Note: This may involve a good deal of number theory, but the formula I found online is only an approximation, so I believe an exact solution requires some sort of iterative calculation by a computer.
My goal is to find an efficient algorithm (in terms of time complexity) to solve the following problem for large values of n:
Let R(a,b) be the amount of steps that the Euclidean algorithm takes to find the GCD of nonnegative integers a and b. That is, R(a,b) = 1 + R(b,a%b), and R(a,0) = 0. Given a natural number n, find the sum of R(a,b) for all 1 <= a,b <= n.
For example, if n = 2, then the solution is R(1,1) + R(1,2) + R(2,1) + R(2,2) = 1 + 2 + 1 + 1 = 5.
Since there are n^2 pairs corresponding to the numbers to be added together, simply computing R(a,b) for every pair can do no better than O(n^2), regardless of the efficiency of R. Thus, to improve the efficiency of the algorithm, a faster method must somehow calculate the sum of R(a,b) over many values at once. There are a few properties that I suspect might be useful:
If a = b, then R(a,b) = 1
If a < b, then R(a,b) = 1 + R(b,a)
R(a,b) = R(ka,kb) where k is some natural number
If b <= a, then R(a,b) = R(a+b,b)
If b <= a < 2b, then R(a,b) = R(2a-b,a)
Because of the first two properties, it is only necessary to find the sum of R(a,b) over pairs where a > b. I tried using this in addition to the third property in a method that computes R(a,b) only for pairs where a and b are also coprime in addition to a being greater than b. The total sum is then n plus the sum of (n / a) * ((2 * R(a,b)) + 1) over all such pairs (using integer division for n / a). This algorithm still had time complexity O(n^2), I discovered, due to Euler's totient function being roughly linear.
I don't need any specific code solution, I just need to figure out the procedure for a more efficient algorithm. But if the programming language matters, my attempts to solve this problem have used C++.
Side note: I have found that a formula has been discovered that nearly solves this problem, but it is only an approximation. Note that the formula calculates the average rather than the sum, so it would just need to be multiplied by n^2. If the formula could be expanded to reduce the error, it might work, but from what I can tell, I'm not sure if this is possible.
Using Stern-Brocot, due to symmetry, we can look at just one of the four subtrees rooted at 1/3, 2/3, 3/2 or 3/1. The time complexity is still O(n^2) but obviously performs less calculations. The version below uses the subtree rooted at 2/3 (or at least that's the one I looked at to think through :). Also note, we only care about the denominators there since the numerators are lower. Also note the code relies on rules 2 and 3 as well.
C++ code (takes about a tenth of a second for n = 10,000):
#include <iostream>
using namespace std;
long g(int n, int l, int mid, int r, int fromL, int turns){
long right = 0;
long left = 0;
if (mid + r <= n)
right = g(n, mid, mid + r, r, 1, turns + (1^fromL));
if (mid + l <= n)
left = g(n, l, mid + l, mid, 0, turns + fromL);
// Multiples
int k = n / mid;
// This subtree is rooted at 2/3
return 4 * k * turns + left + right;
}
long f(int n) {
// 1/1, 2/2, 3/3 etc.
long total = n;
// 1/2, 2/4, 3/6 etc.
if (n > 1)
total += 3 * (n >> 1);
if (n > 2)
// Technically 3 turns for 2/3 but
// we can avoid a subtraction
// per call by starting with 2. (I
// guess that means it could be
// another subtree, but I haven't
// thought it through.)
total += g(n, 2, 3, 1, 1, 2);
return total;
}
int main() {
cout << f(10000);
return 0;
}
I think this is a hard problem. We can avoid division and reduce the space usage to linear at least via the Stern--Brocot tree.
def f(n, a, b, r):
return r if a + b > n else r + f(n, a + b, b, r) + f(n, a + b, a, r + 1)
def R_sum(n):
return sum(f(n, d, d, 1) for d in range(1, n + 1))
def R(a, b):
return 1 + R(b, a % b) if b else 0
def test(n):
print(R_sum(n))
print(sum(R(a, b) for a in range(1, n + 1) for b in range(1, n + 1)))
test(100)
I had an interesting job interview experience a while back. The question started really easy:
Q1: We have a bag containing numbers 1, 2, 3, …, 100. Each number appears exactly once, so there are 100 numbers. Now one number is randomly picked out of the bag. Find the missing number.
I've heard this interview question before, of course, so I very quickly answered along the lines of:
A1: Well, the sum of the numbers 1 + 2 + 3 + … + N is (N+1)(N/2) (see Wikipedia: sum of arithmetic series). For N = 100, the sum is 5050.
Thus, if all numbers are present in the bag, the sum will be exactly 5050. Since one number is missing, the sum will be less than this, and the difference is that number. So we can find that missing number in O(N) time and O(1) space.
At this point I thought I had done well, but all of a sudden the question took an unexpected turn:
Q2: That is correct, but now how would you do this if TWO numbers are missing?
I had never seen/heard/considered this variation before, so I panicked and couldn't answer the question. The interviewer insisted on knowing my thought process, so I mentioned that perhaps we can get more information by comparing against the expected product, or perhaps doing a second pass after having gathered some information from the first pass, etc, but I really was just shooting in the dark rather than actually having a clear path to the solution.
The interviewer did try to encourage me by saying that having a second equation is indeed one way to solve the problem. At this point I was kind of upset (for not knowing the answer before hand), and asked if this is a general (read: "useful") programming technique, or if it's just a trick/gotcha answer.
The interviewer's answer surprised me: you can generalize the technique to find 3 missing numbers. In fact, you can generalize it to find k missing numbers.
Qk: If exactly k numbers are missing from the bag, how would you find it efficiently?
This was a few months ago, and I still couldn't figure out what this technique is. Obviously there's a Ω(N) time lower bound since we must scan all the numbers at least once, but the interviewer insisted that the TIME and SPACE complexity of the solving technique (minus the O(N) time input scan) is defined in k not N.
So the question here is simple:
How would you solve Q2?
How would you solve Q3?
How would you solve Qk?
Clarifications
Generally there are N numbers from 1..N, not just 1..100.
I'm not looking for the obvious set-based solution, e.g. using a bit set, encoding the presence/absence each number by the value of a designated bit, therefore using O(N) bits in additional space. We can't afford any additional space proportional to N.
I'm also not looking for the obvious sort-first approach. This and the set-based approach are worth mentioning in an interview (they are easy to implement, and depending on N, can be very practical). I'm looking for the Holy Grail solution (which may or may not be practical to implement, but has the desired asymptotic characteristics nevertheless).
So again, of course you must scan the input in O(N), but you can only capture small amount of information (defined in terms of k not N), and must then find the k missing numbers somehow.
Here's a summary of Dimitris Andreou's link.
Remember sum of i-th powers, where i=1,2,..,k. This reduces the problem to solving the system of equations
a1 + a2 + ... + ak = b1
a12 + a22 + ... + ak2 = b2
...
a1k + a2k + ... + akk = bk
Using Newton's identities, knowing bi allows to compute
c1 = a1 + a2 + ... ak
c2 = a1a2 + a1a3 + ... + ak-1ak
...
ck = a1a2 ... ak
If you expand the polynomial (x-a1)...(x-ak) the coefficients will be exactly c1, ..., ck - see Viète's formulas. Since every polynomial factors uniquely (ring of polynomials is an Euclidean domain), this means ai are uniquely determined, up to permutation.
This ends a proof that remembering powers is enough to recover the numbers. For constant k, this is a good approach.
However, when k is varying, the direct approach of computing c1,...,ck is prohibitely expensive, since e.g. ck is the product of all missing numbers, magnitude n!/(n-k)!. To overcome this, perform computations in Zq field, where q is a prime such that n <= q < 2n - it exists by Bertrand's postulate. The proof doesn't need to be changed, since the formulas still hold, and factorization of polynomials is still unique. You also need an algorithm for factorization over finite fields, for example the one by Berlekamp or Cantor-Zassenhaus.
High level pseudocode for constant k:
Compute i-th powers of given numbers
Subtract to get sums of i-th powers of unknown numbers. Call the sums bi.
Use Newton's identities to compute coefficients from bi; call them ci. Basically, c1 = b1; c2 = (c1b1 - b2)/2; see Wikipedia for exact formulas
Factor the polynomial xk-c1xk-1 + ... + ck.
The roots of the polynomial are the needed numbers a1, ..., ak.
For varying k, find a prime n <= q < 2n using e.g. Miller-Rabin, and perform the steps with all numbers reduced modulo q.
EDIT: The previous version of this answer stated that instead of Zq, where q is prime, it is possible to use a finite field of characteristic 2 (q=2^(log n)). This is not the case, since Newton's formulas require division by numbers up to k.
You will find it by reading the couple of pages of Muthukrishnan - Data Stream Algorithms: Puzzle 1: Finding Missing Numbers. It shows exactly the generalization you are looking for. Probably this is what your interviewer read and why he posed these questions.
Also see sdcvvc's directly related answer, which also includes pseudocode (hurray! no need to read those tricky math formulations :)) (thanks, great work!).
We can solve Q2 by summing both the numbers themselves, and the squares of the numbers.
We can then reduce the problem to
k1 + k2 = x
k1^2 + k2^2 = y
Where x and y are how far the sums are below the expected values.
Substituting gives us:
(x-k2)^2 + k2^2 = y
Which we can then solve to determine our missing numbers.
As #j_random_hacker pointed out, this is quite similar to Finding duplicates in O(n) time and O(1) space, and an adaptation of my answer there works here too.
Assuming that the "bag" is represented by a 1-based array A[] of size N - k, we can solve Qk in O(N) time and O(k) additional space.
First, we extend our array A[] by k elements, so that it is now of size N. This is the O(k) additional space. We then run the following pseudo-code algorithm:
for i := n - k + 1 to n
A[i] := A[1]
end for
for i := 1 to n - k
while A[A[i]] != A[i]
swap(A[i], A[A[i]])
end while
end for
for i := 1 to n
if A[i] != i then
print i
end if
end for
The first loop initialises the k extra entries to the same as the first entry in the array (this is just a convenient value that we know is already present in the array - after this step, any entries that were missing in the initial array of size N-k are still missing in the extended array).
The second loop permutes the extended array so that if element x is present at least once, then one of those entries will be at position A[x].
Note that although it has a nested loop, it still runs in O(N) time - a swap only occurs if there is an i such that A[i] != i, and each swap sets at least one element such that A[i] == i, where that wasn't true before. This means that the total number of swaps (and thus the total number of executions of the while loop body) is at most N-1.
The third loop prints those indexes of the array i that are not occupied by the value i - this means that i must have been missing.
I asked a 4-year-old to solve this problem. He sorted the numbers and then counted along. This has a space requirement of O(kitchen floor), and it works just as easy however many balls are missing.
Not sure, if it's the most efficient solution, but I would loop over all entries, and use a bitset to remember, which numbers are set, and then test for 0 bits.
I like simple solutions - and I even believe, that it might be faster than calculating the sum, or the sum of squares etc.
I haven't checked the maths, but I suspect that computing Σ(n^2) in the same pass as we compute Σ(n) would provide enough info to get two missing numbers, Do Σ(n^3) as well if there are three, and so on.
The problem with solutions based on sums of numbers is they don't take into account the cost of storing and working with numbers with large exponents... in practice, for it to work for very large n, a big numbers library would be used. We can analyse the space utilisation for these algorithms.
We can analyse the time and space complexity of sdcvvc and Dimitris Andreou's algorithms.
Storage:
l_j = ceil (log_2 (sum_{i=1}^n i^j))
l_j > log_2 n^j (assuming n >= 0, k >= 0)
l_j > j log_2 n \in \Omega(j log n)
l_j < log_2 ((sum_{i=1}^n i)^j) + 1
l_j < j log_2 (n) + j log_2 (n + 1) - j log_2 (2) + 1
l_j < j log_2 n + j + c \in O(j log n)`
So l_j \in \Theta(j log n)
Total storage used: \sum_{j=1}^k l_j \in \Theta(k^2 log n)
Space used: assuming that computing a^j takes ceil(log_2 j) time, total time:
t = k ceil(\sum_i=1^n log_2 (i)) = k ceil(log_2 (\prod_i=1^n (i)))
t > k log_2 (n^n + O(n^(n-1)))
t > k log_2 (n^n) = kn log_2 (n) \in \Omega(kn log n)
t < k log_2 (\prod_i=1^n i^i) + 1
t < kn log_2 (n) + 1 \in O(kn log n)
Total time used: \Theta(kn log n)
If this time and space is satisfactory, you can use a simple recursive
algorithm. Let b!i be the ith entry in the bag, n the number of numbers before
removals, and k the number of removals. In Haskell syntax...
let
-- O(1)
isInRange low high v = (v >= low) && (v <= high)
-- O(n - k)
countInRange low high = sum $ map (fromEnum . isInRange low high . (!)b) [1..(n-k)]
findMissing l low high krange
-- O(1) if there is nothing to find.
| krange=0 = l
-- O(1) if there is only one possibility.
| low=high = low:l
-- Otherwise total of O(knlog(n)) time
| otherwise =
let
mid = (low + high) `div` 2
klow = countInRange low mid
khigh = krange - klow
in
findMissing (findMissing low mid klow) (mid + 1) high khigh
in
findMising 1 (n - k) k
Storage used: O(k) for list, O(log(n)) for stack: O(k + log(n))
This algorithm is more intuitive, has the same time complexity, and uses less space.
A very simple solution to Q2 which I'm surprised nobody answered already. Use the method from Q1 to find the sum of the two missing numbers. Let's denote it by S, then one of the missing numbers is smaller than S/2 and the other is bigger than S/2 (duh). Sum all the numbers from 1 to S/2 and compare it to the formula's result (similarly to the method in Q1) to find the lower between the missing numbers. Subtract it from S to find the bigger missing number.
Wait a minute. As the question is stated, there are 100 numbers in the bag. No matter how big k is, the problem can be solved in constant time because you can use a set and remove numbers from the set in at most 100 - k iterations of a loop. 100 is constant. The set of remaining numbers is your answer.
If we generalise the solution to the numbers from 1 to N, nothing changes except N is not a constant, so we are in O(N - k) = O(N) time. For instance, if we use a bit set, we set the bits to 1 in O(N) time, iterate through the numbers, setting the bits to 0 as we go (O(N-k) = O(N)) and then we have the answer.
It seems to me that the interviewer was asking you how to print out the contents of the final set in O(k) time rather than O(N) time. Clearly, with a bit set, you have to iterate through all N bits to determine whether you should print the number or not. However, if you change the way the set is implemented you can print out the numbers in k iterations. This is done by putting the numbers into an object to be stored in both a hash set and a doubly linked list. When you remove an object from the hash set, you also remove it from the list. The answers will be left in the list which is now of length k.
To solve the 2 (and 3) missing numbers question, you can modify quickselect, which on average runs in O(n) and uses constant memory if partitioning is done in-place.
Partition the set with respect to a random pivot p into partitions l, which contain numbers smaller than the pivot, and r, which contain numbers greater than the pivot.
Determine which partitions the 2 missing numbers are in by comparing the pivot value to the size of each partition (p - 1 - count(l) = count of missing numbers in l and
n - count(r) - p = count of missing numbers in r)
a) If each partition is missing one number, then use the difference of sums approach to find each missing number.
(1 + 2 + ... + (p-1)) - sum(l) = missing #1 and
((p+1) + (p+2) ... + n) - sum(r) = missing #2
b) If one partition is missing both numbers and the partition is empty, then the missing numbers are either (p-1,p-2) or (p+1,p+2)
depending on which partition is missing the numbers.
If one partition is missing 2 numbers but is not empty, then recurse onto that partiton.
With only 2 missing numbers, this algorithm always discards at least one partition, so it retains O(n) average time complexity of quickselect. Similarly, with 3 missing numbers this algorithm also discards at least one partition with each pass (because as with 2 missing numbers, at most only 1 partition will contain multiple missing numbers). However, I'm not sure how much the performance decreases when more missing numbers are added.
Here's an implementation that does not use in-place partitioning, so this example does not meet the space requirement but it does illustrate the steps of the algorithm:
<?php
$list = range(1,100);
unset($list[3]);
unset($list[31]);
findMissing($list,1,100);
function findMissing($list, $min, $max) {
if(empty($list)) {
print_r(range($min, $max));
return;
}
$l = $r = [];
$pivot = array_pop($list);
foreach($list as $number) {
if($number < $pivot) {
$l[] = $number;
}
else {
$r[] = $number;
}
}
if(count($l) == $pivot - $min - 1) {
// only 1 missing number use difference of sums
print array_sum(range($min, $pivot-1)) - array_sum($l) . "\n";
}
else if(count($l) < $pivot - $min) {
// more than 1 missing number, recurse
findMissing($l, $min, $pivot-1);
}
if(count($r) == $max - $pivot - 1) {
// only 1 missing number use difference of sums
print array_sum(range($pivot + 1, $max)) - array_sum($r) . "\n";
} else if(count($r) < $max - $pivot) {
// mroe than 1 missing number recurse
findMissing($r, $pivot+1, $max);
}
}
Demo
For Q2 this is a solution that is a bit more inefficient than the others, but still has O(N) runtime and takes O(k) space.
The idea is to run the original algorithm two times. In the first one you get a total number which is missing, which gives you an upper bound of the missing numbers. Let's call this number N. You know that the missing two numbers are going to sum up to N, so the first number can only be in the interval [1, floor((N-1)/2)] while the second is going to be in [floor(N/2)+1,N-1].
Thus you loop on all numbers once again, discarding all numbers that are not included in the first interval. The ones that are, you keep track of their sum. Finally, you'll know one of the missing two numbers, and by extension the second.
I have a feeling that this method could be generalized and maybe multiple searches run in "parallel" during a single pass over the input, but I haven't yet figured out how.
Here's a solution that uses k bits of extra storage, without any clever tricks and just straightforward. Execution time O (n), extra space O (k). Just to prove that this can be solved without reading up on the solution first or being a genius:
void puzzle (int* data, int n, bool* extra, int k)
{
// data contains n distinct numbers from 1 to n + k, extra provides
// space for k extra bits.
// Rearrange the array so there are (even) even numbers at the start
// and (odd) odd numbers at the end.
int even = 0, odd = 0;
while (even + odd < n)
{
if (data [even] % 2 == 0) ++even;
else if (data [n - 1 - odd] % 2 == 1) ++odd;
else { int tmp = data [even]; data [even] = data [n - 1 - odd];
data [n - 1 - odd] = tmp; ++even; ++odd; }
}
// Erase the lowest bits of all numbers and set the extra bits to 0.
for (int i = even; i < n; ++i) data [i] -= 1;
for (int i = 0; i < k; ++i) extra [i] = false;
// Set a bit for every number that is present
for (int i = 0; i < n; ++i)
{
int tmp = data [i];
tmp -= (tmp % 2);
if (i >= even) ++tmp;
if (tmp <= n) data [tmp - 1] += 1; else extra [tmp - n - 1] = true;
}
// Print out the missing ones
for (int i = 1; i <= n; ++i)
if (data [i - 1] % 2 == 0) printf ("Number %d is missing\n", i);
for (int i = n + 1; i <= n + k; ++i)
if (! extra [i - n - 1]) printf ("Number %d is missing\n", i);
// Restore the lowest bits again.
for (int i = 0; i < n; ++i) {
if (i < even) { if (data [i] % 2 != 0) data [i] -= 1; }
else { if (data [i] % 2 == 0) data [i] += 1; }
}
}
Motivation
If you want to solve the general-case problem, and you can store and edit the array, then Caf's solution is by far the most efficient. If you can't store the array (streaming version), then sdcvvc's answer is the only type of solution currently suggested.
The solution I propose is the most efficient answer (so far on this thread) if you can store the array but can't edit it, and I got the idea from Svalorzen's solution, which solves for 1 or 2 missing items. This solution takes Θ(k*n) time and O(min(k,log(n))) and Ω(log(k)) space. It also works well with parallelism.
Concept
The idea is that if you use the original approach of comparing sums:
sum = SumOf(1,n) - SumOf(array)
... then you take the average of the missing numbers:
average = sum/n_missing_numbers
... which provides a boundary: Of the missing numbers, there's guaranteed to be at least one number less-or-equal to average, and at least one number greater than average. This means that we can split into sub problems that each scan the array [O(n)] and are only concerned with their respective sub-arrays.
Code
C-style solution (don't judge me for the global variables, I'm just trying to make the code readable for non-c folks):
#include "stdio.h"
// Example problem:
const int array [] = {0, 7, 3, 1, 5};
const int N = 8; // size of original array
const int array_size = 5;
int SumOneTo (int n)
{
return n*(n-1)/2; // non-inclusive
}
int MissingItems (const int begin, const int end, int & average)
{
// We consider only sub-array elements with values, v:
// begin <= v < end
// Initialise info about missing elements.
// First assume all are missing:
int n = end - begin;
int sum = SumOneTo(end) - SumOneTo(begin);
// Minus everything that we see (ie not missing):
for (int i = 0; i < array_size; ++i)
{
if ((begin <= array[i]) && (array[i] < end))
{
--n;
sum -= array[i];
}
}
// used by caller:
average = sum/n;
return n;
}
void Find (const int begin, const int end)
{
int average;
if (MissingItems(begin, end, average) == 1)
{
printf(" %d", average); // average(n) is same as n
return;
}
Find(begin, average + 1); // at least one missing here
Find(average + 1, end); // at least one here also
}
int main ()
{
printf("Missing items:");
Find(0, N);
printf("\n");
}
Analysis
Ignoring recursion for a moment, each function call clearly takes O(n) time and O(1) space. Note that sum can equal as much as n(n-1)/2, so requires double the amount of bits needed to store n-1. At most this means than we effectively need two extra elements worth of space, regardless of the size of the array or k, hence it's still O(1) space under the normal conventions.
It's not so obvious how many function calls there are for k missing elements, so I'll provide a visual. Your original sub-array (connected array) is the full array, which has all k missing elements in it. We'll imagine them in increasing order, where -- represent connections (part of same sub-array):
m1 -- m2 -- m3 -- m4 -- (...) -- mk-1 -- mk
The effect of the Find function is to disconnect the missing elements into different non-overlapping sub-arrays. It guarantees that there's at least one missing element in each sub-array, which means breaking exactly one connection.
What this means is that regardless of how the splits occur, it will always take k-1 Find function calls to do the work of finding the sub-arrays that have only one missing element in it.
So the time complexity is Θ((k-1 + k) * n) = Θ(k*n).
For the space complexity, if we divide proportionally each time then we get O(log(k)) space complexity, but if we only separate one at a time it gives us O(k).
See here for a proof as to why the space complexity is O(log(n)). Given that above we've shown that it's also O(k), then we know that it's O(min(k,log(n))).
May be this algorithm can work for question 1:
Precompute xor of first 100 integers(val=1^2^3^4....100)
xor the elements as they keep coming from input stream ( val1=val1^next_input)
final answer=val^val1
Or even better:
def GetValue(A)
val=0
for i=1 to 100
do
val=val^i
done
for value in A:
do
val=val^value
done
return val
This algorithm can in fact be expanded for two missing numbers. The first step remains the same. When we call GetValue with two missing numbers the result will be a a1^a2 are the two missing numbers. Lets say
val = a1^a2
Now to sieve out a1 and a2 from val we take any set bit in val. Lets say the ith bit is set in val. That means that a1 and a2 have different parity at ith bit position.
Now we do another iteration on the original array and keep two xor values. One for the numbers which have the ith bit set and other which doesn't have the ith bit set. We now have two buckets of numbers, and its guranteed that a1 and a2 will lie in different buckets. Now repeat the same what we did for finding one missing element on each of the bucket.
There is a general way to solve streaming problems like this.
The idea is to use a bit of randomization to hopefully 'spread' the k elements into independent sub problems, where our original algorithm solves the problem for us. This technique is used in sparse signal reconstruction, among other things.
Make an array, a, of size u = k^2.
Pick any universal hash function, h : {1,...,n} -> {1,...,u}. (Like multiply-shift)
For each i in 1, ..., n increase a[h(i)] += i
For each number x in the input stream, decrement a[h(x)] -= x.
If all of the missing numbers have been hashed to different buckets, the non-zero elements of the array will now contain the missing numbers.
The probability that a particular pair is sent to the same bucket, is less than 1/u by definition of a universal hash function. Since there are about k^2/2 pairs, we have that the error probability is at most k^2/2/u=1/2. That is, we succeed with probability at least 50%, and if we increase u we increase our chances.
Notice that this algorithm takes k^2 logn bits of space (We need logn bits per array bucket.) This matches the space required by #Dimitris Andreou's answer (In particular the space requirement of polynomial factorization, which happens to also be randomized.)
This algorithm also has constant time per update, rather than time k in the case of power-sums.
In fact, we can be even more efficient than the power sum method by using the trick described in the comments.
Can you check if every number exists? If yes you may try this:
S = sum of all numbers in the bag (S < 5050)
Z = sum of the missing numbers 5050 - S
if the missing numbers are x and y then:
x = Z - y and
max(x) = Z - 1
So you check the range from 1 to max(x) and find the number
You can solve Q2 if you have the sum of both lists and the product of both lists.
(l1 is the original, l2 is the modified list)
d = sum(l1) - sum(l2)
m = mul(l1) / mul(l2)
We can optimise this since the sum of an arithmetic series is n times the average of the first and last terms:
n = len(l1)
d = (n/2)*(n+1) - sum(l2)
Now we know that (if a and b are the removed numbers):
a + b = d
a * b = m
So we can rearrange to:
a = s - b
b * (s - b) = m
And multiply out:
-b^2 + s*b = m
And rearrange so the right side is zero:
-b^2 + s*b - m = 0
Then we can solve with the quadratic formula:
b = (-s + sqrt(s^2 - (4*-1*-m)))/-2
a = s - b
Sample Python 3 code:
from functools import reduce
import operator
import math
x = list(range(1,21))
sx = (len(x)/2)*(len(x)+1)
x.remove(15)
x.remove(5)
mul = lambda l: reduce(operator.mul,l)
s = sx - sum(x)
m = mul(range(1,21)) / mul(x)
b = (-s + math.sqrt(s**2 - (-4*(-m))))/-2
a = s - b
print(a,b) #15,5
I do not know the complexity of the sqrt, reduce and sum functions so I cannot work out the complexity of this solution (if anyone does know please comment below.)
Here is a solution that doesn't rely on complex math as sdcvvc's/Dimitris Andreou's answers do, doesn't change the input array as caf and Colonel Panic did, and doesn't use the bitset of enormous size as Chris Lercher, JeremyP and many others did. Basically, I began with Svalorzen's/Gilad Deutch's idea for Q2, generalized it to the common case Qk and implemented in Java to prove that the algorithm works.
The idea
Suppose we have an arbitrary interval I of which we only know that it contains at least one of the missing numbers. After one pass through the input array, looking only at the numbers from I, we can obtain both the sum S and the quantity Q of missing numbers from I. We do this by simply decrementing I's length each time we encounter a number from I (for obtaining Q) and by decreasing pre-calculated sum of all numbers in I by that encountered number each time (for obtaining S).
Now we look at S and Q. If Q = 1, it means that then I contains only one of the missing numbers, and this number is clearly S. We mark I as finished (it is called "unambiguous" in the program) and leave it out from further consideration. On the other hand, if Q > 1, we can calculate the average A = S / Q of missing numbers contained in I. As all numbers are distinct, at least one of such numbers is strictly less than A and at least one is strictly greater than A. Now we split I in A into two smaller intervals each of which contains at least one missing number. Note that it doesn't matter to which of the intervals we assign A in case it is an integer.
We make the next array pass calculating S and Q for each of the intervals separately (but in the same pass) and after that mark intervals with Q = 1 and split intervals with Q > 1. We continue this process until there are no new "ambiguous" intervals, i.e. we have nothing to split because each interval contains exactly one missing number (and we always know this number because we know S). We start out from the sole "whole range" interval containing all possible numbers (like [1..N] in the question).
Time and space complexity analysis
The total number of passes p we need to make until the process stops is never greater than the missing numbers count k. The inequality p <= k can be proved rigorously. On the other hand, there is also an empirical upper bound p < log2N + 3 that is useful for large values of k. We need to make a binary search for each number of the input array to determine the interval to which it belongs. This adds the log k multiplier to the time complexity.
In total, the time complexity is O(N ᛫ min(k, log N) ᛫ log k). Note that for large k, this is significantly better than that of sdcvvc/Dimitris Andreou's method, which is O(N ᛫ k).
For its work, the algorithm requires O(k) additional space for storing at most k intervals, that is significantly better than O(N) in "bitset" solutions.
Java implementation
Here's a Java class that implements the above algorithm. It always returns a sorted array of missing numbers. Besides that, it doesn't require the missing numbers count k because it calculates it in the first pass. The whole range of numbers is given by the minNumber and maxNumber parameters (e.g. 1 and 100 for the first example in the question).
public class MissingNumbers {
private static class Interval {
boolean ambiguous = true;
final int begin;
int quantity;
long sum;
Interval(int begin, int end) { // begin inclusive, end exclusive
this.begin = begin;
quantity = end - begin;
sum = quantity * ((long)end - 1 + begin) / 2;
}
void exclude(int x) {
quantity--;
sum -= x;
}
}
public static int[] find(int minNumber, int maxNumber, NumberBag inputBag) {
Interval full = new Interval(minNumber, ++maxNumber);
for (inputBag.startOver(); inputBag.hasNext();)
full.exclude(inputBag.next());
int missingCount = full.quantity;
if (missingCount == 0)
return new int[0];
Interval[] intervals = new Interval[missingCount];
intervals[0] = full;
int[] dividers = new int[missingCount];
dividers[0] = minNumber;
int intervalCount = 1;
while (true) {
int oldCount = intervalCount;
for (int i = 0; i < oldCount; i++) {
Interval itv = intervals[i];
if (itv.ambiguous)
if (itv.quantity == 1) // number inside itv uniquely identified
itv.ambiguous = false;
else
intervalCount++; // itv will be split into two intervals
}
if (oldCount == intervalCount)
break;
int newIndex = intervalCount - 1;
int end = maxNumber;
for (int oldIndex = oldCount - 1; oldIndex >= 0; oldIndex--) {
// newIndex always >= oldIndex
Interval itv = intervals[oldIndex];
int begin = itv.begin;
if (itv.ambiguous) {
// split interval itv
// use floorDiv instead of / because input numbers can be negative
int mean = (int)Math.floorDiv(itv.sum, itv.quantity) + 1;
intervals[newIndex--] = new Interval(mean, end);
intervals[newIndex--] = new Interval(begin, mean);
} else
intervals[newIndex--] = itv;
end = begin;
}
for (int i = 0; i < intervalCount; i++)
dividers[i] = intervals[i].begin;
for (inputBag.startOver(); inputBag.hasNext();) {
int x = inputBag.next();
// find the interval to which x belongs
int i = java.util.Arrays.binarySearch(dividers, 0, intervalCount, x);
if (i < 0)
i = -i - 2;
Interval itv = intervals[i];
if (itv.ambiguous)
itv.exclude(x);
}
}
assert intervalCount == missingCount;
for (int i = 0; i < intervalCount; i++)
dividers[i] = (int)intervals[i].sum;
return dividers;
}
}
For fairness, this class receives input in form of NumberBag objects. NumberBag doesn't allow array modification and random access and also counts how many times the array was requested for sequential traversing. It is also more suitable for large array testing than Iterable<Integer> because it avoids boxing of primitive int values and allows wrapping a part of a large int[] for a convenient test preparation. It is not hard to replace, if desired, NumberBag by int[] or Iterable<Integer> type in the find signature, by changing two for-loops in it into foreach ones.
import java.util.*;
public abstract class NumberBag {
private int passCount;
public void startOver() {
passCount++;
}
public final int getPassCount() {
return passCount;
}
public abstract boolean hasNext();
public abstract int next();
// A lightweight version of Iterable<Integer> to avoid boxing of int
public static NumberBag fromArray(int[] base, int fromIndex, int toIndex) {
return new NumberBag() {
int index = toIndex;
public void startOver() {
super.startOver();
index = fromIndex;
}
public boolean hasNext() {
return index < toIndex;
}
public int next() {
if (index >= toIndex)
throw new NoSuchElementException();
return base[index++];
}
};
}
public static NumberBag fromArray(int[] base) {
return fromArray(base, 0, base.length);
}
public static NumberBag fromIterable(Iterable<Integer> base) {
return new NumberBag() {
Iterator<Integer> it;
public void startOver() {
super.startOver();
it = base.iterator();
}
public boolean hasNext() {
return it.hasNext();
}
public int next() {
return it.next();
}
};
}
}
Tests
Simple examples demonstrating the usage of these classes are given below.
import java.util.*;
public class SimpleTest {
public static void main(String[] args) {
int[] input = { 7, 1, 4, 9, 6, 2 };
NumberBag bag = NumberBag.fromArray(input);
int[] output = MissingNumbers.find(1, 10, bag);
System.out.format("Input: %s%nMissing numbers: %s%nPass count: %d%n",
Arrays.toString(input), Arrays.toString(output), bag.getPassCount());
List<Integer> inputList = new ArrayList<>();
for (int i = 0; i < 10; i++)
inputList.add(2 * i);
Collections.shuffle(inputList);
bag = NumberBag.fromIterable(inputList);
output = MissingNumbers.find(0, 19, bag);
System.out.format("%nInput: %s%nMissing numbers: %s%nPass count: %d%n",
inputList, Arrays.toString(output), bag.getPassCount());
// Sieve of Eratosthenes
final int MAXN = 1_000;
List<Integer> nonPrimes = new ArrayList<>();
nonPrimes.add(1);
int[] primes;
int lastPrimeIndex = 0;
while (true) {
primes = MissingNumbers.find(1, MAXN, NumberBag.fromIterable(nonPrimes));
int p = primes[lastPrimeIndex]; // guaranteed to be prime
int q = p;
for (int i = lastPrimeIndex++; i < primes.length; i++) {
q = primes[i]; // not necessarily prime
int pq = p * q;
if (pq > MAXN)
break;
nonPrimes.add(pq);
}
if (q == p)
break;
}
System.out.format("%nSieve of Eratosthenes. %d primes up to %d found:%n",
primes.length, MAXN);
for (int i = 0; i < primes.length; i++)
System.out.format(" %4d%s", primes[i], (i % 10) < 9 ? "" : "\n");
}
}
Large array testing can be performed this way:
import java.util.*;
public class BatchTest {
private static final Random rand = new Random();
public static int MIN_NUMBER = 1;
private final int minNumber = MIN_NUMBER;
private final int numberCount;
private final int[] numbers;
private int missingCount;
public long finderTime;
public BatchTest(int numberCount) {
this.numberCount = numberCount;
numbers = new int[numberCount];
for (int i = 0; i < numberCount; i++)
numbers[i] = minNumber + i;
}
private int passBound() {
int mBound = missingCount > 0 ? missingCount : 1;
int nBound = 34 - Integer.numberOfLeadingZeros(numberCount - 1); // ceil(log_2(numberCount)) + 2
return Math.min(mBound, nBound);
}
private void error(String cause) {
throw new RuntimeException("Error on '" + missingCount + " from " + numberCount + "' test, " + cause);
}
// returns the number of times the input array was traversed in this test
public int makeTest(int missingCount) {
this.missingCount = missingCount;
// numbers array is reused when numberCount stays the same,
// just Fisher–Yates shuffle it for each test
for (int i = numberCount - 1; i > 0; i--) {
int j = rand.nextInt(i + 1);
if (i != j) {
int t = numbers[i];
numbers[i] = numbers[j];
numbers[j] = t;
}
}
final int bagSize = numberCount - missingCount;
NumberBag inputBag = NumberBag.fromArray(numbers, 0, bagSize);
finderTime -= System.nanoTime();
int[] found = MissingNumbers.find(minNumber, minNumber + numberCount - 1, inputBag);
finderTime += System.nanoTime();
if (inputBag.getPassCount() > passBound())
error("too many passes (" + inputBag.getPassCount() + " while only " + passBound() + " allowed)");
if (found.length != missingCount)
error("wrong result length");
int j = bagSize; // "missing" part beginning in numbers
Arrays.sort(numbers, bagSize, numberCount);
for (int i = 0; i < missingCount; i++)
if (found[i] != numbers[j++])
error("wrong result array, " + i + "-th element differs");
return inputBag.getPassCount();
}
public static void strideCheck(int numberCount, int minMissing, int maxMissing, int step, int repeats) {
BatchTest t = new BatchTest(numberCount);
System.out.println("╠═══════════════════════╬═════════════════╬═════════════════╣");
for (int missingCount = minMissing; missingCount <= maxMissing; missingCount += step) {
int minPass = Integer.MAX_VALUE;
int passSum = 0;
int maxPass = 0;
t.finderTime = 0;
for (int j = 1; j <= repeats; j++) {
int pCount = t.makeTest(missingCount);
if (pCount < minPass)
minPass = pCount;
passSum += pCount;
if (pCount > maxPass)
maxPass = pCount;
}
System.out.format("║ %9d %9d ║ %2d %5.2f %2d ║ %11.3f ║%n", missingCount, numberCount, minPass,
(double)passSum / repeats, maxPass, t.finderTime * 1e-6 / repeats);
}
}
public static void main(String[] args) {
System.out.println("╔═══════════════════════╦═════════════════╦═════════════════╗");
System.out.println("║ Number count ║ Passes ║ Average time ║");
System.out.println("║ missimg total ║ min avg max ║ per search (ms) ║");
long time = System.nanoTime();
strideCheck(100, 0, 100, 1, 20_000);
strideCheck(100_000, 2, 99_998, 1_282, 15);
MIN_NUMBER = -2_000_000_000;
strideCheck(300_000_000, 1, 10, 1, 1);
time = System.nanoTime() - time;
System.out.println("╚═══════════════════════╩═════════════════╩═════════════════╝");
System.out.format("%nSuccess. Total time: %.2f s.%n", time * 1e-9);
}
}
Try them out on Ideone
I think this can be done without any complex mathematical equations and theories. Below is a proposal for an in place and O(2n) time complexity solution:
Input form assumptions :
# of numbers in bag = n
# of missing numbers = k
The numbers in the bag are represented by an array of length n
Length of input array for the algo = n
Missing entries in the array (numbers taken out of the bag) are replaced by the value of the first element in the array.
Eg. Initially bag looks like [2,9,3,7,8,6,4,5,1,10].
If 4 is taken out, value of 4 will become 2 (the first element of the array).
Therefore after taking 4 out the bag will look like [2,9,3,7,8,6,2,5,1,10]
The key to this solution is to tag the INDEX of a visited number by negating the value at that INDEX as the array is traversed.
IEnumerable<int> GetMissingNumbers(int[] arrayOfNumbers)
{
List<int> missingNumbers = new List<int>();
int arrayLength = arrayOfNumbers.Length;
//First Pass
for (int i = 0; i < arrayLength; i++)
{
int index = Math.Abs(arrayOfNumbers[i]) - 1;
if (index > -1)
{
arrayOfNumbers[index] = Math.Abs(arrayOfNumbers[index]) * -1; //Marking the visited indexes
}
}
//Second Pass to get missing numbers
for (int i = 0; i < arrayLength; i++)
{
//If this index is unvisited, means this is a missing number
if (arrayOfNumbers[i] > 0)
{
missingNumbers.Add(i + 1);
}
}
return missingNumbers;
}
Thanks for this very interesting question:
It's because you reminded me Newton's work which really can solve this problem
Please refer Newton's Identities
As number of variables to find = number of equations (must for consistency)
I believe for this we should raise power to bag numbers so as to create number of different equations.
I don't know but, I believe if there should a function say f for which we'll add f( xi )
x1 + x2 + ... + xk = z1
x12 + x22 + ... + xk2 = z2
............
............
............
x1k + x2k + ... + xkk = zk
rest is a mathematical work not sure about time and space complexity but Newton's Identities will surely play important role.
Can't we use set theory
.difference_update() or Is there any chance of Linear Algebra in this question method?
You'd probably need clarification on what O(k) means.
Here's a trivial solution for arbitrary k: for each v in your set of numbers, accumulate the sum of 2^v. At the end, loop i from 1 to N. If sum bitwise ANDed with 2^i is zero, then i is missing. (Or numerically, if floor of the sum divided by 2^i is even. Or sum modulo 2^(i+1)) < 2^i.)
Easy, right? O(N) time, O(1) storage, and it supports arbitrary k.
Except that you're computing enormous numbers that on a real computer would each require O(N) space. In fact, this solution is identical to a bit vector.
So you could be clever and compute the sum and the sum of squares and the sum of cubes... up to the sum of v^k, and do the fancy math to extract the result. But those are big numbers too, which begs the question: what abstract model of operation are we talking about? How much fits in O(1) space, and how long does it take to sum up numbers of whatever size you need?
I have read all thirty answers and found the simplest one i.e to use a bit array of 100 to be the best. But as the question said we can't use an array of size N, I would use O(1) space complexity and k iterations i.e O(NK) time complexity to solve this.
To make the explanation simpler, consider I have been given numbers from 1 to 15 and two of them are missing i.e 9 and 14 but I don't know. Let the bag look like this:
[8,1,2,12,4,7,5,10,11,13,15,3,6].
We know that each number is represented internally in the form of bits.
For numbers till 16 we only need 4 bits. For numbers till 10^9, we will need 32 bits. But let's focus on 4 bits and then later we can generalize it.
Now, assume if we had all the numbers from 1 to 15, then internally, we would have numbers like this (if we had them ordered):
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
But now we have two numbers missing. So our representation will look something like this (shown ordered for understanding but can be in any order):
(2MSD|2LSD)
00|01
00|10
00|11
-----
01|00
01|01
01|10
01|11
-----
10|00
missing=(10|01)
10|10
10|11
-----
11|00
11|01
missing=(11|10)
11|11
Now let's make a bit array of size 2 that holds the count of numbers with corresponding 2 most significant digits. i.e
= [__,__,__,__]
00,01,10,11
Scan the bag from left and right and fill the above array such that each of bin of bit array contains the count of numbers. The result will be as under:
= [ 3, 4, 3, 3]
00,01,10,11
If all the numbers would have been present, it would have looked like this:
= [ 3, 4, 4, 4]
00,01,10,11
Thus we know that there are two numbers missing: one whose most 2 significant digits are 10 and one whose most 2 significant bits are 11. Now scan the list again and fill out a bit array of size 2 for the lower 2 significant digits. This time, only consider elements whose most 2 significant digits are 10. We will have the bit array as:
= [ 1, 0, 1, 1]
00,01,10,11
If all numbers of MSD=10 were present, we would have 1 in all the bins but now we see that one is missing. Thus we have the number whose MSD=10 and LSD=01 is missing which is 1001 i.e 9.
Similarly, if we scan again but consider only elements whose MSD=11,we get MSD=11 and LSD=10 missing which is 1110 i.e 14.
= [ 1, 0, 1, 1]
00,01,10,11
Thus, we can find the missing numbers in a constant amount of space. We can generalize this for 100, 1000 or 10^9 or any set of numbers.
References: Problem 1.6 in http://users.ece.utexas.edu/~adnan/afi-samples-new.pdf
Very nice problem. I'd go for using a set difference for Qk. A lot of programming languages even have support for it, like in Ruby:
missing = (1..100).to_a - bag
It's probably not the most efficient solution but it's one I would use in real life if I was faced with such a task in this case (known boundaries, low boundaries). If the set of number would be very large then I would consider a more efficient algorithm, of course, but until then the simple solution would be enough for me.
You could try using a Bloom Filter. Insert each number in the bag into the bloom, then iterate over the complete 1-k set until reporting each one not found. This may not find the answer in all scenarios, but might be a good enough solution.
I'd take a different approach to that question and probe the interviewer for more details about the larger problem he's trying to solve. Depending on the problem and the requirements surrounding it, the obvious set-based solution might be the right thing and the generate-a-list-and-pick-through-it-afterward approach might not.
For example, it might be that the interviewer is going to dispatch n messages and needs to know the k that didn't result in a reply and needs to know it in as little wall clock time as possible after the n-kth reply arrives. Let's also say that the message channel's nature is such that even running at full bore, there's enough time to do some processing between messages without having any impact on how long it takes to produce the end result after the last reply arrives. That time can be put to use inserting some identifying facet of each sent message into a set and deleting it as each corresponding reply arrives. Once the last reply has arrived, the only thing to be done is to remove its identifier from the set, which in typical implementations takes O(log k+1). After that, the set contains the list of k missing elements and there's no additional processing to be done.
This certainly isn't the fastest approach for batch processing pre-generated bags of numbers because the whole thing runs O((log 1 + log 2 + ... + log n) + (log n + log n-1 + ... + log k)). But it does work for any value of k (even if it's not known ahead of time) and in the example above it was applied in a way that minimizes the most critical interval.
This might sound stupid, but, in the first problem presented to you, you would have to see all the remaining numbers in the bag to actually add them up to find the missing number using that equation.
So, since you get to see all the numbers, just look for the number that's missing. The same goes for when two numbers are missing. Pretty simple I think. No point in using an equation when you get to see the numbers remaining in the bag.
You can motivate the solution by thinking about it in terms of symmetries (groups, in math language). No matter the order of the set of numbers, the answer should be the same. If you're going to use k functions to help determine the missing elements, you should be thinking about what functions have that property: symmetric. The function s_1(x) = x_1 + x_2 + ... + x_n is an example of a symmetric function, but there are others of higher degree. In particular, consider the elementary symmetric functions. The elementary symmetric function of degree 2 is s_2(x) = x_1 x_2 + x_1 x_3 + ... + x_1 x_n + x_2 x_3 + ... + x_(n-1) x_n, the sum of all products of two elements. Similarly for the elementary symmetric functions of degree 3 and higher. They are obviously symmetric. Furthermore, it turns out they are the building blocks for all symmetric functions.
You can build the elementary symmetric functions as you go by noting that s_2(x,x_(n+1)) = s_2(x) + s_1(x)(x_(n+1)). Further thought should convince you that s_3(x,x_(n+1)) = s_3(x) + s_2(x)(x_(n+1)) and so on, so they can be computed in one pass.
How do we tell which items were missing from the array? Think about the polynomial (z-x_1)(z-x_2)...(z-x_n). It evaluates to 0 if you put in any of the numbers x_i. Expanding the polynomial, you get z^n-s_1(x)z^(n-1)+ ... + (-1)^n s_n. The elementary symmetric functions appear here too, which is really no surprise, since the polynomial should stay the same if we apply any permutation to the roots.
So we can build the polynomial and try to factor it to figure out which numbers are not in the set, as others have mentioned.
Finally, if we are concerned about overflowing memory with large numbers (the nth symmetric polynomial will be of the order 100!), we can do these calculations mod p where p is a prime bigger than 100. In that case we evaluate the polynomial mod p and find that it again evaluates to 0 when the input is a number in the set, and it evaluates to a non-zero value when the input is a number not in the set. However, as others have pointed out, to get the values out of the polynomial in time that depends on k, not N, we have to factor the polynomial mod p.
I believe I have a O(k) time and O(log(k)) space algorithm, given that you have the floor(x) and log2(x) functions for arbitrarily big integers available:
You have an k-bit long integer (hence the log8(k) space) where you add the x^2, where x is the next number you find in the bag: s=1^2+2^2+... This takes O(N) time (which is not a problem for the interviewer). At the end you get j=floor(log2(s)) which is the biggest number you're looking for. Then s=s-j and you do again the above:
for (i = 0 ; i < k ; i++)
{
j = floor(log2(s));
missing[i] = j;
s -= j;
}
Now, you usually don't have floor and log2 functions for 2756-bit integers but instead for doubles. So? Simply, for each 2 bytes (or 1, or 3, or 4) you can use these functions to get the desired numbers, but this adds an O(N) factor to time complexity
Try to find the product of numbers from 1 to 50:
Let product, P1 = 1 x 2 x 3 x ............. 50
When you take out numbers one by one, multiply them so that you get the product P2. But two numbers are missing here, hence P2 < P1.
The product of the two mising terms, a x b = P1 - P2.
You already know the sum, a + b = S1.
From the above two equations, solve for a and b through a quadratic equation. a and b are your missing numbers.