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I have an array a[n]. The number n is entered by us. I need to find the minimal product of a[i] and a[j] if:
1) abs(i - j) > k
2) a[i] * a[j] is minimised
Here is my solution (very naive):
#include <iostream>
using namespace std;
#define ll long long
int main() {
ll n,k; cin >> n >> k;
ll a[n]; for(ll i=0;i<n;i++) cin >> a[i];
ll mn; bool first = true;
for(ll i=0;i<n;i++) {
for(ll j=0;j<n;j++) {
if(i!=j)
if(abs(i-j) > k) {
if(first) {
mn = a[i]*a[j];
first = false;
} else if(a[i]*a[j] < mn) mn = a[i]*a[j];
}
}
}
cout << mn << endl;
}
But I want to know if there is any faster way to find a minimal product with distance?
Assuming there is at least one pair of elements satisfying the conditions and no multiplication of two elements in it overflows, this can be done in Theta(n-k) time and Theta(1) space worst- and best-case, with something like this:
auto back_max = a[0];
auto back_min = a[0];
auto best = a[0]*a[k+1];
for(std::size_t i=1; i<n-(k+1); ++i) {
back_max = std::max(back_max, a[i]);
back_min = std::min(back_min, a[i]);
best = std::min(best, std::min(a[i+k+1]*back_max, a[i+k+1]*back_min));
}
return best;
This is optimal in terms of asymptotic worst-case complexity for both time and space because the optimal product may be a[0] with any of the n-(k+1) elements in distance at least k+1, so at least n-(k+1) integers need to be read by any algorithm solving the problem.
The idea behind the algorithm is as follows:
The optimal product uses two elements of a, assume these are a[r] and a[s]. Without loss of generality we can assume that s > r since the product is commutative.
Due to the restriction abs(s-r) > k this implies that s >= k+1. Now s could be each of the indices satisfying this condition, so we iterate over these indices. That is the iteration over i in the shown code, but it is shifted by k+1 for convenience (doesn't really matter). For each iteration we need to find the optimal product involving i+k+1 as largest index and compare it with the previous best guess.
The possible indices to pair i+k+1 with are all indices smaller or equal i due to the distance requirement. We would need to iterate over all of these as well, but that is unnecessary because the minimum of a[i+k+1]*a[j] over j at fixed i is equal to min(a[i+k+1]*max(a[j]), a[i+k+1]*min(a[j])) due to monotonicity of the product (taking the minimum with respect to both the minimum and maximum over a[j] accounts for the two possible signs of a[i+k+1] or equivalently the two possible directions of monotonicity.)
Since the set of a[j] values over which we optimize here is just {a[0], ..., a[i]}, which simply growths by one element (a[i]) in each iteration of i, we can simply keep track of max(a[j]) and min(a[j]) with single variables by updating them if a[i] is larger or smaller than the previous optimal values. This is done with back_max and back_min in the code example.
The first step of the iteration (i=0) is skipped in the loop and instead performed as initialization of the variables.
Not sure about fastest.
For the simpler problem without i < j - k, the minimal product is among the products of pairs from the two smallest and largest elements.
So, (the following is too complicated, see walnut's answer)
( • balk if k ≤ n
• initialise minProduct to a[0]*a[k+1])
keep two dynamic minmax data structures upToI and beyondIplusK
starting with { } and { a[j] | k ≤ j }
for each i from 0 to n - k - 1
add a[i] to upToI
remove a[i+k] from beyondIplusK
check for new minimal product among
min(upToI)×min(beyondIplusK), min(upToI)×max(beyondIplusK),
max(upToI)×min(beyondIplusK) and max(upToI)×max(beyondIplusK)
For "minimum magnitude"
Find the 2 "smallest magnitude" elements, then (after you've either found two zeros or searched the whole array), multiply them.
For "lowest value" without the abs(i - j) > k part
There are 3 possibilities:
the two highest (smallest magnitude) negative numbers
the two lowest (smallest magnitude) non-negative numbers
the lowest (largest magnitude) negative number and the highest (largest magnitude) non-negative number
You could search for all 6 values and figure out the products and which is best at end.
However; as soon as you see a zero you know you don't need to know any more about the first 2 possibilities; and as soon as you see one negative number and one non-negative number you know that you only care about the third possibility.
This leads to a finite state machine with 3 states - "care about all 3 possibilities", "answer is zero unless a negative number is seen" and "only care about the last possibility". This can be implemented as a set of 3 loops, where 2 of the loops jump into (goto) the middle of another loop when the state (of the finite state machine) changes.
Specifically, it might looks something vaguely like (untested):
// It could be any possibility
for(ll i=0;i<n;i++) {
if(a[i] >= 0) {
if(a[i] < lowestNonNegative1) {
lowestNonNegative2 = lowestNonNegative1;
lowestNonNegative1 = a[i];
}
if(lowestNonNegative2 == 0) {
goto state2;
}
} else {
if(a[i] > highestNegative1) {
highestNegative2 = highestNegative1;
highestNegative1= a[i];
}
if(lowestNonNegative1 < LONG_MAX) {
goto state3;
}
}
}
if(lowestNonNegative2 * lowestNonNegative1 < highestNegative2 * highestNegative1) {
cout << lowestNonNegative2 * lowestNonNegative1;
} else {
cout << highestNegative2 * highestNegative1;
}
return;
// It will be zero, or a negative and a non-negative
for(ll i=0;i<n;i++) {
state2:
if(a[i] < 0) {
goto state3;
}
}
cout << "0";
return;
// It will be a negative and a non-negative
for(ll i=0;i<n;i++) {
state3:
if(a[i] < lowestNegative) {
lowestNegative = a[i];
} else if(a[i] > highestNonNegative) {
highestNonNegative = a[i];
}
}
cout << lowestNegative * highestNonNegative;
return;
For "lowest value" with the abs(i - j) > k part
In this case you still have the 3 possibilities; and could make it work with the same "3 loops with finite state machine" approach but it gets too messy/ugly. For this case a better alternative is likely to pre-scan the array to determine if there are any zeros and if they're all negative or all positive; so that after the pre-scan you can either know the answer is zero or select a loop designed for the specific possibility alone.
Hexagonal grid is represented by a two-dimensional array with R rows and C columns. First row always comes "before" second in hexagonal grid construction (see image below). Let k be the number of turns. Each turn, an element of the grid is 1 if and only if the number of neighbours of that element that were 1 the turn before is an odd number. Write C++ code that outputs the grid after k turns.
Limitations:
1 <= R <= 10, 1 <= C <= 10, 1 <= k <= 2^(63) - 1
An example with input (in the first row are R, C and k, then comes the starting grid):
4 4 3
0 0 0 0
0 0 0 0
0 0 1 0
0 0 0 0
Simulation: image, yellow elements represent '1' and blank represent '0'.
This problem is easy to solve if I simulate and produce a grid each turn, but with big enough k it becomes too slow. What is the faster solution?
EDIT: code (n and m are used instead R and C) :
#include <cstdio>
#include <cstring>
using namespace std;
int old[11][11];
int _new[11][11];
int n, m;
long long int k;
int main() {
scanf ("%d %d %lld", &n, &m, &k);
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) scanf ("%d", &old[i][j]);
}
printf ("\n");
while (k) {
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
int count = 0;
if (i % 2 == 0) {
if (i) {
if (j) count += old[i-1][j-1];
count += old[i-1][j];
}
if (j) count += (old[i][j-1]);
if (j < m-1) count += (old[i][j+1]);
if (i < n-1) {
if (j) count += old[i+1][j-1];
count += old[i+1][j];
}
}
else {
if (i) {
if (j < m-1) count += old[i-1][j+1];
count += old[i-1][j];
}
if (j) count += old[i][j-1];
if (j < m-1) count += old[i][j+1];
if (i < n-1) {
if (j < m-1) count += old[i+1][j+1];
count += old[i+1][j];
}
}
if (count % 2) _new[i][j] = 1;
else _new[i][j] = 0;
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) old[i][j] = _new[i][j];
}
k--;
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
printf ("%d", old[i][j]);
}
printf ("\n");
}
return 0;
}
For a given R and C, you have N=R*C cells.
If you represent those cells as a vector of elements in GF(2), i.e, 0s and 1s where arithmetic is performed mod 2 (addition is XOR and multiplication is AND), then the transformation from one turn to the next can be represented by an N*N matrix M, so that:
turn[i+1] = M*turn[i]
You can exponentiate the matrix to determine how the cells transform over k turns:
turn[i+k] = (M^k)*turn[i]
Even if k is very large, like 2^63-1, you can calculate M^k quickly using exponentiation by squaring: https://en.wikipedia.org/wiki/Exponentiation_by_squaring This only takes O(log(k)) matrix multiplications.
Then you can multiply your initial state by the matrix to get the output state.
From the limits on R, C, k, and time given in your question, it's clear that this is the solution you're supposed to come up with.
There are several ways to speed up your algorithm.
You do the neighbour-calculation with the out-of bounds checking in every turn. Do some preprocessing and calculate the neighbours of each cell once at the beginning. (Aziuth has already proposed that.)
Then you don't need to count the neighbours of all cells. Each cell is on if an odd number of neighbouring cells were on in the last turn and it is off otherwise.
You can think of this differently: Start with a clean board. For each active cell of the previous move, toggle the state of all surrounding cells. When an even number of neighbours cause a toggle, the cell is on, otherwise the toggles cancel each other out. Look at the first step of your example. It's like playing Lights Out, really.
This method is faster than counting the neighbours if the board has only few active cells and its worst case is a board whose cells are all on, in which case it is as good as neighbour-counting, because you have to touch each neighbours for each cell.
The next logical step is to represent the board as a sequence of bits, because bits already have a natural way of toggling, the exclusive or or xor oerator, ^. If you keep the list of neigbours for each cell as a bit mask m, you can then toggle the board b via b ^= m.
These are the improvements that can be made to the algorithm. The big improvement is to notice that the patterns will eventually repeat. (The toggling bears resemblance with Conway's Game of Life, where there are also repeating patterns.) Also, the given maximum number of possible iterations, 2⁶³ is suspiciously large.
The playing board is small. The example in your question will repeat at least after 2¹⁶ turns, because the 4×4 board can have at most 2¹⁶ layouts. In practice, turn 127 reaches the ring pattern of the first move after the original and it loops with a period of 126 from then.
The bigger boards may have up to 2¹⁰⁰ layouts, so they may not repeat within 2⁶³ turns. A 10×10 board with a single active cell near the middle has ar period of 2,162,622. This may indeed be a topic for a maths study, as Aziuth suggests, but we'll tacke it with profane means: Keep a hash map of all previous states and the turns where they occurred, then check whether the pattern has occurred before in each turn.
We now have:
a simple algorithm for toggling the cells' state and
a compact bitwise representation of the board, which allows us to create a hash map of the previous states.
Here's my attempt:
#include <iostream>
#include <map>
/*
* Bit representation of a playing board, at most 10 x 10
*/
struct Grid {
unsigned char data[16];
Grid() : data() {
}
void add(size_t i, size_t j) {
size_t k = 10 * i + j;
data[k / 8] |= 1u << (k % 8);
}
void flip(const Grid &mask) {
size_t n = 13;
while (n--) data[n] ^= mask.data[n];
}
bool ison(size_t i, size_t j) const {
size_t k = 10 * i + j;
return ((data[k / 8] & (1u << (k % 8))) != 0);
}
bool operator<(const Grid &other) const {
size_t n = 13;
while (n--) {
if (data[n] > other.data[n]) return true;
if (data[n] < other.data[n]) return false;
}
return false;
}
void dump(size_t n, size_t m) const {
for (size_t i = 0; i < n; i++) {
for (size_t j = 0; j < m; j++) {
std::cout << (ison(i, j) ? 1 : 0);
}
std::cout << '\n';
}
std::cout << '\n';
}
};
int main()
{
size_t n, m, k;
std::cin >> n >> m >> k;
Grid grid;
Grid mask[10][10];
for (size_t i = 0; i < n; i++) {
for (size_t j = 0; j < m; j++) {
int x;
std::cin >> x;
if (x) grid.add(i, j);
}
}
for (size_t i = 0; i < n; i++) {
for (size_t j = 0; j < m; j++) {
Grid &mm = mask[i][j];
if (i % 2 == 0) {
if (i) {
if (j) mm.add(i - 1, j - 1);
mm.add(i - 1, j);
}
if (j) mm.add(i, j - 1);
if (j < m - 1) mm.add(i, j + 1);
if (i < n - 1) {
if (j) mm.add(i + 1, j - 1);
mm.add(i + 1, j);
}
} else {
if (i) {
if (j < m - 1) mm.add(i - 1, j + 1);
mm.add(i - 1, j);
}
if (j) mm.add(i, j - 1);
if (j < m - 1) mm.add(i, j + 1);
if (i < n - 1) {
if (j < m - 1) mm.add(i + 1, j + 1);
mm.add(i + 1, j);
}
}
}
}
std::map<Grid, size_t> prev;
std::map<size_t, Grid> pattern;
for (size_t turn = 0; turn < k; turn++) {
Grid next;
std::map<Grid, size_t>::const_iterator it = prev.find(grid);
if (1 && it != prev.end()) {
size_t start = it->second;
size_t period = turn - start;
size_t index = (k - turn) % period;
grid = pattern[start + index];
break;
}
prev[grid] = turn;
pattern[turn] = grid;
for (size_t i = 0; i < n; i++) {
for (size_t j = 0; j < m; j++) {
if (grid.ison(i, j)) next.flip(mask[i][j]);
}
}
grid = next;
}
for (size_t i = 0; i < n; i++) {
for (size_t j = 0; j < m; j++) {
std::cout << (grid.ison(i, j) ? 1 : 0);
}
std::cout << '\n';
}
return 0;
}
There is probably room for improvement. Especially, I'm not so sure how it fares for big boards. (The code above uses an ordered map. We don't need the order, so using an unordered map will yield faster code. The example above with a single active cell on a 10×10 board took significantly longer than a second with an ordered map.)
Not sure about how you did it - and you should really always post code here - but let's try to optimize things here.
First of all, there is not really a difference between that and a quadratic grid. Different neighbor relationships, but I mean, that is just a small translation function. If you have a problem there, we should treat this separately, maybe on CodeReview.
Now, the naive solution is:
for all fields
count neighbors
if odd: add a marker to update to one, else to zero
for all fields
update all fields by marker of former step
this is obviously in O(N). Iterating twice is somewhat twice the actual run time, but should not be that bad. Try not to allocate space every time that you do that but reuse existing structures.
I'd propose this solution:
at the start:
create a std::vector or std::list "activated" of pointers to all fields that are activated
each iteration:
create a vector "new_activated"
for all items in activated
count neighbors, if odd add to new_activated
for all items in activated
set to inactive
replace activated by new_activated*
for all items in activated
set to active
*this can be done efficiently by putting them in a smart pointer and use move semantics
This code only works on the activated fields. As long as they stay within some smaller area, this is far more efficient. However, I have no idea when this changes - if there are activated fields all over the place, this might be less efficient. In that case, the naive solution might be the best one.
EDIT: after you now posted your code... your code is quite procedural. This is C++, use classes and use representation of things. Probably you do the search for neighbors right, but you can easily make mistakes there and therefore should isolate that part in a function, or better method. Raw arrays are bad and variables like n or k are bad. But before I start tearing your code apart, I instead repeat my recommendation, put the code on CodeReview, having people tear it apart until it is perfect.
This started off as a comment, but I think it could be helpful as an answer in addition to what has already been stated.
You stated the following limitations:
1 <= R <= 10, 1 <= C <= 10
Given these restrictions, I'll take the liberty to can represent the grid/matrix M of R rows and C columns in constant space (i.e. O(1)), and also check its elements in O(1) instead of O(R*C) time, thus removing this part from our time-complexity analysis.
That is, the grid can simply be declared as bool grid[10][10];.
The key input is the large number of turns k, stated to be in the range:
1 <= k <= 2^(63) - 1
The problem is that, AFAIK, you're required to perform k turns. This makes the algorithm be in O(k). Thus, no proposed solution can do better than O(k)[1].
To improve the speed in a meaningful way, this upper-bound must be lowered in some way[1], but it looks like this cannot be done without altering the problem constraints.
Thus, no proposed solution can do better than O(k)[1].
The fact that k can be so large is the main issue. The most anyone can do is improve the rest of the implementation, but this will only improve by a constant factor; you'll have to go through k turns regardless of how you look at it.
Therefore, unless some clever fact and/or detail is found that allows this bound to be lowered, there's no other choice.
[1] For example, it's not like trying to determine if some number n is prime, where you can check all numbers in the range(2, n) to see if they divide n, making it a O(n) process, or notice that some improvements include only looking at odd numbers after checking n is not even (constant factor; still O(n)), and then checking odd numbers only up to √n, i.e., in the range(3, √n, 2), which meaningfully lowers the upper-bound down to O(√n).
I'm going through some dynamic programming articles on geeksforgeeks and ran across the Longest Common Subsequence problem. I did not come up with an implementation of the exponential naive solution on my own, however after working out some examples of the problem on paper I came up with what I thought was a successful implementation of an O(n*m) version . However, an OJ proved me wrong. My algorithm fails with the input strings:
"LRBBMQBHCDARZOWKKYHIDDQSCDXRJMOWFRXSJYBLDBEFSARCBYNECDYGGXXPKLORELLNMPAPQFWKHOPKMCO"
"QHNWNKUEWHSQMGBBUQCLJJIVSWMDKQTBXIXMVTRRBLJPTNSNFWZQFJMAFADRRWSOFSBCNUVQHFFBSAQXWPQCAC"
My thought process for the algorithm is as follows. I want to maintain a DP array whose length is the length of string a where a is the smaller of the input strings. dpA[i] would be the Longest Common Subsequence ending in a[i]. To do this I need to iterate through string a from index 0 => length-1 and see if a[i] exists in b. If a[i] exists in b it will be at position pos.
First mark dp[i] as 1 if dp[i] was 0
To know that a[i] is an extension of an existing subsequence we must go through a and find the first character behind i that matches a value in b behind pos. Let's call the indices of these matching values j and k respectively. This value is guaranteed to be a value we've seen before since we've covered all of a[0...i-1] and have filled out dpA[0...i-1]. When we find the first match, dpA[i] = dpA[j]+1 because we're extending the previous subsequence that ends in a[j]. Rinse repeat.
Obviously this method is not perfect or I wouldn't be asking this question, but I can't quite seem to see the problem with the algorithm. I've been looking at it so long I can hardly think about it anymore but any ideas on how to fix it would be greatly appreciated!
int longestCommonSubsequenceString(const string& x, const string& y) {
string a = (x.length() < y.length()) ? x : y;
string b = (x.length() >= y.length()) ? x : y;
vector<int> dpA(a.length(), 0);
int pos;
bool breakFlag = false;
for (int i = 0; i < a.length(); ++i) {
pos = b.find_last_of(a[i]);
if (pos != string::npos) {
if (!dpA[i]) dpA[i] = 1;
for (int j = i-1; j >= 0; --j) {
for (int k = pos-1; k >= 0; --k) {
if (a[j] == b[k]) {
dpA[i] = dpA[j]+1;
breakFlag = true;
break;
}
if (breakFlag) break;
}
}
}
breakFlag = false;
}
return *max_element(dpA.begin(), dpA.end());
}
EDIT
I think the complexity might actually be O(n*n*m)
I met a very simple interview question, but my solution is incorrect. Any helps on this? 1)any bugs in my solution? 2)any good idea for time complexity O(n)?
Question:
Given an int array A[], define X=A[i]+A[j]+(j-i), j>=i. Find max value of X?
My solution is:
int solution(vector<int> &A){
if(A.empty())
return -1;
long long max_dis=-2000000000, cur_dis;
int size = A.size();
for(int i=0;i<size;i++){
for(int j=i;j<size;j++){
cur_dis=A[j]+A[i]+(j-i);
if(cur_dis > max_dis)
max_dis=cur_dis;
}
}
return max_dis;
}
The crucial insight is that it can be done in O(n) only if you track where potentially useful values are even before you're certain they'll prove usable.
Start with best_i = best_j = max_i = 0. The first two track the i and j values to use in the solution. The next one will record the index with the highest contributing factor for i, i.e. where A[i] - i is highest.
Let's call the value of X for some values of i and j "Xi,j", and start by recording our best solution so far ala Xbest = X0,0
Increment n along the array...
whenever the value at [n] gives a better "i" contribution for A[i] - i than max_i, update max_i.
whenever using n as the "j" index yields Xmax_i,n greater than Xbest, best_i = max_i, best_j = n.
Discussion - why/how it works
j_random_hacker's comment suggests I sketch a proof, but honestly I've no idea where to start. I'll try to explain as best I can - if someone else has a better explanation please chip in....
Restating the problem: greatest Xi,j where j >= i. Given we can set an initial Xbest of X0,0, the problem is knowing when to update it and to what. As we contemplate successive indices in the array as potential values for j, we want to generate Xi,j=n for some i (discussed next) to compare with Xbest. But, what i value to use? Well, given any index from 0 to n is <= j, the j >= i constraint isn't relevant if we pick the best i value from the indices we've already visited. We work out the best i value by separating the i-related contribution to X from the j-related contribution - A[i] - i - so in preparation for considering whether we've a new best solution with j=n we must maintain the best_i variable too as we go.
A way to approach the problem
For whatever it's worth - when I was groping around for a solution, I wrote down on paper some imaginary i and j contributions that I could see covered the interesting cases... where Ci and Cj are the contributions related to n's use as i and j respectively, something like
n 0 1 2 3 4
Ci 4 2 8 3 1
Cj 12 4 3 5 9
You'll notice I didn't bother picking values where Ci could be A[i] - i while Cj was A[j] + j... I could see the emerging solution should work for any formulas, and that would have just made it harder to capture the interesting cases. So - what's the interesting case? When n = 2 the Ci value is higher than anything we've seen in earlier elements, but given only knowledge of those earlier elements we can't yet see a way to use it. That scenario is the single "great" complication of the problem. What's needed is a Cj value of at least 9 so Xbest is improved, which happens to come along when n = 4. If we'd found an even better Ci at [3] then we'd of course want to use that. best_i tracks where that waiting-on-a-good-enough-Cj value index is.
Longer version of my comment: what about iterating the array from both ends, trying to find the highest number, while decreasing it by the distance from the appripriate end. Would that find the correct indexes (and thus the correct X)?
#include <vector>
#include <algorithm>
#include <iostream>
#include <random>
#include <climits>
long long brutal(const std::vector<int>& a) {
long long x = LLONG_MIN;
for(int i=0; i < a.size(); i++)
for(int j=i; j < a.size(); j++)
x = std::max(x, (long long)a[i] + a[j] + j-i);
return x;
}
long long smart(const std::vector<int>& a) {
if(a.size() == 0) return LLONG_MIN;
long long x = LLONG_MIN, y = x;
for(int i = 0; i < a.size(); i++)
x = std::max(x, (long long)a[i]-i);
for(int j = 0; j < a.size(); j++)
y = std::max(y, (long long)a[j]+j);
return x + y;
}
int main() {
std::random_device rd;
std::uniform_int_distribution<int> rlen(0, 1000);
std::uniform_int_distribution<int> rnum(INT_MIN,INT_MAX);
std::vector<int> v;
for(int loop = 0; loop < 10000; loop++) {
v.resize(rlen(rd));
for(int i = 0; i < v.size(); i++)
v[i] = rnum(rd);
if(brutal(v) != smart(v)) {
std::cout << "bad" << std::endl;
return -1;
}
}
std::cout << "good" << std::endl;
}
I'll write in pseudo code because I don't have much time, but this should be the most performing way using recursion
compare(array, left, right)
val = array[left] + array[right] + (right - left);
if (right - left) > 1
val1 = compare(array, left, right-1);
val2 = compare(array, left+1, right);
val = Max(Max(val1,val2),val);
end if
return val
and than you call simply
compare(array,0,array.length);
I think I found a incredibly faster solution but you need to check it:
you need to rewrite your array as follow
Array[i] = array[i] + (MOD((array.lenght / 2) - i));
Then you just find the 2 highest value of the array and sum them, that should be your solution, almost O(n)
wait maybe I'm missing something... I have to check.
Ok you get the 2 highest value from this New Array, and save the positions i, and j. Then you need to calculate from the original array your result.
------------ EDIT
This should be an implementation of the method suggested by Tony D (in c#) that I tested.
int best_i, best_j, max_i, currentMax;
best_i = 0;
best_j = 0;
max_i = 0;
currentMax = 0;
for (int n = 0; n < array.Count; n++)
{
if (array[n] - n > array[max_i] - max_i) max_i = n;
if (array[n] + array[max_i] - (n - max_i) > currentMax)
{
best_i = max_i;
best_j = n;
currentMax = array[n] + array[max_i] - (n - max_i);
}
}
return currentMax;
Question:
Given an int array A[], define X=A[i]+A[j]+(j-i), j>=i. Find max value of X?
Answer O(n):
lets rewrite the formula: X = A[i]-i + A[j]+j
we can track the highest A[i]-i we got and the highest A[j]+j we got. We loop over the array once and update both of our max values. After looping once we return the sum of A[i]-i + A[j]+j, which equals X.
We absolutely don't care about the j>=i constraint, because it is always true when we maximize both A[i]-i and A[j]+j
Code:
int solution(vector<int> &A){
if(A.empty()) return -1;
long long max_Ai_part =-2000000000;
long long max_Aj_part =-2000000000;
int size = A.size();
for(int i=0;i<size;i++){
if(max_Ai_part < A[i] - i)
max_Ai_part = A[i] - i;
if(max_Aj_part < A[j] + j)
max_Ai_part = A[j] - j;
}
return max_Ai_part + max_Aj_part;
}
Bonus:
most people get confused with the j>=i constraint. If you have a feeling for numbers, you should be able to see that i should tend to be lower than j.
Assume we have our formula, it is maximized and i > j. (this is impossible, but lets check it out)
we define x1 := j-i and x2 = i-j
A[i]+A[j]+j-i = A[i]+A[j] + x1, x1 < 0
we could then swap i with j and end up with this:
A[j]+A[i]+i-j = A[i]+A[j] + x2, x2 > 0
it is basically the same formula, but now because i > j the second formula will be greater than the first. In other words we could increase the maximum by swapping i and j which can't be true if we already had the maximum.
If we ever find a maximum, i cannot be greater than j.
int s_dynamic(int n,int k) {
int maxj = n-k;
int *arr = new int[maxj+1];
for (int i = 0; i <= maxj; ++i)
arr[i] = 1;
for (int i = 1; i <= k; ++i)
for(int j = 1; j <= maxj; ++j)
arr[j] += i*arr[j-1];
return arr[maxj];
}
Here's my attempt at determining Stirling numbers using Dynamic Programming.
It is defined as follows:
S(n,k) = S(n-1,k-1) + k S(n-1,k), if 1 < k < n
S(n,k) = 1, if k=1 ou k=n
Seems ok, right? Except when I run my unit test...
partitioningTest ..\src\Test.cpp:44 3025 == s_dynamic(9,3) expected: 3025 but was: 4414
Can anyone see what I'm doing wrong?
Thanks!
BTW, here's the recursive solution:
int s_recursive(int n,int k) {
if (k == 1 || k == n)
return 1;
return s_recursive(n-1,k-1) + k*s_recursive(n-1,k);
}
Found the bug.
You already computed your dynamic array of Stirlings numbers for k=1 (S(n,1)=1 for all n).
You should start computing S(n,2) - that is:
for (int i = 2; i <= k; ++i) //i=2, not 1
for(int j = 1; j <= maxj; ++j)
arr[j] += i*arr[j-1];
Your approach is just fine, except you seem to have made a simple indexing error. If you think about what indexes i and j represent, and what the inner loop transforms arr[j] to, you'll see it easy enough (I lie, it took me a good half hour to figure out what was what :)).
From what I can decode, i represents the value k during calculations, and arr[j] is transformed from S(i+j, i-1) to S(i+1+j, i). The topmost for loop that initializes arr sets it up as S(1+j, 1). According to these loops, the calculations look just fine. Except for one thing: The very first i loop assumes that arr[j] contains S(0+j, 0), and so it is where your problem lies. If you change the starting value of i from 1 to 2, all should be OK (you may need an if or two for edge cases). The initial i=2 will transform arr[j] from S(1+j, 1) to S(2+j, 2), and the rest of the transformations will be just fine.
Alternatively, you could have initialized arr[j] to S(0+j, 0) if it were defined, but unfortunately, Stirling's numbers are undefined at k=0.
EDIT: Apparently I was wrong in my last comment. If you initialize arr to {1, 0, 0, ...}, you can leave starting value of i as 1. For this, you use the initial values S(0, 0)=1, and S(n, 0)=0, n>0 instead.