even, then its wrong !!
Create a map, add the elements of the input array to it.
Divide each element in the input array by 2 and check if that value exists in the map. If it exists, add it to the array (slice) orig.
There is a chance we have added duplicate values to this original array, clean it!!
Here is a sample go code:
https://go.dev/play/p/w4mm-rloHyi
I am sure we can optimize this code in a lot of ways for space complexities. But its O(n) time complexity.
I am trying to pick one number from multiple arraylists and find all possible ways to pick the numbers such that the sum of those numbers is greater than a given number. I can only think of brute force implementation.
For example, I have five arraylists such as
A = [2, 6, 7]
B = [6, 9]
C = [4]
D = [4, 7]
E = [8, 10, 15]
and a given number is 40.
Then after picking one number from each list, all possible ways could be
[7, 9, 4, 7, 15]
[6, 9, 4, 7, 15]
So, these are the two possible ways to pick numbers greater than or equal to 40. In case the given number is small then there could be many solutions. So how can I count them without brute force? Even with brute force how do I devise the solution in Java.
Below is my implementation. It works fine for small numbers but if the numbers are large then it gives me runtime error since the program runs for too long.
public static void numberOfWays(List<Integer> A, List<Integer> B, List<Integer> C, List<Integer> D,
List<Integer> E, int k){
long ways = 0;
for(Integer a:A){
for(Integer b:B){
for(Integer c:C){
for(Integer d:D){
for(Integer e:E){
int sum = a+b+c+d+e;
//System.out.println(a+" "+b+" "+c+" "+d+" "+e+" "+sum);
if(sum > k)
ways++;
}
}
}
}
}
System.out.println(ways);
}
The list can contain up to 1000 elements and the elements can range from 1 to 1000. The threshold value k can range from 1 to 10^9.
I am not a java programmer.But I think its a logical problem.So,I have solved it for you in python.I am pretty sure you can convert it into java.
Here is the code:
x = input('Enter the number:')
a = [2, 6, 7]
b = [6, 9]
c = [4]
d = [4, 7]
e = [8, 10, 15]
i = 0
z = 0
final_list = []
while i <= int(x):
try:
i += a[z]
final_list.append(a[z])
except BaseException:
pass
try:
i += b[z]
final_list.append(b[z])
except BaseException:
pass
try:
i += c[z]
final_list.append(c[z])
except BaseException:
pass
try:
i += d[z]
final_list.append(d[z])
except BaseException:
pass
try:
i += e[z]
final_list.append(e[z])
except BaseException:
pass
z += 1
print(final_list)
One way is this. There has to be at least one solution where you pick one number from each array and add them up to be greater than or equal to another.
Considering the fact that arrays might have random numbers in any order, first use this sort function to have them in decreasing order (largest number first, smallest number last) :
Arrays.sort(<array name>, Collections.reverseOrder());
Then pick the 1st element in the array :
v = A[0]
w = B[0]
x = C[0]
y = D[0]
z = E[0]
Then you can print them like this : v,w,x,y,z
Now your output will be :
7,9,4,7,15
Since it took the largest number of each array, it has to be equal to or greater than the given number, unless the number is greater than all of these combined in which case it is impossible.
Edit : I think I got the question wrong. If you want to know how many of the possible solutions there are, that is much easier.
First create a variable to store the possibilities
var total = 0
Use the rand function to get a random number. In your array say something like :
v=A[Math.random(0,A[].length)]
Do the same thing for all arrays, then add them up
var sum = v+w+x+y+z
Now you have an if statement to see if the sum is greater than or equal to the number given (lets say the value is stored in the variable "given")
if(sum >= given){
total+=1
}else{
<repeat the random function to restart the process and generate a new sum>
}
Finally, you need to repeat this multiple times as incase there are multiple solutions, the code will only find one and give you a false total.
To solve this, create a for loop and put all of this code inside it :
//create a variable outside to store the total number of elements in all the arrays
var elements = A[].length + B[].length + C[].length + D[].length + E[].length
for(var i = 0; i <= elements; i++){
<The code is inside here, except for "total" as otherwise the value will keep resetting>
}
The end result should look something like this :
var total = 0
var elements = A[].length + B[].length + C[].length + D[].length + E[].length
for(var i = 0; i <= elements; i++){
v=A[Math.random(0,A[].length)]
w=B[Math.random(0,B[].length)]
x=C[Math.random(0,C[].length)]
y=D[Math.random(0,D[].length)]
z=E[Math.random(0,E[].length)]
var sum = v+w+x+y+z
if(sum >= given){
total+=1
}else{
v=A[Math.random(0,A[].length)]
w=B[Math.random(0,B[].length)]
x=C[Math.random(0,C[].length)]
y=D[Math.random(0,D[].length)]
z=E[Math.random(0,E[].length)]
}
}
At the end just print the total once the entire cycle is over or just do
console.log(total)
This is just for reference and the code might not work, it probably has a bunch of bugs in it, this was just my 1st draft attempt at it. I have to test it out on my own but i hope you see where I'm coming from. Just look at the process, make your own amendments and this should work fine.
I have not deleted the first part of my answer even though it isn't the answer to this question just so that if you're having trouble in that part as well, where you select the highest possible number, it might help you
Good luck!
I was wondering if anyone would be willing to give me some feedback on my solution to today's question from the Daily Coding problem email list. It's an interesting question about finding the first value that's NOT present in a list, as opposed to finding a value that IS there.
This problem was asked by Stripe.
Given an array of integers, find the first missing positive integer in linear time and
constant space.
In other words, find the lowest positive integer that does not exist in the array.
The array can contain duplicates and negative numbers as well.
For example, the input [3, 4, -1, 1] should give 2. The input [1, 2, 0] should give 3.
You can modify the input array in-place.
My main question is that I'm not sure if I've met the O(n) time and O(1) space requirements. Here is my solution using Groovy:
def find_first( Xs ) {
ys = []
szXs = Xs.size()
// First, transfer Xs list to a set ys.
szXs.times { // this loop is O(n)
if ( (hd = Xs.head()) > 0 ) { ys << hd; } // We only need to keep val if it's positive.
Xs = Xs.drop(1); // Remove val from Xs in order to meet O(1) space requirement.
}
ys = ys.toSet() // converting list to set is O(n)
sz = ys.size() // ys.size() might be same as Xs.size(), but it could be smaller, so give it a new var.
for (int i = 1; i < sz; ++i) { // this loop is O(n)
if ( !ys.contains(i) ) return i // Testing a hash set for set membership is O(1)
}
return sz + 1 // if this point is reached, ys had all values [1..sz] already
} // end find_first
def test(Xs) {
println "\nstarting with Xs = " + Xs
ans1 = find_first( Xs )
println "first missing = " + ans1
}
xs1 = [ 3,4,-1,1 ]
xs2 = [ 1, 2, 0 ]
xs3 = [ 7, 8 ]
xs4 = [ 1,1,2,1 ]
xs5 = [ -4, -1 ]
test(xs1)
test(xs2)
test(xs3)
test(xs4)
test(xs5)
And here is the output:
starting with Xs = [3, 4, -1, 1]
first missing = 2
starting with Xs = [1, 2, 0]
first missing = 3
starting with Xs = [7, 8]
first missing = 1
starting with Xs = [1, 1, 2, 1]
first missing = 3
starting with Xs = [-4, -1]
first missing = 1
So it's correct on that limited set of values, just want to be sure that it's O(n) time and O(1) space. Any advice is appreciated!
Hank
I am not familiar with groovy, but it seems like your code is O(n) space, since you create an additional hash set (ys).
Note that the question has a hint:
You can modify the input array in-place.
This can be done in O(n) time with O(1) space, by first sorting the array inplace with radix sort, and then iterating over the values to find first missing one.
I have the following code which implements a recursive solution for this problem, instead of using the reference variable 'x' to store overall max, How can I or can I return the result from recursion so I don't have to use the 'x' which would help memoization?
// Test Cases:
// Input: {1, 101, 2, 3, 100, 4, 5} Output: 106
// Input: {3, 4, 5, 10} Output: 22
int sum(vector<int> seq)
{
int x = INT32_MIN;
helper(seq, seq.size(), x);
return x;
}
int helper(vector<int>& seq, int n, int& x)
{
if (n == 1) return seq[0];
int maxTillNow = seq[0];
int res = INT32_MIN;
for (int i = 1; i < n; ++i)
{
res = helper(seq, i, x);
if (seq[i - 1] < seq[n - 1] && res + seq[n - 1] > maxTillNow) maxTillNow = res + seq[n - 1];
}
x = max(x, maxTillNow);
return maxTillNow;
}
First, I don't think this implementation is correct. For this input {5, 1, 2, 3, 4} it gives 14 while the correct result is 10.
For writing a recursive solution for this problem, you don't need to pass x as a parameter, as x is the result you expect to get from the function itself. Instead, you can construct a state as the following:
Current index: this is the index you're processing at the current step.
Last taken number: This is the value of the last number you included in your result subsequence so far. This is to make sure that you pick larger numbers in the following steps to keep the result subsequence increasing.
So your function definition is something like sum(current_index, last_taken_number) = the maximum increasing sum from current_index until the end, given that you have to pick elements greater than last_taken_number to keep it an increasing subsequence, where the answer that you desire is sum(0, a small value) since it calculates the result for the whole sequence. by a small value I mean smaller than any other value in the whole sequence.
sum(current_index, last_taken_number) could be calculated recursively using smaller substates. First assume the simple cases:
N = 0, result is 0 since you don't have a sequence at all.
N = 1, the sequence contains only one number, the result is either that number or 0 in case the number is negative (I'm considering an empty subsequence as a valid subsequence, so not taking any number is a valid answer).
Now to the tricky part, when N >= 2.
Assume that N = 2. In this case you have two options:
Either ignore the first number, then the problem can be reduced to the N=1 version where that number is the last one in the sequence. In this case the result is the same as sum(1,MIN_VAL), where current_index=1 since we already processed index=0 and decided to ignore it, and MIN_VAL is the small value we mentioned above
Take the first number. Assume the its value is X. Then the result is X + sum(1, X). That means the solution includes X since you decided to include it in the sequence, plus whatever the result is from sum(1,X). Note that we're calling sum with MIN_VAL=X since we decided to take X, so the following values that we pick have to be greater than X.
Both decisions are valid. The result is whatever the maximum of these two. So we can deduce the general recurrence as the following:
sum(current_index, MIN_VAL) = max(
sum(current_index + 1, MIN_VAL) // ignore,
seq[current_index] + sum(current_index + 1, seq[current_index]) // take
).
The second decision is not always valid, so you have to make sure that the current element > MIN_VAL in order to be valid to take it.
This is a pseudo code for the idea:
sum(current_index, MIN_VAL){
if(current_index == END_OF_SEQUENCE) return 0
if( state[current_index,MIN_VAL] was calculated before ) return the perviously calculated result
decision_1 = sum(current_index + 1, MIN_VAL) // ignore case
if(sequence[current_index] > MIN_VAL) // decision_2 is valid
decision_2 = sequence[current_index] + sum(current_index + 1, sequence[current_index]) // take case
else
decision_2 = INT_MIN
result = max(decision_1, decision_2)
memorize result for the state[current_index, MIN_VAL]
return result
}
The Problem
I need an algorithm that does this:
Find all the unique ways to partition a given sum across 'buckets' not caring about order
I hope I was clear reasonably coherent in expressing myself.
Example
For the sum 5 and 3 buckets, what the algorithm should return is:
[5, 0, 0]
[4, 1, 0]
[3, 2, 0]
[3, 1, 1]
[2, 2, 1]
Disclaimer
I'm sorry if this question might be a dupe, but I don't know exactly what these sort of problems are called. Still, I searched on Google and SO using all wordings that I could think of, but only found results for distributing in the most even way, not all unique ways.
Its bit easier for me to code few lines than writing a 5-page essay on algorithm.
The simplest version to think of:
vector<int> ans;
void solve(int amount, int buckets, int max){
if(amount <= 0) { printAnswer(); return;}
if(amount > buckets * max) return; // we wont be able to fulfill this request anymore
for(int i = max; i >= 1; i--){
ans.push_back(i);
solve(amount-i, buckets-1, i);
ans.pop_back();
}
}
void printAnswer(){
for(int i = 0; i < ans.size(); i++) printf("%d ", ans[i]);
for(int i = 0; i < all_my_buckets - ans.size(); i++) printf("0 ");
printf("\n");
}
Its also worth improving to the point where you stack your choices like solve( amount-k*i, buckets-k, i-1) - so you wont create too deep recurrence. (As far as I know the stack would be of size O(sqrt(n)) then.
Why no dynamic programming?
We dont want to find count of all those possibilities, so even if we reach the same point again, we would have to print every single number anyway, so the complexity will stay the same.
I hope it helps you a bit, feel free to ask me any question
Here's something in Haskell that relies on this answer:
import Data.List (nub, sort)
parts 0 = []
parts n = nub $ map sort $ [n] : [x:xs | x <- [1..n`div`2], xs <- parts(n - x)]
partitions n buckets =
let p = filter (\x -> length x <= buckets) $ parts n
in map (\x -> if length x == buckets then x else addZeros x) p
where addZeros xs = xs ++ replicate (buckets - length xs) 0
OUTPUT:
*Main> partitions 5 3
[[5,0,0],[1,4,0],[1,1,3],[1,2,2],[2,3,0]]
If there are only three buckets this wud be the simplest code.
for(int i=0;i<=5;i++){
for(int j=0;j<=5-i&&j<=i;j++){
if(5-i-j<=i && 5-i-j<=j)
System.out.println("["+i+","+j+","+(5-i-j)+"]");
}
}
A completely different method, but if you don't care about efficiency or optimization, you could always use the old "bucket-free" partition algorithms. Then, you could filter the search by checking the number of zeroes in the answers.
For example [1,1,1,1,1] would be ignored since it has more than 3 buckets, but [2,2,1,0,0] would pass.
This is called an integer partition.
Fast Integer Partition Algorithms is a comprehensive paper describing all of the fastest algorithms for performing an integer partition.
Just adding my approach here along with the others'. It's written in Python, so it's practically like pseudocode.
My first approach worked, but it was horribly inefficient:
def intPart(buckets, balls):
return uniqify(_intPart(buckets, balls))
def _intPart(buckets, balls):
solutions = []
# base case
if buckets == 1:
return [[balls]]
# recursive strategy
for i in range(balls + 1):
for sol in _intPart(buckets - 1, balls - i):
cur = [i]
cur.extend(sol)
solutions.append(cur)
return solutions
def uniqify(seq):
seen = set()
sort = [list(reversed(sorted(elem))) for elem in seq]
return [elem for elem in sort if str(elem) not in seen and not seen.add(str(elem))]
Here's my reworked solution. It completely avoids the need to 'uniquify' it by the tracking the balls in the previous bucket using the max_ variable. This sorts the lists and prevents any dupes:
def intPart(buckets, balls, max_ = None):
# init vars
sols = []
if max_ is None:
max_ = balls
min_ = max(0, balls - max_)
# assert stuff
assert buckets >= 1
assert balls >= 0
# base cases
if (buckets == 1):
if balls <= max_:
sols.append([balls])
elif balls == 0:
sol = [0] * buckets
sols.append(sol)
# recursive strategy
else:
for there in range(min_, balls + 1):
here = balls - there
ways = intPart(buckets - 1, there, here)
for way in ways:
sol = [here]
sol.extend(way)
sols.append(sol)
return sols
Just for comprehensiveness, here's another answer stolen from MJD written in Perl:
#!/usr/bin/perl
sub part {
my ($n, $b, $min) = #_;
$min = 0 unless defined $min;
# base case
if ($b == 0) {
if ($n == 0) { return ([]) }
else { return () }
}
my #partitions;
for my $first ($min .. $n) {
my #sub_partitions = part($n - $first, $b-1, $first);
for my $sp (#sub_partitions) {
push #partitions, [$first, #$sp];
}
}
return #partitions;
}