Related
I'm working on the following task.
Given an array of n integers and two integer numbers m and k.
You can add any positive integer to any element of the array such that
the total value does not exceed k.
The task is to maximize the
multiples of m in the resultant array.
Consider the following example.
Input:
n = 5, m = 2, k = 2, arr[] = [1, 2, 3, 4, 5]
Let's add 1 to the element arr[0] and 1 to arr[2] then the final array would be:
[2, 2, 4, 4, 5]
Now there are four (4) elements which are multiples of m (2).
I am not getting correct output.
My code:
public class Main {
public static void main(String[] args) {
int n = 5;
int m = 4;
int k = 3;
int count = 0;
int[] arr = {17, 8, 9, 1, 4};
for (int i = 0; i < n; i++) {
for (int j = 0; j <= k; j++) {
// check initial
if (arr[i] % m == 0) {
break;
}
// add
arr[i] = arr[i] + j;
// check again
if (arr[i] % m == 0) {
count++;
break;
}
}
}
System.out.println("Final Array : " + Arrays.toString(arr));
System.out.println("Count : " + count);
}
}
This task boils down to a well-known Dynamic programming algorithm called Knapsack problem after a couple of simple manipulations with the given array.
This approach doesn't require sorting and would be advantages when k is much smaller n.
We can address the problem in the following steps:
Iterate over the given array and count all the numbers that are already divisible by m (this number is stored in the variable count in the code below).
While iterating, for every element of the array calculate the difference between m and remainder from the division of this element by m. Which would be equal to m - currentElement % m. If the difference is smaller or equal to k (it can cave this difference) it should be added to the list (differences in the code below) and also accumulated in a variable which is meant to store the total difference (totalDiff). All the elements which produce difference that exceeds k would be omitted.
If the total difference is less than or equal to k - we are done, the return value would be equal to the number of elements divisible by m plus the size of the list of differences.
Otherwise, we need to apply the logic of the Knapsack problem to the list of differences.
The idea behind the method getBestCount() (which is an implementation Knapsack problem) boils down to generating the "2D" array (a nested array of length equal to the size of the list of differences +1, in which every inner array having the length of k+1) and populating it with maximum values that could be achieved for various states of the Knapsack.
Each element of this array would represent the maximum total number of elements which can be adjusted to make them divisible by m for the various sizes of the Knapsack, i.e. number of items available from the list of differences, and different number of k (in the range from 0 to k inclusive).
The best way to understand how the algorithm works is to draw a table on a piece of paper and fill it with numbers manually (follow the comments in the code, some intermediate variables were introduced only for the purpose of making it easier to grasp, and also see the Wiki article linked above).
For instance, if the given array is [1, 8, 3, 9, 5], k=3 and m=3. We can see 2 elements divisible by m - 3 and 9. Numbers 1, 8, 5 would give the following list of differences [2, 1, 1]. Applying the logic of the Knapsack algorithm, we should get the following table:
[0, 0, 0, 0]
[0, 0, 1, 1]
[0, 1, 1, 2]
[0, 1, 2, 2]
We are interested in the value right most column of the last row, which is 2 plus 2 (number of elements divisible by 3) would give us 4.
Note: that code provided below can dial only with positive numbers. I don't want to shift the focus from the algorithm to such minor details. If OP or reader of the post are interested in making the code capable to work with negative number as well, I'm living the task of adjusting the code for them as an exercise. Hint: only a small change in the countMultiplesOfM() required for that purpose.
That how it might be implemented:
public static int countMultiplesOfM(int[] arr, int k, int m) {
List<Integer> differences = new ArrayList<>();
int count = 0;
long totalDiff = 0; // counter for the early kill - case when `k >= totalDiff`
for (int next : arr) {
if (next % m == 0)
count++; // number is already divisible by `m` we can increment the count and from that moment we are no longer interested in it
else if (m - next % m <= k) {
differences.add(m - next % m);
totalDiff += m - next % m;
}
}
if (totalDiff <= k) { // early kill - `k` is large enough to adjust all numbers in the `differences` list
return count + differences.size();
}
return count + getBestCount(differences, k); // fire the rest logic
}
// Knapsack Algorithm implementation
public static int getBestCount(List<Integer> differences, int knapsackSize) {
int[][] tab = new int[differences.size() + 1][knapsackSize + 1];
for (int numItemAvailable = 1; numItemAvailable < tab.length; numItemAvailable++) {
int next = differences.get(numItemAvailable - 1); // next available item which we're trying to place to knapsack to Maximize the current total
for (int size = 1; size < tab[numItemAvailable].length; size++) {
int prevColMax = tab[numItemAvailable][size - 1]; // maximum result for the current size - 1 in the current row of the table
int prevRowMax = tab[numItemAvailable - 1][size]; // previous maximum result for the current knapsack's size
if (next <= size) { // if it's possible to fit the next item in the knapsack
int prevRowMaxWithRoomForNewItem = tab[numItemAvailable - 1][size - next] + 1; // maximum result from the previous row for the size = `current size - next` (i.e. the closest knapsack size which guarantees that there would be a space for the new item)
tab[numItemAvailable][size] = Math.max(prevColMax, prevRowMaxWithRoomForNewItem);
} else {
tab[numItemAvailable][size] = Math.max(prevRowMax, prevColMax); // either a value in the previous row or a value in the previous column of the current row
}
}
}
return tab[differences.size()][knapsackSize];
}
main()
public static void main(String[] args) {
System.out.println(countMultiplesOfM(new int[]{17, 8, 9, 1, 4}, 3, 4));
System.out.println(countMultiplesOfM(new int[]{1, 2, 3, 4, 5}, 2, 2));
System.out.println(countMultiplesOfM(new int[]{1, 8, 3, 9, 5}, 3, 3));
}
Output:
3 // input array [17, 8, 9, 1, 4], m = 4, k = 3
4 // input array [1, 2, 3, 4, 5], m = 2, k = 2
4 // input array [1, 8, 3, 9, 5], m = 3, k = 3
A link to Online Demo
You must change 2 line in your code :
if(arr[i]%m==0)
{
count++; // add this line
break;
}
// add
arr[i]=arr[i]+1; // change j to 1
// check again
if(arr[i]%m==0)
{
count++;
break;
}
The first is because the number itself is divisible.
and The second is because you add a number to it each time.That is wrong.
for example chenge your arr to :
int[] arr ={17,8,10,2,4};
your output is :
Final Array : [20, 8, 16, 8, 4]
and That is wrong because 16-10=6 and is bigger than k=3.
I believe the problem is that you aren't processing the values in ascending order of the amount by which to adjust.
To solve this I started by using a stream to preprocess the array. This could be done using other methods.
map the values to the amount to make each one, when added, divisible by m.
filter out those that equal to m' (already divisible by m`)
sort in ascending order.
Once that is done. Intialize max to the difference between the original array length and the processed length. This is the number already divisible by m.
As the list is iterated
check to see if k > amount needed. If so, subtract from k and increment max
otherwise break out of the loop (because of the sort, no value remaining can be less than k)
public static int maxDivisors(int m, int k, int[] arr) {
int[] need = Arrays.stream(arr).map(v -> m - v % m)
.filter(v -> v != m).sorted().toArray();
int max = arr.length - need.length;
for (int val : need) {
if (k >= val) {
k -= val;
max++;
} else {
break;
}
}
return max;
}
int m = 4;
int k = 3;
int[] arr ={17,8,9,1,4};
int count = maxDivisors(m, k, arr);
System.out.println(count);
prints
3
Given two arrays of ints, a and b, try to create an arithmetic sequence by adding ints from b into a. Return the maximum length of a or -1 if there does not exist an arithmetic sequence e.g. a = [2, 4, 8], b = [1, 6, 10, 12] -> a = [2, 4, 6, 8, 10, 12] -> return 6
I tried creating a new array and merging both a and b and counting the longest subsequence but the count could remove elements from a which should not be touched
static int maxSeq(int[] arr1, int[] arr2){
if(arr1.length ==0)return 0;
int n =arr1.length, m = arr2.length;
int[] arr = new int[n+m];
System.arraycopy(arr1,0,arr,0,n);
System.arraycopy(arr2,0,arr,n,m);
Arrays.sort(arr);
int result =0;
Map<Integer,Integer>[]d = new HashMap[n+m];
for(int i =0; i < arr.length;i++){
d[i] = new HashMap<Integer, Integer>();
}
for(int i =1; i < arr.length; ++i){
for(int j = 0; j<i;++j ){
int diff = arr[i]-arr[j];
int len =2;
if(d[j].containsKey(diff)){
len = d[j].get(diff) +1;
}
d[i].put(diff,len);
result = Math.max(result,d[i].get(diff));
}
}
return result;
}
a = [2, 4, 8], b = [1, 6, 10, 12] -> a = [2, 4, 6, 8, 10, 12] -> return 6
int[] a = {5,7,13,14}, b = {9,11,15}; return -1 not 6
I think you should try to fix your code.
if(d[j].containsKey(diff)){ len = d[j].get(diff) +1; }
Here you are looking for differences in a map of some index j, and there should be only one map of key value paires, not array of maps.
The key here is to fill in array A with numbers from array B so A become an arithmetic sequence.
Solution:
First find the minimum gap between 2 consequence number in A
With the given "gap", try to see if an arithmetic sequence can be build by looping through the 2 arrays, and find out if numbers in B can fill in array A so A become arithmetic with step = "gap". Take count of the length if success.
If it can be found, try to see if smaller gap can build longer arithmetic sequence by looping through all the divisor of the original "gap" as the new "gap" and test again. (example: A = [ 1,5], B= [3,7,9] => Previous step see that we can build arithmetic sequence [1,5,9], but the final answer should be [1,3,5,7,9].
I'd like to share my solution (probably is not optimized and I did not wrote many tests, but it worked on your two sample tests):
Idea:
Notice the common difference d of arithmetic sequence is upperbounded by min(a[i] - a[i-1]), otherwise we won't be able to visit all elements in a
We iterate on common difference d to check the length of each potential list, and find the max length
Complete code in Python:
(Suppose a, b are both sorted)
def max_arithmetic_length(a, b):
min_diff = float('inf') # common difference d is upper bounded by min_diff
for i in range(1, len(a)):
min_diff = min(min_diff, a[i] - a[i-1])
d_a = {x : True for x in a}
d_b = {x : True for x in b}
max_cnt = 0
for d in range(1, min_diff + 1):
skip_current_d = False # a switch to skip outer loop
for x in a:
if (x - a[0]) % d != 0: # must exist: a[i] - a[0] = kd
skip_current_d = True
break
if skip_current_d:
continue
cur = a[0]
cnt = 0
visited = {}
while cur in d_a or cur in d_b:
cnt += 1
visited[cur] = True
cur += d
if a[-1] not in visited: # if the last element in a is visited, then every element in a is visited
continue
# check those smaller than a[0] (may only exist in b)
cur = a[0] - d
while cur in b:
cnt += 1
cur -= d
max_cnt = max(cnt, max_cnt)
return max_cnt if max_cnt else -1
a = [2, 4, 8]
b = [1, 6, 10, 12]
print(max_arithmetic_length(a,b)) # return 6
a = [5,7,13,14]
b = [9,11,15]
print(max_arithmetic_length(a,b)) # return -1
Given an array of integers, I'm trying to find the longest subset (powerset) with sum equal to k using the lease possible time complexity.
e.g. if inputArr= [1, 2, 8, 1, 1, 7] and k= 10, then the output should be 4 since the longest subset with sum equals to 10 is [1, 1, 1, 7].
Edit: I might've forgotten an important detail; the elements of the array are all positive and non-zero.
I used this algorithm that I found on geeksforgeeks:
https://www.geeksforgeeks.org/finding-all-subsets-of-a-given-set-in-java/
The code works fine, but the only problem that I have is with the execution time. I am supposed to submit this online, and when I submit it the execution terminates due to timeout.
int maxSubLength=0;
for (int i = 1; i < (1<<n); i++) //n is the length of inputArr
{
int sum=0, length=0;
for (int j = 0; j < n; j++)
if ((i & (1 << j)) > 0)
{
sum+=inputArr[j];
length++;
if (sum>k)
break;
}
if (sum==k)
maxSubLength=Math.max(maxSubLength, length);
}
Is there any faster algorithm? I tried a recursive one and it didn't help.
We can solve this with dynamic programming in O(n*k) time and O(k) space. JavaScript code:
function f(A, K){
let m = new Array(K + 1).fill(0)
for (let a of A){
for (let k=K; k>=a; k--)
if (m[k - a])
m[k] = Math.max(m[k], 1 + m[k - a])
m[a] = Math.max(m[a], 1)
}
return m[K]
}
var A = [1, 2, 8, 1, 1, 7]
var K = 10
console.log(f(A, K))
I'm trying to write the contents of an array after a method call.
EX:
int[] a1 = {7, 5};
mystery(a1);
int[] a3 = {0, 0, 0, 1, 1, 1};
mystery(a3);
int[] a5 = {-1, 1, -2, 16, 3, -3, -4, 5);
mystery(a5);
I've tried to write the output on paper going over each function for each iteration. But unable to make sense of how the method works.
public static void mystery(int[] a) {
int[] b = new int[2];
for (int i = 0; i < a.length; i++) {
if (a[i] >= 0) {
int x = a[i] % 2;
b[x]++;
a[i] = b[x];
}
}
}
Expected output of array a1 is {1, 2}, but I'm trying to make sense of the process behind reaching these values in the new array.
This loop modifies all the non-negative elements of the array a:
x determines if a[i] is even (0) or odd (1)
b[x]++ increments the counter of even (0) or odd (1) elements
if a[i] is the n'th even number, it is assigned the number n
if a[i] is the m'th odd number, it is assigned the number m
Hence, for an input array of two odd numbers - {7,5} - the output would be {1,2}, since 7 is the 1st non-negative odd number and 5 is the 2nd odd non-negative number.
The method works in the following manner:
It keeps a count of all non-negative even numbers in b[0] and all non-negative odd numbers in b[1]
Now, it modifies the non-negative number in the array "a", such that it now represents as follows:
if a[i] were an even number then it now represents it is Nth even number
if a[i] were an odd number then it now represents it is Mth odd number
Or in other words
b[0] gives the count of all non-negative even numbers in the array "a"
b[1] gives the count of all non-negative odd numbers in the array "a"
a[i] is now the count of all non-negative even/odd numbers encountered before and including a[i].
Whether a[i] gives the occurrence of even or odd number can't be determined as the existing values are overwritten. You need to have a copy of original a[i] to determine that!
Please note it makes no changes to the "a[i]" if it were a negative number
Closed. This question needs debugging details. It is not currently accepting answers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 7 years ago.
Improve this question
Right now I'm trying to write a function that takes an array and an integer n, and gives a list of each size n combination (so a list of int arrays). I am able to write it using n nested loops, but this only works for a specific size of subset. I can't figure out how to generalize it to work for any size of combination. I think I need to use recursion?
This is the code for all combinations of 3 elements, and I need an algorithm for any number of elements.
import java.util.List;
import java.util.ArrayList;
public class combinatorics{
public static void main(String[] args) {
List<int[]> list = new ArrayList<int[]>();
int[] arr = {1,2,3,4,5};
combinations3(arr,list);
listToString(list);
}
static void combinations3(int[] arr, List<int[]> list){
for(int i = 0; i<arr.length-2; i++)
for(int j = i+1; j<arr.length-1; j++)
for(int k = j+1; k<arr.length; k++)
list.add(new int[]{arr[i],arr[j],arr[k]});
}
private static void listToString(List<int[]> list){
for(int i = 0; i<list.size(); i++){ //iterate through list
for(int j : list.get(i)){ //iterate through array
System.out.printf("%d ",j);
}
System.out.print("\n");
}
}
}
This is a well-studied problem of generating all k-subsets, or k-combinations, which can be easily done without recursion.
The idea is to have array of size k keeping sequence of indices of elements from the input array (which are numbers from 0 to n - 1) in increasing order. (Subset then can be created by taking items by these indices from the initial array.) So we need to generate all such index sequences.
First index sequence will be [0, 1, 2, ... , k - 1], on the second step it switches to [0, 1, 2,..., k], then to [0, 1, 2, ... k + 1] and so on. The last possible sequence will be [n - k, n - k + 1, ..., n - 1].
On each step, algorithm looks for the closest to the end item which can be incremented, increments it and fills up items right to that item.
To illustrate, consider n = 7 and k = 3. First index sequence is [0, 1, 2], then [0, 1, 3] and so on... At some point we have [0, 5, 6]:
[0, 5, 6] <-- scan from the end: "6" cannot be incremented, "5" also, but "0" can be
[1, ?, ?] <-- "0" -> "1"
[1, 2, 3] <-- fill up remaining elements
next iteration:
[1, 2, 3] <-- "3" can be incremented
[1, 2, 4] <-- "3" -> "4"
Thus, [0, 5, 6] is followed by [1, 2, 3], then goes [1, 2, 4] etc.
Code:
int[] input = {10, 20, 30, 40, 50}; // input array
int k = 3; // sequence length
List<int[]> subsets = new ArrayList<>();
int[] s = new int[k]; // here we'll keep indices
// pointing to elements in input array
if (k <= input.length) {
// first index sequence: 0, 1, 2, ...
for (int i = 0; (s[i] = i) < k - 1; i++);
subsets.add(getSubset(input, s));
for(;;) {
int i;
// find position of item that can be incremented
for (i = k - 1; i >= 0 && s[i] == input.length - k + i; i--);
if (i < 0) {
break;
}
s[i]++; // increment this item
for (++i; i < k; i++) { // fill up remaining items
s[i] = s[i - 1] + 1;
}
subsets.add(getSubset(input, s));
}
}
// generate actual subset by index sequence
int[] getSubset(int[] input, int[] subset) {
int[] result = new int[subset.length];
for (int i = 0; i < subset.length; i++)
result[i] = input[subset[i]];
return result;
}
If I understood your problem correctly, this article seems to point to what you're trying to do.
To quote from the article:
Method 1 (Fix Elements and Recur)
We create a temporary array ‘data[]’ which stores all outputs one by
one. The idea is to start from first index (index = 0) in data[], one
by one fix elements at this index and recur for remaining indexes. Let
the input array be {1, 2, 3, 4, 5} and r be 3. We first fix 1 at index
0 in data[], then recur for remaining indexes, then we fix 2 at index
0 and recur. Finally, we fix 3 and recur for remaining indexes. When
number of elements in data[] becomes equal to r (size of a
combination), we print data[].
Method 2 (Include and Exclude every element)
Like the above method, We create a temporary array data[]. The idea
here is similar to Subset Sum Problem. We one by one consider every
element of input array, and recur for two cases:
The element is included in current combination (We put the element in data[] and increment next available index in data[])
The element is excluded in current combination (We do not put the element and do not change index)
When number of elements in data[] become equal to r (size of a
combination), we print it.