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you have given array of integers
for example
5,3,4,3,9,4,4,6,8,6,5,7
you have to find largest interval (i,j) such that all number between i and j are greater than equal to number at i but less than equal to number at j. numbers need not to be in sorted order. Also number at i should be less than number at j.
in above example there are two such intervals
3,4,3,9 and 4,4,6,8
In 3,4,3,9 - middle two numbers (4,3) are greater than equal to starting number(3) but less than equal to ending number(9) . Also starting number (3) is less than ending number(9).
In 4,4,6,8 - middle two numbers (4,6) are greater than equal to starting number(4) but less than equal to ending number(8). Also starting number(4) is less than ending number(8).
One way to solve this problem is check all intervals of size n, then size n-1 , so on till we get such interval. But I want efficient algo either by Divide or Conquer /DP/or some other approach
We can solve this in O(n log n) by considering each element as the potential leftmost element in the interval. For each such candidate, (1) find the first element to the right that's lower than it - that's a bound on the right for the possible interval since including such an element would invalidate the constraints; (2) find the leftmost instance of the maximum element in the interval (or rightmost if A[j] is allowed to be equal to an element on its left) - that's the farthest right the interval could extend without invalidating the constraints.
We can store the answer for (1) for all elements in O(n), using a stack. We can preprocess a structure for range-maximum-query that would answer the first part of (2) in O(log n) time. Once the maximum for the interval is found, if we have an ordered list of its instances in the array (which we can preprocess in O(n) with a hash table), we can find the rightmost or leftmost occurrence of it in the interval in O(log n) time.
A simple answer by DP, hope it could help you:
public class Test {
public static void main(String[] args){
int[] input = {5,3,4,3,9,4,4,6,8,6,5,7};
//
// lenDp array to store the max possible interval sizes starting at every index.
//
int[] lenDp = new int[input.length];
//
// minNumIdxs array to store the indexes of minimum numbers in all sub arrays with the same length
//
int[] minNumIdxs = new int[input.length];
//
// for a start, every number is an interval with size of 1,
// and there are as many sub arrays with length of 1 as input.length,
// the minimum number in a sub array is the first number:
//
for(int i = 0; i < input.length; i++){
lenDp[i] = 1;
minNumIdxs[i] = i;
}
int maxLen = 1;
//
// check all possible intervals of size from 2 to input.length
//
for(int len = 2; len <= input.length; len++){
//
// for every possible interval with size of `len`, the starting index can be in range of [0, input.length - len]
//
for(int i = 0, end = input.length - len; i <= end ; i++ ){
//
// assume the interval of size `len` is possible for index i,then the number at input[i] has to be the minimum number,
// and the number at input[i + len - 1] has to be greater than or equal to
// the largest number ( at input[i + dp[i] - 1] ) of the current interval starting at index i.
// in the corner case where dp[i] == 1 and len > 2, it wouldn't be valid, for example:
// when len = 5, i = 0, dp[i] = dp[0] = 1, then input[i + len - 1] = input[0 + 5 - 1] = input[4] = 9;
// 9 is greater than 5 (input[0]), but it's invalid.
// So we also need to check (len == 2 || dp[i] > 1)
//
if(minNumIdxs[i] == i && input[i + len - 1] >= input[i + lenDp[i] - 1] &&
(len == 2 || lenDp[i] > 1)){
lenDp[i] = len;
maxLen = len;
}
//
// update the index of the minimum number in the sub array starting at i
//
if(input[i + len - 1] < input[minNumIdxs[i]])minNumIdxs[i] = i + len - 1;
}
}
//
// print out the final result
//
for(int i = 0 ; i < lenDp.length ; i++){
if(lenDp[i] == maxLen){
System.out.println(Arrays.toString(Arrays.copyOfRange(input, i, i + maxLen)));
}
}
}
}
output:
[3, 4, 3, 9]
[4, 4, 6, 8]
Given array : 8 3 5 2 10 6 7 9 5 2
So the o/p will be Yes.
as: {8,3,5} {10,6} {9,5,2} they all have same sum value i.e. 16.
But for this array : 1 4 9 6 2 12
o/p will be No.
as: No contiguous slide have same sum value
I was thinking to go with SubSetSum Algorithm / Kadane Maximum SubArray Algorithm but later I end up as all of the algorithms requires a target sum which is predefined.
But here we don't know the target sum
If desired sum is given, and all subarrays should be contiguous, then it's easily can be done in O(n).
Run a loop over array and maintain boundaries of slices (left and right indexes) and currentSum.
Start with first element as a 0. Boundaries will be [0, 0] (for simplicity we include right). Then in a loop you have three conditions.
If sum is less than desired, add right element to the sum and advance right index
If sum is greater than desired, remove left element from the sum and advance left index
If sum is equal to given, print the slice. To avoid this slice in next iteration, advance left index and adjust the sum.
Translated to code
public static void main(String[] args) {
int givenSum = 16;
int[] a = new int[] {8, 3, 5, 2, 10, 6, 7, 9, 5, 2};
// boundaries of slice
int left = 0; // defines position of slice
int right = 0; // exclusive
int currentSum = 0;
while (right < a.length) {
if (currentSum < givenSum) { // sum is not enough, add from the right
currentSum += a[right];
right++;
}
if (currentSum > givenSum) { // sum exceeds given, remove from the left
currentSum -= a[left];
left++;
}
if (currentSum == givenSum) { // boundaries of given sum found, print it
System.out.println(Arrays.toString(Arrays.copyOfRange(a, left, right)));
// remove the left element, so we can process next sums
currentSum -= a[left];
left++;
}
}
}
For your case it prints 4 slices which yields sum 16
[8, 3, 5]
[10, 6]
[7, 9]
[9, 5, 2]
EDIT:
As OP clarified, no given sum available, the goal is to check if there are at least two different contiguous subarrays present which yields equal sum.
The most straightforward algorithm is to generate all possible sums and check if there are duplicates
int[] a = new int[] {1, 4, 9, 6, 2, 12};
HashSet<Integer> sums = new HashSet<>();
int numOfSums = 0;
for (int left = 0; left < a.length - 1; left++) {
for (int right = left; right < a.length; right++) {
// sum from left to right
int sum = 0;
for (int k = left; k <= right; k++) {
sum += a[k];
}
numOfSums++;
sums.add(sum);
}
}
System.out.println(sums.size() == numOfSums);
Complexity of this is O(n^3), not a good one, but works.
Hint: One trick could be explored to boost it to O(n^2), you don't need to calculate sum for every pair of slices!
You can do it in the following way
You have the total sum = 48
Now the each subset would have a sum which would be equal to a factor of 48. The smaller the factor the more number of subsets you can break it into
For all factors of the sum, check if the answer is possible for that factor or not. This can be done in O(n) by simply traversing the array.
Time Complexity would be O(n * factors(sum))
Use dynamic programming to find all sub-sums of the array, then find the sub array with same sum. The complexity should be O(n2).
void subsum(int n, int* arr, int** sum) {
for (int i = 0; i < n; ++i) {
sum[i][i] = arr[i];
}
for (int l = 2; l <= n; ++l) {
for (int i = 0; i < n - l + 1; ++i) {
sum[i][i + l - 1] = sum[i][i + l - 2] + arr[i + l -1];
}
}
}
I have a following problem.
Given N numbers, in range -100..100.
It is required to rearrange elements to have maximum sum of product value.
Sum of Product in this task is defined as A1*A2+A2*A3...AN-1*AN
For example, given numbers 10 20 50 40 30.
Then, we can rearrange them following way:
10, 30, 50, 40, 20 from the left to have maximum 10×30+30×50+50×40+40×20=4600
The idea is to sort the sequence, and then put max number in the middle of new sequence, then put next max number to the right, then to the left, and so on.
But, regarding negative numbers this is not working.
I have tried following algorithm:
1) sort initial sequence
2) process positive numbers and zero values how described above
3) process negative numbers how described above
4) find minimum number from positive sequence, it would be either left or right element and add after of before this number processed negative sequence.
For example, given sequence:
1,-2,3,-4,5,-6,7,-8,9,10,11,12,13,14,15,-16
Expected maximum sum of product is 1342.
My algorithm gives next rearrangements:
3,7,10,12,14,15,13,11,9,5,1,-4,-8,-16,-6,-2
Sum of product is 1340.
This seem to work, but it does not.
Could you please advise?
Your approach is sound, but you have to separate the positive and negative numbers.
Sort the array and split it into left and right parts, one containing all the negative numbers, and one containing all the non-negative numbers. Rearrange them as you were doing before, with the largest (absolute) values in the middle and decreasing values placed alternately on either side, but make sure that the smallest values in each part are at opposite ends.
Specifically, the negative number with the smallest absolute value should be the last element of the left part, and the non-negative value with the smallest value should be the first element of the right part.
Then concatenate the two parts and calculate the sum of adjacent products.
Here's a worked example:
arr = [2, 3, 5, -6, -2, -5]
arr.sort() = [-6, -5, -2, 2, 3, 5]
left, right = [-5, -6, -2], [2, 5, 3]
max_sum_of_product = -5*-6 + -6*-2 + -2*2 + 2*5 + 5*3 = 63
I don't have a formal proof of correctness, but this method gives the same results as a brute force search over all permutations of the input array:
def max_sum_of_products(arr):
from itertools import permutations
n = len(arr)
###### brute force method
max1 = max([sum([a[x-1]*a[x] for x in range(1,n)]) for a in permutations(arr)])
###### split method
lo, hi = [x for x in arr if x<0], [x for x in arr if x>=0]
lo.sort()
hi.sort()
lo_ordered, hi_ordered = [], []
t = (len(lo)%2 == 1)
for x in lo:
if t:
lo_ordered = lo_ordered + [x]
else:
lo_ordered = [x] + lo_ordered
t = not t
t = (len(hi)%2 == 0)
for x in hi[::-1]:
if t:
hi_ordered = hi_ordered + [x]
else:
hi_ordered = [x] + hi_ordered
t = not t
arr = lo_ordered + hi_ordered
max2 = sum([arr[x-1]*arr[x] for x in range(1,n)])
return (max1, max2)
def test():
from random import randint
for i in range(10):
a = []
for j in range(randint(4,9)):
a = a + [randint(-10,10)]
print a,
(max1,max2) = max_sum_of_products(a)
if max2!=max1:
print "bad result :-("
else:
print max1
test()
I have written a method in java that will take the array as an input and return the maximum sum of product pairs as output.
First I compute the negative part, then the positive part and then return their computed sum.
While computing the negative part, if the number of elements are odd, then the remaining element needs to be avoided (as it can be multiplied by 0 and nullified), we do this so that that negative addition will lower the sum.
All other negative items are needed to multiplied in pair and summed.
Then coming to second positive part, when we see 1 we need to add it if number of elements are odd, otherwise simply multiply and go forward.
public static long sum(int arr[]) {
Arrays.sort(arr);
long ans = 0;
long ans1 = 0;
boolean flag = false;
boolean flag2 = false;
int[] arr1 = new int[arr.length];
int[] arr2 = new int[arr.length];
int i = 0;
while (arr[i] < 0) {
arr1[i] = arr[i];
i++;
}
if (arr[i] == 0) flag = true;
if (i % 2 == 0) { //even -6,-5,-3,-2,-1
for (int j = 0; j < i - 1; j += 2) {
ans = arr1[j] * arr1[j + 1];
}
} else {
if (flag) {
for (int j = 0; j < i - 2; j += 2) {
ans = arr1[j] * arr1[j + 1];
}
}
}
int j = 0;
while (i<arr.length) {
arr2[j] = arr[i];
i++;
j++;
}
if (arr2[j] == 1) flag2 = true;
if (i % 2 == 0) {
for (int k=i-1; k>0; k-=2) {
ans1 = arr2[k] * arr2[k-1];
}
if (flag2) ans1 = ans1 + 1;
} else {
for (int k=arr2.length-1; k>1; k-=2) {
ans1 = arr2[k] * arr2[k-1];
}
ans1 = ans1 + arr2[0];
}
return ans + ans1;
}
There is an MxN size matrix which has integers in it. We need to find the largest size sub-matrix which has same integers in it. For example :
1 2 2 4 5
1 2 2 8 7
3 2 2 6 1
Here the largest submatrix is of 3x2 size which has all 2's in it. My idea was to check for each element arr[i][j] , whether arr[i][j+1] , arr[i+1]arr[j+1] and arr[i+1][j] are equal or not. If they are equal then we can somehow update maximum size of matrix. But I wasn't able to arrive at the exact solution.
I was wondering if somehow we can make use of Dynamic Programming. This was an interview question.
Let's assume that a bottom row of the submatrix is fixed. Then each column can represented a as a pair (value, height), where value is the value of the element in this row and column and height is how many consecutive elements in this column are equal to it. For example, if the matrix is
3 1 2 3
1 2 2 1
3 2 2 1
2 2 2 2
and the bottom row we are looking at is 2(in zero-based indexing), the values are (3, 1), (2, 2), (2, 3) and (1, 2), respectively.
Let's split the columns into groups grouping adjacent elements with the same value together(for the previous example it would be {(3, 1)}, {(2, 2), (2, 3)} and {(1, 2)}. Now we can solve a standard problem: given an array of values h[i], find the largest value of min(h[i], h[i + 1], ..., h[j]) * (j - i + 1) for all i and j within each group(there is a linear solution that uses stacks, I can elaborate on this one if necessary). It allows us to process one row in linear time.
The last thing we need is to compute the array of (value, height) for each row efficiently. For the first row it is trivial. So we can iterate over all rows and them one by one(when one element is added to the bottom of the column, (value, height) pair changes in two possible ways: it becomes either (new_value, 1) or (value, height + 1)).
These observation allow us to process one row in O(M) time. The total time complexity is O(N * M).
Here is some code(it is not a full implementation and it can contain bugs):
int solve() {
int res = 0;
Pair[] valueForColumn = new Pair[rowsCount];
for (int col = 0; col < columnsCount; col++)
valueForColunm[col] = new Pair(matrix[0][col], 1);
res = Math.max(res, solveForRow(valueForColumn);
for (int row = 1; row < rowsCount; row++) {
for (int col = 0; col < columnsCount; col++)
if (matrix[row][col] == matrix[row - 1][col]) {
valueForColumn[col].second++;
} else {
valueForColumn[col].first = matrix[row][col];
valueForColumn[col].second = 1;
}
}
res = Math.max(res, solveForRow(valueForColumn));
}
return res;
}
int solveForRow(Pair[] valueForColumn) {
List<Integer> group = new ArrayList<>();
int res = 0;
for (int i = 0; i < valueForColumn.length; i++) {
group.add(valueForColumn[i].second);
if (i == valueForColumn.length - 1
|| valueForColumn[i].first != valueForColumn[i + 1].first) {
res = Math.max(res, solveForGroup(group));
group.clear();
}
}
return res;
}
int solveForGroup(List<Integer> heights) {
// a stack-based algorithm
// with linear time complexity mentioned in 2. goes here
}
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