Given an array of positive integers. How to find a subsequence of length L with max sum which has the distance between any two of its neighboring elements that do not exceed K
I have the following solution but don't know how to take into account length L.
1 <= N <= 100000, 1 <= L <= 200, 1 <= K <= N
f[i] contains max sum of the subsequence that ends in i.
for i in range(K, N)
f[i] = INT_MIN
for j in range(1, K+1)
f[i] = max(f[i], f[i-j] + a[i])
return max(f)
(edit: slightly simplified non-recursive solution)
You can do it like this, just for each iteration consider if the item should be included or excluded.
def f(maxK,K, N, L, S):
if L == 0 or not N or K == 0:
return S
#either element is included
included = f(maxK,maxK, N[1:], L-1, S + N[0] )
#or excluded
excluded = f(maxK,K-1, N[1:], L, S )
return max(included, excluded)
assert f(2,2,[10,1,1,1,1,10],3,0) == 12
assert f(3,3,[8, 3, 7, 6, 2, 1, 9, 2, 5, 4],4,0) == 30
If N is very long you can consider changing to a table version, you could also change the input to tuples and use memoization.
Since OP later included the information that N can be 100 000, we can't really use recursive solutions like this. So here is a solution that runs in O(nKL), with same memory requirement:
import numpy as np
def f(n,K,L):
t = np.zeros((len(n),L+1))
for l in range(1,L+1):
for i in range(len(n)):
t[i,l] = n[i] + max( (t[i-k,l-1] for k in range(1,K+1) if i-k >= 0), default = 0 )
return np.max(t)
assert f([10,1,1,1,1,10],2,3) == 12
assert f([8, 3, 7, 6, 2, 1, 9],3,4) == 30
Explanation of the non recursive solution. Each cell in the table t[ i, l ] expresses the value of max subsequence with exactly l elements that use the element in position i and only elements in position i or lower where elements have at most K distance between each other.
subsequences of length n (those in t[i,1] have to have only one element, n[i] )
Longer subsequences have the n[i] + a subsequence of l-1 elements that starts at most k rows earlier, we pick the one with the maximal value. By iterating this way, we ensure that this value is already calculated.
Further improvements in memory is possible by considering that you only look at most K steps back.
Here is a bottom up (ie no recursion) dynamic solution in Python. It takes memory O(l * n) and time O(l * n * k).
def max_subseq_sum(k, l, values):
# table[i][j] will be the highest value from a sequence of length j
# ending at position i
table = []
for i in range(len(values)):
# We have no sum from 0, and i from len 1.
table.append([0, values[i]])
# By length of previous subsequence
for subseq_len in range(1, l):
# We look back up to k for the best.
prev_val = None
for last_i in range(i-k, i):
# We don't look back if the sequence was not that long.
if subseq_len <= last_i+1:
# Is this better?
this_val = table[last_i][subseq_len]
if prev_val is None or prev_val < this_val:
prev_val = this_val
# Do we have a best to offer?
if prev_val is not None:
table[i].append(prev_val + values[i])
# Now we look for the best entry of length l.
best_val = None
for row in table:
# If the row has entries for 0...l will have len > l.
if l < len(row):
if best_val is None or best_val < row[l]:
best_val = row[l]
return best_val
print(max_subseq_sum(2, 3, [10, 1, 1, 1, 1, 10]))
print(max_subseq_sum(3, 4, [8, 3, 7, 6, 2, 1, 9, 2, 5, 4]))
If I wanted to be slightly clever I could make this memory O(n) pretty easily by calculating one layer at a time, throwing away the previous one. It takes a lot of cleverness to reduce running time to O(l*n*log(k)) but that is doable. (Use a priority queue for your best value in the last k. It is O(log(k)) to update it for each element but naturally grows. Every k values you throw it away and rebuild it for a O(k) cost incurred O(n/k) times for a total O(n) rebuild cost.)
And here is the clever version. Memory O(n). Time O(n*l*log(k)) worst case, and average case is O(n*l). You hit the worst case when it is sorted in ascending order.
import heapq
def max_subseq_sum(k, l, values):
count = 0
prev_best = [0 for _ in values]
# i represents how many in prev subsequences
# It ranges from 0..(l-1).
for i in range(l):
# We are building subsequences of length i+1.
# We will have no way to find one that ends
# before the i'th element at position i-1
best = [None for _ in range(i)]
# Our heap will be (-sum, index). It is a min_heap so the
# minimum element has the largest sum. We track the index
# so that we know when it is in the last k.
min_heap = [(-prev_best[i-1], i-1)]
for j in range(i, len(values)):
# Remove best elements that are more than k back.
while min_heap[0][-1] < j-k:
heapq.heappop(min_heap)
# We append this value + (best prev sum) using -(-..) = +.
best.append(values[j] - min_heap[0][0])
heapq.heappush(min_heap, (-prev_best[j], j))
# And now keep min_heap from growing too big.
if 2*k < len(min_heap):
# Filter out elements too far back.
min_heap = [_ for _ in min_heap if j - k < _[1]]
# And make into a heap again.
heapq.heapify(min_heap)
# And now finish this layer.
prev_best = best
return max(prev_best)
Extending the code for itertools.combinations shown at the docs, I built a version that includes an argument for the maximum index distance (K) between two values. It only needed an additional and indices[i] - indices[i-1] < K check in the iteration:
def combinations_with_max_dist(iterable, r, K):
# combinations('ABCD', 2) --> AB AC AD BC BD CD
# combinations(range(4), 3) --> 012 013 023 123
pool = tuple(iterable)
n = len(pool)
if r > n:
return
indices = list(range(r))
yield tuple(pool[i] for i in indices)
while True:
for i in reversed(range(r)):
if indices[i] != i + n - r and indices[i] - indices[i-1] < K:
break
else:
return
indices[i] += 1
for j in range(i+1, r):
indices[j] = indices[j-1] + 1
yield tuple(pool[i] for i in indices)
Using this you can bruteforce over all combinations with regards to K, and then find the one that has the maximum value sum:
def find_subseq(a, L, K):
return max((sum(values), values) for values in combinations_with_max_dist(a, L, K))
Results:
print(*find_subseq([10, 1, 1, 1, 1, 10], L=3, K=2))
# 12 (10, 1, 1)
print(*find_subseq([8, 3, 7, 6, 2, 1, 9, 2, 5, 4], L=4, K=3))
# 30 (8, 7, 6, 9)
Not sure about the performance if your value lists become very long though...
Algorithm
Basic idea:
Iteration on input array, choose each index as the first taken element.
Then Recursion on each first taken element, mark the index as firstIdx.
The next possible index would be in range [firstIdx + 1, firstIdx + K], both inclusive.
Loop on the range to call each index recursively, with L - 1 as the new L.
Optionally, for each pair of (firstIndex, L), cache its max sum, for reuse.
Maybe this is necessary for large input.
Constraints:
array length <= 1 << 17 // 131072
K <= 1 << 6 // 64
L <= 1 << 8 // 256
Complexity:
Time: O(n * L * K)
Since each (firstIdx , L) pair only calculated once, and that contains a iteration of K.
Space: O(n * L)
For cache, and method stack in recursive call.
Tips:
Depth of recursion is related to L, not array length.
The defined constraints are not the actual limit, it could be larger, though I didn't test how large it can be.
Basically:
Both array length and K actually could be of any size as long as there are enough memory, since they are handled via iteration.
L is handled via recursion, thus it does has a limit.
Code - in Java
SubSumLimitedDistance.java:
import java.util.HashMap;
import java.util.Map;
public class SubSumLimitedDistance {
public static final long NOT_ENOUGH_ELE = -1; // sum that indicate not enough element, should be < 0,
public static final int MAX_ARR_LEN = 1 << 17; // max length of input array,
public static final int MAX_K = 1 << 6; // max K, should not be too long, otherwise slow,
public static final int MAX_L = 1 << 8; // max L, should not be too long, otherwise stackoverflow,
/**
* Find max sum of sum array.
*
* #param arr
* #param K
* #param L
* #return max sum,
*/
public static long find(int[] arr, int K, int L) {
if (K < 1 || K > MAX_K)
throw new IllegalArgumentException("K should be between [1, " + MAX_K + "], but get: " + K);
if (L < 0 || L > MAX_L)
throw new IllegalArgumentException("L should be between [0, " + MAX_L + "], but get: " + L);
if (arr.length > MAX_ARR_LEN)
throw new IllegalArgumentException("input array length should <= " + MAX_ARR_LEN + ", but get: " + arr.length);
Map<Integer, Map<Integer, Long>> cache = new HashMap<>(); // cache,
long maxSum = NOT_ENOUGH_ELE;
for (int i = 0; i < arr.length; i++) {
long sum = findTakeFirst(arr, K, L, i, cache);
if (sum == NOT_ENOUGH_ELE) break; // not enough elements,
if (sum > maxSum) maxSum = sum; // larger found,
}
return maxSum;
}
/**
* Find max sum of sum array, with index of first taken element specified,
*
* #param arr
* #param K
* #param L
* #param firstIdx index of first taken element,
* #param cache
* #return max sum,
*/
private static long findTakeFirst(int[] arr, int K, int L, int firstIdx, Map<Integer, Map<Integer, Long>> cache) {
// System.out.printf("findTakeFirst(): K = %d, L = %d, firstIdx = %d\n", K, L, firstIdx);
if (L == 0) return 0; // done,
if (firstIdx + L > arr.length) return NOT_ENOUGH_ELE; // not enough elements,
// check cache,
Map<Integer, Long> map = cache.get(firstIdx);
Long cachedResult;
if (map != null && (cachedResult = map.get(L)) != null) {
// System.out.printf("hit cache, cached result = %d\n", cachedResult);
return cachedResult;
}
// cache not exists, calculate,
long maxRemainSum = NOT_ENOUGH_ELE;
for (int i = firstIdx + 1; i <= firstIdx + K; i++) {
long remainSum = findTakeFirst(arr, K, L - 1, i, cache);
if (remainSum == NOT_ENOUGH_ELE) break; // not enough elements,
if (remainSum > maxRemainSum) maxRemainSum = remainSum;
}
if ((map = cache.get(firstIdx)) == null) cache.put(firstIdx, map = new HashMap<>());
if (maxRemainSum == NOT_ENOUGH_ELE) { // not enough elements,
map.put(L, NOT_ENOUGH_ELE); // cache - as not enough elements,
return NOT_ENOUGH_ELE;
}
long maxSum = arr[firstIdx] + maxRemainSum; // max sum,
map.put(L, maxSum); // cache - max sum,
return maxSum;
}
}
SubSumLimitedDistanceTest.java:
(test case, via TestNG)
import org.testng.Assert;
import org.testng.annotations.BeforeClass;
import org.testng.annotations.Test;
import java.util.concurrent.ThreadLocalRandom;
public class SubSumLimitedDistanceTest {
private int[] arr;
private int K;
private int L;
private int maxSum;
private int[] arr2;
private int K2;
private int L2;
private int maxSum2;
private int[] arrMax;
private int KMax;
private int KMaxLargest;
private int LMax;
private int LMaxLargest;
#BeforeClass
private void setUp() {
// init - arr,
arr = new int[]{10, 1, 1, 1, 1, 10};
K = 2;
L = 3;
maxSum = 12;
// init - arr2,
arr2 = new int[]{8, 3, 7, 6, 2, 1, 9, 2, 5, 4};
K2 = 3;
L2 = 4;
maxSum2 = 30;
// init - arrMax,
arrMax = new int[SubSumLimitedDistance.MAX_ARR_LEN];
ThreadLocalRandom rd = ThreadLocalRandom.current();
long maxLongEle = Long.MAX_VALUE / SubSumLimitedDistance.MAX_ARR_LEN;
int maxEle = maxLongEle > Integer.MAX_VALUE ? Integer.MAX_VALUE : (int) maxLongEle;
for (int i = 0; i < arrMax.length; i++) {
arrMax[i] = rd.nextInt(maxEle);
}
KMax = 5;
LMax = 10;
KMaxLargest = SubSumLimitedDistance.MAX_K;
LMaxLargest = SubSumLimitedDistance.MAX_L;
}
#Test
public void test() {
Assert.assertEquals(SubSumLimitedDistance.find(arr, K, L), maxSum);
Assert.assertEquals(SubSumLimitedDistance.find(arr2, K2, L2), maxSum2);
}
#Test(timeOut = 6000)
public void test_veryLargeArray() {
run_printDuring(arrMax, KMax, LMax);
}
#Test(timeOut = 60000) // takes seconds,
public void test_veryLargeArrayL() {
run_printDuring(arrMax, KMax, LMaxLargest);
}
#Test(timeOut = 60000) // takes seconds,
public void test_veryLargeArrayK() {
run_printDuring(arrMax, KMaxLargest, LMax);
}
// run find once, and print during,
private void run_printDuring(int[] arr, int K, int L) {
long startTime = System.currentTimeMillis();
long sum = SubSumLimitedDistance.find(arr, K, L);
long during = System.currentTimeMillis() - startTime; // during in milliseconds,
System.out.printf("arr length = %5d, K = %3d, L = %4d, max sum = %15d, running time = %.3f seconds\n", arr.length, K, L, sum, during / 1000.0);
}
#Test
public void test_corner_notEnoughEle() {
Assert.assertEquals(SubSumLimitedDistance.find(new int[]{1}, 2, 3), SubSumLimitedDistance.NOT_ENOUGH_ELE); // not enough element,
Assert.assertEquals(SubSumLimitedDistance.find(new int[]{0}, 1, 3), SubSumLimitedDistance.NOT_ENOUGH_ELE); // not enough element,
}
#Test
public void test_corner_ZeroL() {
Assert.assertEquals(SubSumLimitedDistance.find(new int[]{1, 2, 3}, 2, 0), 0); // L = 0,
Assert.assertEquals(SubSumLimitedDistance.find(new int[]{0}, 1, 0), 0); // L = 0,
}
#Test(expectedExceptions = IllegalArgumentException.class)
public void test_invalid_K() {
// SubSumLimitedDistance.find(new int[]{1, 2, 3}, 0, 2); // K = 0,
// SubSumLimitedDistance.find(new int[]{1, 2, 3}, -1, 2); // K = -1,
SubSumLimitedDistance.find(new int[]{1, 2, 3}, SubSumLimitedDistance.MAX_K + 1, 2); // K = SubSumLimitedDistance.MAX_K+1,
}
#Test(expectedExceptions = IllegalArgumentException.class)
public void test_invalid_L() {
// SubSumLimitedDistance.find(new int[]{1, 2, 3}, 2, -1); // L = -1,
SubSumLimitedDistance.find(new int[]{1, 2, 3}, 2, SubSumLimitedDistance.MAX_L + 1); // L = SubSumLimitedDistance.MAX_L+1,
}
#Test(expectedExceptions = IllegalArgumentException.class)
public void test_invalid_tooLong() {
SubSumLimitedDistance.find(new int[SubSumLimitedDistance.MAX_ARR_LEN + 1], 2, 3); // input array too long,
}
}
Output of test case for large input:
arr length = 131072, K = 5, L = 10, max sum = 20779205738, running time = 0.303 seconds
arr length = 131072, K = 64, L = 10, max sum = 21393422854, running time = 1.917 seconds
arr length = 131072, K = 5, L = 256, max sum = 461698553839, running time = 9.474 seconds
I'm completing a google foobar challenge, but for the life of me, I can't pass the 4th test case, here is the challenge:
You need to figure out which sets of panels in any given array you can take offline to repair while still maintaining the maximum amount of power output per array, and to do THAT, you'll first need to figure out what the maximum output of each array actually is. Write a function answer(xs) that takes a list of integers representing the power output levels of each panel in an array, and returns the maximum product of some non-empty subset of those numbers. So for example, if an array contained panels with power output levels of [2, -3, 1, 0, -5], then the maximum product would be found by taking the subset:
xs[0] = 2, xs[1] = -3, xs[4] = -5, giving the product 2*(-3)*(-5) = 30. So answer([2,-3,1,0,-5]) will be "30".
Each array of solar panels contains at least 1 and no more than 50 panels, and each panel will have a power output level whose absolute value is no greater than 1000 (some panels are malfunctioning so badly that they're draining energy, but you know a trick with the panels' wave stabilizer that lets you combine two negative-output panels to produce the positive output of the multiple of their power values). The final products may be very large, so give the answer as a string representation of the number.
Here are some given test cases:
Test cases
Inputs:
(int list) xs = [2, 0, 2, 2, 0]
Output:
(string) "8"
Inputs:
(int list) xs = [-2, -3, 4, -5]
Output:
(string) "60"
And here is my code:
def product_of_values(lst):
product = 1
if len(lst) > 0:
for x in lst:
product *= x
return product
def answer(xs):
# two seperate list for positive and negative values
positive_list = [x for x in xs if x > 0]
negative_list = [x for x in xs if x < 0]
pos_product = product_of_values(pos_list)
# multiplication of an even number of negatives == positive value
if len(negative_list) % 2 == 0:
negative_product = product_of_values(negative_list)
# if length of negative_list is odd, pop value closest to zero
else:
neg_list.remove(max(neg_list))
neg_product = product_of_values(neg_list)
# If there is only one negative value in the negative_list, return 0.
if len(pos_list) < 1 and len(neg_list) <= 1:
return '0'
else:
return str(neg_product * pos_product)
am I missing something obvious?
#edi_allen I am also facing the same problem. I have written the code in java, visible test cases are passing in my compiler, but failing at foobar. How did you overcome this? Below is my code:
public static String solution(int[] xs) {
if(xs.length < 1 || xs.length > 50) {
return "0";
}
Arrays.parallelSort(xs);
for(int i = 1; i < xs.length ; i++) {
if(xs[i] < 0 && xs[i-1] < 0) {
xs[i-1] = Math.abs(xs[i-1]);
xs[i] = Math.abs(xs[i]);
}
}
BigInteger prod = null;
for(int i = 0; i < xs.length ; i++) {
if(Math.abs(xs[i]) > 1000) {
return "0";
}
else if(xs[i] <= 0) {
continue;
}
else if(prod == null){
prod = new BigInteger(String.valueOf(xs[i]));
}
else {
prod = prod.multiply(new BigInteger(String.valueOf(xs[i])));
}
}
if(prod == null) {
return "0";
}
return prod.toString();
}
Maybe its the time complexity that is problem here.
For the product try this -
from operator import mul
reduce(mul, list, 1)
I need to convert a list of ints to a string containing all the ranges in the list.
So for example, the output should be as follows:
getIntRangesFromList([1,3,7,2,11,8,9,11,12,15]) -> "1-3,7-9,11-12,15"
So the input is not sorted and there can be duplicate values. The lists range in size from one element to 4k elements. The minimum and maximum values are 1 and 4094.
This is part of a performance critical piece of code. I have been trying to optimize this, but I can't find a way to get this faster. This is my current code:
def _getIntRangesFromList(list):
if (list==[]):
return ''
list.sort()
ranges = [[list[0],list[0]]] # ranges contains the start and end values of each range found
for val in list:
r = ranges[-1]
if val==r[1]+1:
r[1] = val
elif val>r[1]+1:
ranges.append([val,val])
return ",".join(["-".join([str(y) for y in x]) if x[0]!=x[1] else str(x[0]) for x in ranges])
Any idea on how to get this faster?
This could be a task for the itertools module.
import itertools
list_num = [1, 2, 3, 7, 8, 9, 11, 12, 15]
groups = (list(x) for _, x in
itertools.groupby(list_num, lambda x, c=itertools.count(): x - next(c)))
print(', '.join('-'.join(map(str, (item[0], item[-1])[:len(item)])) for item in groups))
This will give you 1-3, 7-9, 11-12, 15.
To understand what's going on you might want to check the content of groups.
import itertools
list_num = [1, 2, 3, 7, 8, 9, 11, 12, 15]
groups = (list(x) for _, x in
itertools.groupby(list_num, lambda x, c=itertools.count(): x - next(c)))
for element in groups:
print('element={}'.format(element))
This will give you the following output.
element=[1, 2, 3]
element=[7, 8, 9]
element=[11, 12]
element=[15]
The basic idea is to have a counter running parallel to the numbers. groupby will create individual groups for numbers with the same numerical distance to the current value of the counter.
I don't know whether this is faster on your version of Python. You'll have to check this yourself. In my setting it's slower with this data set, but faster with a bigger number of elements.
The fastest one I could come up, which tests about 10% faster than your solution on my machine (according to timeit):
def _ranges(l):
if l:
l.sort()
return ''.join([(str(l[i]) + ('-' if l[i] + 1 == l[i + 1] else ','))
for i in range(0, len(l) - 1) if l[i - 1] + 2 != l[i + 1]] +
[str(l[-1])])
else: return ''
The above code assumes that the values in the list are unique. If they aren't, it's easy to fix but there's a subtle hack which will no longer work and the end result will be slightly slower.
I actually timed _ranges(u[:]) because of the sort; u is 600 randomly selected integers from range(1000) comprising 235 subsequences; 83 are singletons and 152 contain at least two numbers. If the list is sorted, quite a lot of time is saved.
def _to_range(l, start, stop, idx, result):
if idx == len(l):
result.append((start, stop))
return result
if l[idx] - stop > 1:
result.append((start, stop))
return _to_range(l, l[idx], l[idx], idx + 1, result)
return _to_range(l, start, l[idx], idx + 1, result)
def get_range(l):
if not l:
return []
return _to_range(l, start = l[0], stop = l[0], idx = 0, result = [])
l = [1, 2, 3, 7, 8, 9, 11, 12, 15]
result = get_range(l)
print(result)
>>> [(1, 3), (7, 9), (11, 12), (15, 15)]
# I think it's better to fetch the data as it is and if needed, change it
# with
print(','.join('-'.join([str(start), str(stop)]) for start, stop in result))
>>> 1-3,7-9,11-12,15-15
Unless you don't care at all about the data, then u can just append str(start) + '-' + str(stop) in _to_range function so later there will be no need to type extra '-'.join method.
I'll concentrate on the performance that is your main issue. I'll give 2 solutions:
1) If the boundaries of the integers stored is between A and B, and you can create an array of booleans(even you can choose an array of bits for expanding the range you can storage) with (B - A + 2) elements, e.g. A = 0 and B = 1 000 000, we can do this (i'll write it in C#, sorry XD). This run in O(A - B) and is a good solution if A - B is less than the number of numbers:
public string getIntRangesFromList(int[] numbers)
{
//You can change this 2 constants
const int A = 0;
const int B = 1000000;
//Create an array with all its values in false by default
//Last value always will be in false in propourse, as you can see it storage 1 value more than needed for 2nd cycle
bool[] apparitions = new bool[B - A + 2];
int minNumber = B + 1;
int maxNumber = A - 1;
int pos;
for (int i = 0; i < numbers.Length; i++)
{
pos = numbers[i] - A;
apparitions[pos] = true;
if (minNumber > pos)
{
minNumber = pos;
}
if (maxNumber < pos)
{
maxNumber = pos;
}
}
//I will mantain the concatenation simple, but you can make it faster to improve performance
string result = "";
bool isInRange = false;
bool isFirstRange = true;
int firstPosOfRange = 0; //Irrelevant what is its initial value
for (int i = minNumber; i <= maxNumber + 1; i++)
{
if (!isInRange)
{
if (apparitions[i])
{
if (!isFirstRange)
{
result += ",";
}
else
{
isFirstRange = false;
}
result += (i + A);
isInRange = true;
firstPosOfRange = i;
}
}
else
{
if (!apparitions[i])
{
if (i > firstPosOfRange + 1)
{
result += "-" + (i + A - 1);
}
isInRange = false;
}
}
}
return result;
}
2) O(N * log N)
public string getIntRangesFromList2(int[] numbers)
{
string result = "";
if (numbers.Length > 0)
{
numbers.OrderBy(x => x); //sorting and making the algorithm complexity O(N * log N)
result += numbers[0];
int countNumbersInRange = 1;
for (int i = 1; i < numbers.Length; i++)
{
if (numbers[i] != numbers[i - 1] + 1)
{
if (countNumbersInRange > 1)
{
result += "-" + numbers[i - 1];
}
result += "," + numbers[i];
countNumbersInRange = 1;
}
else
{
countNumbersInRange++;
}
}
}
return result;
}
I have been given a set S, of n integers, and have to print the size of a maximal subset S' of S where the sum of any 2 numbers in S' are not evenly divisible by k.
Input Format
The first line contains 2 space-separated integers, n and k, respectively.
The second line contains n space-separated integers describing the unique values of the set.
My Code :
import sys
n,k = raw_input().strip().split(' ')
n,k = [int(n),int(k)]
a = map(int,raw_input().strip().split(' '))
count = 0
for i in range(len(a)):
for j in range(len(a)):
if (a[i]+a[j])%k != 0:
count = count+1
print count
Input:
4 3
1 7 2 4
Expected Output:
3
My Output:
10
What am i doing wrong? Anyone?
You can solve it in O(n) time using the following approach:
L = [0]*k
for x in a:
L[x % k] += 1
res = 0
for i in range(k//2+1):
if i == 0 or k == i*2:
res += bool(L[i])
else:
res += max(L[i], L[k-i])
print(res)
Yes O(n) solution for this problem is very much possible. Like planetp rightly pointed out its pretty much the same solution I have coded in java. Added comments for better understanding.
import java.io.; import java.util.;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n=in.nextInt();
int k=in.nextInt();
int [] arr = new int[k];
Arrays.fill(arr, 0);
Map<Integer,Integer> mp=new HashMap<>();
Storing the values in a map considering there are no duplicates. You can store them in array list if there are duplicates. Only then you have different results.
for(int i=0;i
int res=0;
for(int i=0;i<=(k/2);i++)
{
if(i==0 || k==i*2)
{
if(arr[i]!=0)
res+=1;
}
If the no. is divisible by k we can have only one and if the no is exactly half of k then we can have only 1. Rational if a & b are divisble by k then a+b is also divisible by k. Similarly if c%k=k/2 then if we have more than one such no. their combination is divisible by k. Hence we restrict them to 1 value each.
else
{
int p=arr[i];
int q=arr[k-i];
if(p>=q)
res+=p;
else
res+=q;
}
This is simple figure out which is more from a list of 0 to k/2 in the list if a[x]>a[k-x] get the values which is greater. i.e. if we have k=4 and we have no. 1,3,5,7,9,13,17. Then a[1]=4 and a[3]=2 thus pick a[1] because 1,5,13,17 can be kept together.
}
System.out.println(res);
}
}
# given k, n and a as per your input.
# Will return 0 directly if n == 1
def maxsize(k, n, a):
import itertools
while n > 1:
sets = itertools.combinations(a, n)
for set_ in sets:
if all((u+v) % k for (u, v) in itertools.combinations(set_, 2)):
return n
n -= 1
return 0
Java solution
public class Solution {
static PrintStream out = System.out;
public static void main(String[] args) {
/* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
Scanner in = new Scanner (System.in);
int n = in.nextInt();
int k = in.nextInt();
int[] A = new int[n];
for(int i=0;i<n;i++){
A[i]=in.nextInt();
}
int[] R = new int[k];
for(int i=0;i<n;i++)
R[A[i] % k]+=1;
int res=0;
for(int i=0;i<k/2+1;i++){
if(i==0 || k==i*2)
res+= (R[i]!=0)?1:0;
else
res+= Math.max(R[i], R[k-i]);
}
out.println(res);
}
}
I am studying queuing theory in which I am frequently presented with the following situation.
Let x, y both be n-tuples of nonnegative integers (depicting lengths of the n queues). In addition, x and y each have distinguished queue called their "prime queue". For example,
x = [3, 6, 1, 9, 5, 2] with x' = 1
y = [6, 1, 5, 9, 5, 5] with y' = 5
(In accordance with Python terminology I am counting the queues 0-5.)
How can I implement/construct the following permutation f on {0,1,...,5} efficiently?
first set f(x') = y'. So here f(1) = 5.
then set f(i) = i for any i such that x[i] == y[i]. Clearly there is no need to consider the indices x' and y'. So here f(3) = 3 (both length 9) and f(4) = 4 (both length 5).
there are now equally sized sets of queues unpaired in x and in y. So here in x this is {0,2,5} and in y this is {0,1,2}.
rank these from from 1 to s, where s is the common size of the sets, by length with 1 == lowest rank == shortest queue and s == highest rank == longest queue. So here, s = 3, and in x rank(0) = 1, rank(2) = 3 and rank(5) = 2, and in y rank(0) = 1, rank(1) = 3, rank(2) = 2. If there is a tie, give the queue with the larger index the higher rank.
pair these s queues off by rank. So here f(0) = 0, f(2) = 1, f(5) = 2.
This should give the permutation [0, 5, 1, 3, 4, 2].
My solution consists of tracking the indices and loops over x and y multiple times, and is terribly inefficient. (Roughly looking at n >= 1,000,000 in my application.)
Any help would be most appreciated.
Since you must do the ranking, you can't get linear and will need to sort. So it looks pretty straightforward. You do 1. in O(1) and 2. in O(n) by just going over the n-tuples. At the same time, you can construct the copy of x and y with only those that are left for 3. but do not include only the value, but instead use tuple of value and its index in the original.
In your example, x-with-tuples-left would be [[3,0],[1,2],[2,5]] and y-with-tuples-left would be [[6,0],[1,1],[5,2]].
Then just sort both x-with-tuples-left and y-with-tuples-left (it will be O(n.log n)), and read the permutation from the second element of the corresponding tuples.
In your example, sorted x-with-... would be [[1,2],[2,5],[3,0]] and sorted y-with-... would be [[1,1],[5,2],[6,0]]. Now, you nicely see 5. from the second elements: f(2)=1, f(5)=2, f(0)=0.
EDIT: Including O(n+L) in Javascript:
function qperm (x, y, xprime, yprime) {
var i;
var n = x.length;
var qperm = new Array(n);
var countsx = [], countsy = []; // same as new Array()
qperm[xprime] = yprime; // doing 1.
for (i = 0; i < n; ++i) {
if (x[i] == y[i] && i != xprime && i != yprime) { // doing 2.
qperm[i] = i; }
else { // preparing for 4. below
if (i != xprime) {
if (countsx[x[i]]) countsx[x[i]]++; else countsx[x[i]] = 1; }
if (i != yprime) {
if (countsy[y[i]]) countsy[y[i]]++; else countsy[y[i]] = 1; } }
// finishing countsx and countsy
var count, sum;
for (i = 0, count = 0; i < countsx.length; ++i) {
if (countsx[i]) {
sum = count + countsx[i];
countsx[i] = count;
count = sum; }
for (i = 0, count = 0; i < countsy.length; ++i) {
if (countsy[i]) {
sum = count + countsy[i];
countsy[i] = count;
count = sum; }
var yranked = new Array(count);
for (i = 0; i < n; ++i) {
if (i != yprime && (x[i] != y[i] || i == xprime)) { // doing 4. for y
yranked[countsy[y[i]]] = y[i];
countsy[y[i]]++; } }
for (i = 0; i < n; ++i) {
if (i != xprime && (x[i] != y[i] || i == yprime)) { // doing 4. for x and 5. at the same time
// this was here but was not right: qperm[x[i]] = yranked[countsx[x[i]]];
qperm[i] = yranked[countsx[x[i]]];
// this was here but was not right: countsy[y[i]]++; } } }
countsx[x[i]]++; } }
return qperm; }
Hopefully it's correct ;-)