I believe that the runtime of func2 is O(n * log(n)).
But some people have told me that it's not.
int func2(int* arr, int n){
int i, j;
for (i = 1; i <= n; i *= 2)
reverseArray(arr, i);
}
}
void reverseArray(int* arr, int n){
int left, right, temp;
for (left = 0, right = n-1; left <= right; left++, right--){
temp = arr[left];
arr[left] = arr[right];
arr[left] = temp;
}
}
func2 runs in linear time, i.e., O(n)
Explanation:
It's easy to say time complexity of reverseArray method is linear, i.e., big-Theta(n)
let's assume func2 is called with n = 8;
The calls to reverseArray will be:-
reverseArray(arr, 1)
reverseArray(arr, 2)
reverseArray(arr, 4)
reverseArray(arr, 8)
So total runtime = 1 + 2 + 4 + 8 = 15, i.e. 2*8 - 1
So, if n was a power of 2, total run time = O(2*n - 1)
If n was not a power of 2, total run time would be = O(2*K - 1), where K is the highest power of 2 less than n. We can safely say, O(2*K - 1) = O(2*n - 1) [since, O is upperbound]
O(2*n - 1) = O(n)
For theta notation, lower bound is O(2*K - 1), where K is the highest power of 2 less than n.
Therefore, time complexity = Theta(2^(floor(log(n)base2)+1))
so I was wondering how would I work out the time complexity (T(n)) of a piece of code, for example, the one below, in terms of the number of operations.
for( int i = n; i > 0; i /= 2 ) {
for( int j = 1; j < n; j *= 2 ) {
for( int k = 0; k < n; k += 2 ) {
... // constant number of operations
}
}
}
I'm sure its simple but this concept wasn't taught very well by my lecturer and I really want to know how to work out the time complexity!
Thank you in advance!
To approach this, one method is to breakdown the complexity of your three loops individually.
A key observation we can make is that:
(P): The number of steps in each of the loop does not depend on the value of the "index" of its parent loop.
Let's call
f(n) the number of operations aggregated in the outer loop (1)
g(n) in the intermediate inner loop (2)
h(n) in the most inner loop (3).
for( int i = n; i > 0; i /= 2 ) { // (1): f(n)
for( int j = 1; j < n; j *= 2 ) { // (2): g(n)
for( int k = 0; k < n; k += 2 ) { // (3): h(n)
// constant number of operations // => (P)
}
}
}
Loop (1)
Number of steps
i gets the values n, n/2, n/4, ... etc. until it reaches n/2^k where 2^k is greater than n (2^k > n), such that n/2^k = 0, at which point you exit the loop.
Another way to say it is that you have step 1 (i = n), step 2 (i = n/2), step 3 (i = n/4), ... step k - 1 (i = n/2^(k-1)), then you exit the loop. These are k steps.
Now what is the value of k? Observe that n - 1 <= 2^k < n <=> log2(n - 1) <= k < log2(n) <= INT(log2(n - 1)) <= k <= INT(log2(n)). This makes k = INT(log2(n)) or loosely speaking k = log2(n).
Cost of each step
Now how many operations do you have for each individual step?
At step i, it is g(i) = g(n) according to the notations we chose and the property (P).
Loop (2)
Number of steps
You have step (1) (j = 1), step (2) (j = 2), step (3) (j = 4), etc. until you reach step (p) (j = 2^p) where p is defined as the smallest integer such that 2^p > n, or loosely speaking log2(n).
Cost of each step
The cost of step j is h(j) = h(n) according to the notations we chose and the property (P).
Loop (3)
Number of steps
Again, let's count the steps: (1):k = 0, (1):k = 2, (2):k = 4, ..., k = n - 1 or k = n - 2. This amounts to n / 2 steps.
Cost of each step
Because of (P), it is constant. Let's call this constant K.
All loops altogether
The number of aggregated operations is
T(n) = f(n) = sum(i = 0, i < log2(n), g(i))
= sum(i = 0, i < log2(n), g(n))
= log2(n).g(n)
= log2(n).sum(j = 0, j < log2(n), h(j))
= log2(n).log2(n).h(n)
= log2(n).log2(n).(n/2).K
So T(n) = (K/2).(log2(n))^2.n
Write a method, add a counter, return the result:
int nIterator (int n) {
int counter = 0;
for( int i = n; i > 0; i /= 2 ) {
for( int j = 1; j < n; j *= 2 ) {
for( int k = 0; k < n; k += 2 ) {
++counter;
}
}
}
return counter;
}
Protocol for fast increasing N and document in a readable manner the results:
int old = 0;
for (int i = 0, j=1; i < 18; ++i, j*=2) {
int res = nIterator (j);
double quote = (old == 0) ? 0.0 : (res*1.0)/old;
System.out.printf ("%6d %10d %3f\n", j, res, quote);
old=res;
}
Result:
1 0 0,000000
2 2 0,000000
4 12 6,000000
8 48 4,000000
16 160 3,333333
32 480 3,000000
64 1344 2,800000
128 3584 2,666667
256 9216 2,571429
512 23040 2,500000
1024 56320 2,444444
2048 135168 2,400000
4096 319488 2,363636
8192 745472 2,333333
16384 1720320 2,307692
32768 3932160 2,285714
65536 8912896 2,266667
131072 20054016 2,250000
So n is increasing by factor 2, the counter increases in the beginning with more than 2², but then decreases rapidly towards something, not much higher than 2. This should help you find the way.
I've just done the following Codility Peaks problem. The problem is as follows:
A non-empty zero-indexed array A consisting of N integers is given.
A peak is an array element which is larger than its neighbors. More precisely, it is an index P such that 0 < P < N − 1, A[P − 1] < A[P] and A[P] > A[P + 1].
For example, the following array A:
A[0] = 1
A[1] = 2
A[2] = 3
A[3] = 4
A[4] = 3
A[5] = 4
A[6] = 1
A[7] = 2
A[8] = 3
A[9] = 4
A[10] = 6
A[11] = 2
has exactly three peaks: 3, 5, 10.
We want to divide this array into blocks containing the same number of elements. More precisely, we want to choose a number K that will yield the following blocks:
A[0], A[1], ..., A[K − 1],
A[K], A[K + 1], ..., A[2K − 1],
...
A[N − K], A[N − K + 1], ..., A[N − 1].
What's more, every block should contain at least one peak. Notice that extreme elements of the blocks (for example A[K − 1] or A[K]) can also be peaks, but only if they have both neighbors (including one in an adjacent blocks).
The goal is to find the maximum number of blocks into which the array A can be divided.
Array A can be divided into blocks as follows:
one block (1, 2, 3, 4, 3, 4, 1, 2, 3, 4, 6, 2). This block contains three peaks.
two blocks (1, 2, 3, 4, 3, 4) and (1, 2, 3, 4, 6, 2). Every block has a peak.
three blocks (1, 2, 3, 4), (3, 4, 1, 2), (3, 4, 6, 2). Every block has a peak.
Notice in particular that the first block (1, 2, 3, 4) has a peak at A[3], because A[2] < A[3] > A[4], even though A[4] is in the adjacent block.
However, array A cannot be divided into four blocks, (1, 2, 3), (4, 3, 4), (1, 2, 3) and (4, 6, 2), because the (1, 2, 3) blocks do not contain a peak. Notice in particular that the (4, 3, 4) block contains two peaks: A[3] and A[5].
The maximum number of blocks that array A can be divided into is three.
Write a function:
class Solution { public int solution(int[] A); }
that, given a non-empty zero-indexed array A consisting of N integers, returns the maximum number of blocks into which A can be divided.
If A cannot be divided into some number of blocks, the function should return 0.
For example, given:
A[0] = 1
A[1] = 2
A[2] = 3
A[3] = 4
A[4] = 3
A[5] = 4
A[6] = 1
A[7] = 2
A[8] = 3
A[9] = 4
A[10] = 6
A[11] = 2
the function should return 3, as explained above.
Assume that:
N is an integer within the range [1..100,000];
each element of array A is an integer within the range [0..1,000,000,000].
Complexity:
expected worst-case time complexity is O(N*log(log(N)))
expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).
Elements of input arrays can be modified.
My Question
So I solve this with what to me appears to be the brute force solution – go through every group size from 1..N, and check whether every group has at least one peak. The first 15 minutes I was trying to solve this I was trying to figure out some more optimal way, since the required complexity is O(N*log(log(N))).
This is my "brute-force" code that passes all the tests, including the large ones, for a score of 100/100:
public int solution(int[] A) {
int N = A.length;
ArrayList<Integer> peaks = new ArrayList<Integer>();
for(int i = 1; i < N-1; i++){
if(A[i] > A[i-1] && A[i] > A[i+1]) peaks.add(i);
}
for(int size = 1; size <= N; size++){
if(N % size != 0) continue;
int find = 0;
int groups = N/size;
boolean ok = true;
for(int peakIdx : peaks){
if(peakIdx/size > find){
ok = false;
break;
}
if(peakIdx/size == find) find++;
}
if(find != groups) ok = false;
if(ok) return groups;
}
return 0;
}
My question is how do I deduce that this is in fact O(N*log(log(N))), as it's not at all obvious to me, and I was surprised I pass the test cases. I'm looking for even the simplest complexity proof sketch that would convince me of this runtime. I would assume that a log(log(N)) factor means some kind of reduction of a problem by a square root on each iteration, but I have no idea how this applies to my problem. Thanks a lot for any help
You're completely right: to get the log log performance the problem needs to be reduced.
A n.log(log(n)) solution in python [below]. Codility no longer test 'performance' on this problem (!) but the python solution scores 100% for accuracy.
As you've already surmised:
Outer loop will be O(n) since it is testing whether each size of block is a clean divisor
Inner loop must be O(log(log(n))) to give O(n log(log(n))) overall.
We can get good inner loop performance because we only need to perform d(n), the number of divisors of n. We can store a prefix sum of peaks-so-far, which uses the O(n) space allowed by the problem specification. Checking whether a peak has occurred in each 'group' is then an O(1) lookup operation using the group start and end indices.
Following this logic, when the candidate block size is 3 the loop needs to perform n / 3 peak checks. The complexity becomes a sum: n/a + n/b + ... + n/n where the denominators (a, b, ...) are the factors of n.
Short story: The complexity of n.d(n) operations is O(n.log(log(n))).
Longer version:
If you've been doing the Codility Lessons you'll remember from the Lesson 8: Prime and composite numbers that the sum of harmonic number operations will give O(log(n)) complexity. We've got a reduced set, because we're only looking at factor denominators. Lesson 9: Sieve of Eratosthenes shows how the sum of reciprocals of primes is O(log(log(n))) and claims that 'the proof is non-trivial'. In this case Wikipedia tells us that the sum of divisors sigma(n) has an upper bound (see Robin's inequality, about half way down the page).
Does that completely answer your question? Suggestions on how to improve my python code are also very welcome!
def solution(data):
length = len(data)
# array ends can't be peaks, len < 3 must return 0
if len < 3:
return 0
peaks = [0] * length
# compute a list of 'peaks to the left' in O(n) time
for index in range(2, length):
peaks[index] = peaks[index - 1]
# check if there was a peak to the left, add it to the count
if data[index - 1] > data[index - 2] and data[index - 1] > data[index]:
peaks[index] += 1
# candidate is the block size we're going to test
for candidate in range(3, length + 1):
# skip if not a factor
if length % candidate != 0:
continue
# test at each point n / block
valid = True
index = candidate
while index != length:
# if no peak in this block, break
if peaks[index] == peaks[index - candidate]:
valid = False
break
index += candidate
# one additional check since peaks[length] is outside of array
if index == length and peaks[index - 1] == peaks[index - candidate]:
valid = False
if valid:
return length / candidate
return 0
Credits:
Major kudos to #tmyklebu for his SO answer which helped me a lot.
I'm don't think that the time complexity of your algorithm is O(Nlog(logN)).
However, it is certainly much lesser than O(N^2). This is because your inner loop is entered only k times where k is the number of factors of N. The number of factors of an integer can be seen in this link: http://www.cut-the-knot.org/blue/NumberOfFactors.shtml
I may be inaccurate but from the link it seems,
k ~ logN * logN * logN ...
Also, the inner loop has a complexity of O(N) since the number of peaks can be N/2 in the worst case.
Hence, in my opinion, the complexity of your algorithm is O(NlogN) at best but it must be sufficient to clear all test cases.
#radicality
There's at least one point where you can optimize the number of passes in the second loop to O(sqrt(N)) -- collect divisors of N and iterate through them only.
That will make your algo a little less "brute force".
Problem definition allows for O(N) space complexity. You can store divisors without violating this condition.
This is my solution based on prefix sums. Hope it helps:
class Solution {
public int solution(int[] A) {
int n = A.length;
int result = 1;
if (n < 3)
return 0;
int[] prefixSums = new int[n];
for (int i = 1; i < n-1; i++)
if (A[i] > A[i-1] && A[i] > A[i+1])
prefixSums[i] = prefixSums[i-1] + 1;
else
prefixSums[i] = prefixSums[i-1];
prefixSums[n-1] = prefixSums[n-2];
if (prefixSums[n-1] <= 1)
return prefixSums[n-1];
for (int i = 2; i <= prefixSums[n-2]; i++) {
if (n % i != 0)
continue;
int prev = 0;
boolean containsPeak = true;
for (int j = n/i - 1; j < n; j += n/i) {
if (prefixSums[j] == prev) {
containsPeak = false;
break;
}
prev = prefixSums[j];
}
if (containsPeak)
result = i;
}
return result;
}
}
def solution(A):
length = len(A)
if length <= 2:
return 0
peek_indexes = []
for index in range(1, length-1):
if A[index] > A[index - 1] and A[index] > A[index + 1]:
peek_indexes.append(index)
for block in range(3, int((length/2)+1)):
if length % block == 0:
index_to_check = 0
temp_blocks = 0
for peek_index in peek_indexes:
if peek_index >= index_to_check and peek_index < index_to_check + block:
temp_blocks += 1
index_to_check = index_to_check + block
if length/block == temp_blocks:
return temp_blocks
if len(peek_indexes) > 0:
return 1
else:
return 0
print(solution([1, 2, 3, 4, 3, 4, 1, 2, 3, 4, 6, 2, 1, 2, 5, 2]))
I just found the factors at first,
then just iterated in A and tested all number of blocks to see which is the greatest block division.
This is the code that got 100 (in java)
https://app.codility.com/demo/results/training9593YB-39H/
A javascript solution with complexity of O(N * log(log(N))).
function solution(A) {
let N = A.length;
if (N < 3) return 0;
let peaks = 0;
let peaksTillNow = [ 0 ];
let dividers = [];
for (let i = 1; i < N - 1; i++) {
if (A[i - 1] < A[i] && A[i] > A[i + 1]) peaks++;
peaksTillNow.push(peaks);
if (N % i === 0) dividers.push(i);
}
peaksTillNow.push(peaks);
if (peaks === 0) return 0;
let blocks;
let result = 1;
for (blocks of dividers) {
let K = N / blocks;
let prevPeaks = 0;
let OK = true;
for (let i = 1; i <= blocks; i++) {
if (peaksTillNow[i * K - 1] > prevPeaks) {
prevPeaks = peaksTillNow[i * K - 1];
} else {
OK = false;
break;
}
}
if (OK) result = blocks;
}
return result;
}
Solution with C# code
public int GetPeaks(int[] InputArray)
{
List<int> lstPeaks = new List<int>();
lstPeaks.Add(0);
for (int Index = 1; Index < (InputArray.Length - 1); Index++)
{
if (InputArray[Index - 1] < InputArray[Index] && InputArray[Index] > InputArray[Index + 1])
{
lstPeaks.Add(1);
}
else
{
lstPeaks.Add(0);
}
}
lstPeaks.Add(0);
int totalEqBlocksWithPeaks = 0;
for (int factor = 1; factor <= InputArray.Length; factor++)
{
if (InputArray.Length % factor == 0)
{
int BlockLength = InputArray.Length / factor;
int BlockCount = factor;
bool isAllBlocksHasPeak = true;
for (int CountIndex = 1; CountIndex <= BlockCount; CountIndex++)
{
int BlockStartIndex = CountIndex == 1 ? 0 : (CountIndex - 1) * BlockLength;
int BlockEndIndex = (CountIndex * BlockLength) - 1;
if (!(lstPeaks.GetRange(BlockStartIndex, BlockLength).Sum() > 0))
{
isAllBlocksHasPeak = false;
}
}
if (isAllBlocksHasPeak)
totalEqBlocksWithPeaks++;
}
}
return totalEqBlocksWithPeaks;
}
There is actually an O(n) runtime complexity solution for this task, so this is a humble attempt to share that.
The trick to go from the proposed O(n * loglogn) solutions to O(n) is to calculate the maximum gap between any two peaks (or a leading or trailing peak to the corresponding endpoint).
This can be done while building the peak hash in the first O(n) loop.
Then, if the gap is 'g' between two consecutive peaks, then the minimum group size must be 'g/2'. It will simply be 'g' between start and first peak, or last peak and end. Also, there will be at least one peak in any group from group size 'g', so the range to check for is: g/2, 1+g/2, 2+g/2, ... g.
Therefore, the runtime is the sum over d = g/2, g/2+1, ... g) * n/d where 'd' is the divisor'.
(sum over d = g/2, 1 + g/2, ... g) * n/d = n/(g/2) + n/(1 + g/2) + ... + (n/g)
if g = 5, this n/5 + n/6 + n/7 + n/8 + n/9 + n/10 = n(1/5+1/6+1/7+1/8+1/9+1/10)
If you replace each item with the largest element, then you get sum <= n * (1/5 + 1/5 + 1/5 + 1/5 + 1/5) = n
Now, generalising this, every element is replaced with n / (g/2).
The number of items from g/2 to g is 1 + g/2 since there are (g - g/2 + 1) items.
So, the whole sum is: n/(g/2) * (g/2 + 1) = n + 2n/g < 3n.
Therefore, the bound on the total number of operations is O(n).
The code, implementing this in C++, is here:
int solution(vector<int> &A)
{
int sizeA = A.size();
vector<bool> hash(sizeA, false);
int min_group_size = 2;
int pi = 0;
for (int i = 1, pi = 0; i < sizeA - 1; ++i) {
const int e = A[i];
if (e > A[i - 1] && e > A[i + 1]) {
hash[i] = true;
int diff = i - pi;
if (pi) diff /= 2;
if (diff > min_group_size) min_group_size = diff;
pi = i;
}
}
min_group_size = min(min_group_size, sizeA - pi);
vector<int> hash_next(sizeA, 0);
for (int i = sizeA - 2; i >= 0; --i) {
hash_next[i] = hash[i] ? i : hash_next[i + 1];
}
for (int group_size = min_group_size; group_size <= sizeA; ++group_size) {
if (sizeA % group_size != 0) continue;
int number_of_groups = sizeA / group_size;
int group_index = 0;
for (int peak_index = 0; peak_index < sizeA; peak_index = group_index * group_size) {
peak_index = hash_next[peak_index];
if (!peak_index) break;
int lower_range = group_index * group_size;
int upper_range = lower_range + group_size - 1;
if (peak_index > upper_range) {
break;
}
++group_index;
}
if (number_of_groups == group_index) return number_of_groups;
}
return 0;
}
var prev, curr, total = 0;
for (var i=1; i<A.length; i++) {
if (curr == 0) {
curr = A[i];
} else {
if(A[i] != curr) {
if (prev != 0) {
if ((prev < curr && A[i] < curr) || (prev > curr && A[i] > curr)) {
total += 1;
}
} else {
prev = curr;
total += 1;
}
prev = curr;
curr = A[i];
}
}
}
if(prev != curr) {
total += 1;
}
return total;
I agree with GnomeDePlume answer... the piece on looking for the divisors on the proposed solution is O(N), and that could be decreased to O(sqrt(N)) by using the algorithm provided on the lesson text.
So just adding, here is my solution using Java that solves the problem on the required complexity.
Be aware, it has way more code then yours - some cleanup (debug sysouts and comments) would always be possible :-)
public int solution(int[] A) {
int result = 0;
int N = A.length;
// mark accumulated peaks
int[] peaks = new int[N];
int count = 0;
for (int i = 1; i < N -1; i++) {
if (A[i-1] < A[i] && A[i+1] < A[i])
count++;
peaks[i] = count;
}
// set peaks count on last elem as it will be needed during div checks
peaks[N-1] = count;
// check count
if (count > 0) {
// if only one peak, will need the whole array
if (count == 1)
result = 1;
else {
// at this point (peaks > 1) we know at least the single group will satisfy the criteria
// so set result to 1, then check for bigger numbers of groups
result = 1;
// for each divisor of N, check if that number of groups work
Integer[] divisors = getDivisors(N);
// result will be at least 1 at this point
boolean candidate;
int divisor, startIdx, endIdx;
// check from top value to bottom - stop when one is found
// for div 1 we know num groups is 1, and we already know that is the minimum. No need to check.
// for div = N we know it's impossible, as all elements would have to be peaks (impossible by definition)
for (int i = divisors.length-2; i > 0; i--) {
candidate = true;
divisor = divisors[i];
for (int j = 0; j < N; j+= N/divisor) {
startIdx = (j == 0 ? j : j-1);
endIdx = j + N/divisor-1;
if (peaks[startIdx] == peaks[endIdx]) {
candidate = false;
break;
}
}
// if all groups had at least 1 peak, this is the result!
if (candidate) {
result = divisor;
break;
}
}
}
}
return result;
}
// returns ordered array of all divisors of N
private Integer[] getDivisors(int N) {
Set<Integer> set = new TreeSet<Integer>();
double sqrt = Math.sqrt(N);
int i = 1;
for (; i < sqrt; i++) {
if (N % i == 0) {
set.add(i);
set.add(N/i);
}
}
if (i * i == N)
set.add(i);
return set.toArray(new Integer[]{});
}
Thanks,
Davi
int main() {
int i, a[N];
// initialize the array
for(i = 2; i < N; i++) a[i] = 1;
for(i = 2; i < N; i++)
if(a[i])
for(int j = i; j*i < N; j++) a[i*j] =0;
// pirnt the primes less then N
for(i = 2; i < N; i++)
if(a[i]) cout << " " << i;
cout << endl;
}
It was given in algorithm book i am reading running time of above program is proportional to N+N/2+N/3+N/5+N/7+N/11+...,
Please help me in understanding how author came up with above equation from the program.
Thanks!
Venkata
This is the "Sieve of Eratosthenes" method for finding primes. For each prime, the if(a[i]) test succeeds and the inner loop gets executed. Consider how this inner loop terminates at each step (remember, the condition is j*i < N, or equivalently, j < N/i):
i = 2 -> j = 2, 3, 4, ..., N/2
i = 3 -> j = 3, 4, 5, ..., N/3
i = 4 -> not prime
i = 5 -> j = 5, 6, 7, ..., N/5
...
Summing the total number of operations (including initialising the array/extracting the primes) gives the runtime mentioned in the book.
See this question for more, including a discussion of how, in terms of bit operations, this turns into an expansion of O(n(log n)(log log n)) as per the Wikipedia article.
This algorithm is called the Sieve of Eratosthenes. This image explains everything:
(from Wikipedia)
The inner loop (inside if(a[i])) is executed for prime is only. I.e., for i equal to 2, 3, 5, 7, 11, ... And for single i, this loop has approximately N/i iterations. Thus, we have N/2 + N/3 + N/5 + N/7 + N/11 + ... iterations overall.