I am studying some code to try to learn Swift 3. I came across this for loop and can't decipher what exactly it is doing. If you understand it, could you try to break it down with an explanation?
var size = 0
var candidate = (value: 0, index: 0)
for i in 0..<count {
if size == 0 {
candidate = (A[i], i)
size += 1
} else {
if candidate.value != A[i] {
size -= 1
} else {
size += 1
}
}
}
// set variable "size" to the value of zero
var size = 0
// set variable "candidate" to be a NAMED TUPLE with initial values (0,0)
var candidate = (value: 0, index: 0)
// start a FOR loop from 0 to "count" which by the way has not been defined yet, so that might be a problem.
for i in 0..<count {
// if size is zero, then we set candidate equal to ARRAY "A" of index "i" and increment 'size'... and by the way, this Array "A" has not been defined anywhere, so that might be a problem.
if size == 0 {
candidate = (A[i], i)
size += 1
// otherwise, we do one of two things: 1) if the 'candidate' tuple's (value:) parameter is NOT equal to the value of Array 'A''s index of 'i', then we decrement 'size' ... by the way this Array 'A' has not yet been defined so that might be a problem... or 2) increment the value of 'size.'
} else {
if candidate.value != A[i] {
size -= 1
} else {
size += 1
}
}
}
So it looks like there should be an array somewhere called 'A'
and 'count' should really be A.count
Basically the loop is (seems to be trying to) going through the Array 'A' and comparing adjacent values and adjusting the 'size' variable as a result of each of these comparisons.
Related
Let's say I have a square boolean grid (2D array) of size N. Some of the values are true and some are false (the <true values> / <false values> ratio is unspecified). I want to randomly choose an indice (x, y) so that grid[x][y] is true. If I wanted a time-efficient solution, I'd do something like this (Python):
x, y = random.choice([(x, y) for x in range(N) for y in range(N) if grid[x][y]])
But this is O(N^2), which is more than sufficient for, say, a tic-tac-toe game implementation, but I'm guessing it would get much more memory-consuming for large N.
If I wanted something that's not memory consuming, I'd do:
x, y = 0, 0
t = N - 1
while True:
x = random.randint(0, t)
y = random.randint(0, t)
if grid[x][y]:
break
But the issue is, if I have a grid of size of order 10^4 and there is only one or two true values in it, it could take forever to "guess" which indice is the one I'm interested in. How should I go about making this algorithm optimal?
If the grid is static or doesn't change much, or you have time to do some preprocessing, you could store an array that holds the number of true values per row, the total number of true values, and a list of the non-zero rows (all of which you could keep updated if the grid changes):
grid per row
0 1 0 0 1 0 2
0 0 0 0 0 0 0
0 0 1 0 0 0 1
0 0 0 0 1 0 1
0 0 0 0 0 0 0
1 0 1 1 1 0 4
total = 8
non-zero rows: [0, 2, 3, 5]
To select a random index, choose a random value r up to the total number of true values, iterate over the array with the number of true values per non-zero row, adding them up until you know what row the r-th true value is in, and then iterate over that row to find the location of the r-th true value.
(You could simply pick a non-empty row first, and then pick a true value from that row, but that would create non-uniform probabilities.)
For an N×N-sized grid, the pre-processing would take N×N time and 2×N space, but the worst case look-up time would be N. In practice, using the JavaScript code example below, the pre-processing and look-up times (in ms) are in the order of:
grid size pre-processing look-up
10000 x 10000 5000 2.2
1000 x 1000 50 0.22
100 x 100 0.5 0.022
As you can see, look-up is more than 2000 times faster than pre-processing for a large grid, so if you need to randomly select several positions on the same (or slightly altered) grid, pre-processing makes a lot of sense.
function random2D(grid) {
this.grid = grid;
this.num = this.grid.map(function(elem) { // number of true values per row
return elem.reduce(function(sum, val) {
return sum + (val ? 1 : 0);
}, 0);
});
this.total = this.num.reduce(function(sum, val) { // total number of true values
return sum + val;
}, 0);
this.update = function(row, col, val) { // change value in grid
var prev = this.grid[row][col];
this.grid[row][col] = val;
if (prev ^ val) {
this.num[row] += val ? 1 : -1;
this.total += val ? 1 : -1;
}
}
this.select = function() { // select random index
var row = 0, col = 0;
var rnd = Math.floor(Math.random() * this.total) + 1;
while (rnd > this.num[row]) { // find row
rnd -= this.num[row++];
}
while (rnd) { // find column
if (this.grid[row][col]) --rnd;
if (rnd) ++col;
}
return {x: col, y: row};
}
}
var grid = [], size = 1000, prob = 0.01; // generate test data
for (var i = 0; i < size; i++) {
grid[i] = [];
for (var j = 0; j < size; j++) {
grid[i][j] = Math.random() < prob;
}
}
var rnd = new random2D(grid); // pre-process grid
document.write(JSON.stringify(rnd.select())); // get random index
Keeping a list of the rows which contain at least one true value only makes sense for very sparsely populated grids, where many rows contain no true values, so I haven't implemented it in the code example. If you do implement it, the look-up time for very sparse arrays is reduced to less than 1µs.
You can go with a dictionary implemented as a binary tree with logarithmic depth. This takes O(N^2) space and allows you to search/delete in O(log(N^2)) = O(logN) time. You can for example use Red-Black Tree.
The algorithm to find a random value might be:
t = tree.root
if (t == null)
throw Exception("No more values");
// logarithmic serach
while t.left != null or t.right != null
pick a random value k from range(0, 1, 2)
if (k == 0)
break;
if (k == 1)
if (t.left == null)
break
t = t.left
if (k == 2)
if (t.right == null)
break
t = t.right
result = t.value
// logarithmic delete
tree.delete(t)
return result
Of course, you can represent (i, j) indices as i * N + j.
Without additional memory you can't track changes to the state of cells. And in my opinion you can't get better than O(N^2) (iterating through the array).
LeetCode medium 120. Triangle (Dynamic Programming)
Question:
Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.
For example, given the following triangle
[
[2],
[3,4],
[6,5,7],
[4,1,8,3]
]
//The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11).
//Note:
//Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle.
I always get
fatal error: Can't form Range with end < start
on "for i in (row-1)...0".
Thank you so much! Appreciate your time!
class Solution
{
func minimumTotal(triangle: [[Int]]) -> Int
{
if triangle.count == 0
{
return 0
}
if triangle.count == 1
{
return triangle[0][0]
}
var arr = [Int](count: triangle.last!.count, repeatedValue: 0)
let row = triangle.count
for i in (row-1)...0
{
let col = triangle[i].count
for j in 0...col-1
{
if i == row-1
{
arr[i] = triangle[i][j]
continue
}
arr[j] = min(arr[j], arr[j+1]) + triangle[i][j]
}
}
return arr[0]
}
}
var test1 = Solution()
//var input = [[10]]
//var input = [[1],[2,3]]
var input = [[-1],[2,3],[1,-1,-3]]
var result = test1.minimumTotal(input)
print(result)
for in (0...row-1).reverse()
Swift can't read row-1...0
It's a bad idea to create a range where the start is higher than the end: your code will compile, but it will crash at runtime, so use stride instead of ranage
for i in (row-1).stride(to: 0, by: 1) { }
Imagine you have 3 buckets, but each of them has a hole in it. I'm trying to fill a bath tub. The bath tub has a minimum level of water it needs and a maximum level of water it can contain. By the time you reach the tub with the bucket it is not clear how much water will be in the bucket, but you have a range of possible values.
Is it possible to adequately fill the tub with water?
Pretty much you have 3 ranges (min,max), is there some sum of them that will fall within a 4th range?
For example:
Bucket 1 : 5-10L
Bucket 2 : 15-25L
Bucket 3 : 10-50L
Bathtub 100-150L
Is there some guaranteed combination of 1 2 and 3 that will fill the bathtub within the requisite range? Multiples of each bucket can be used.
EDIT: Now imagine there are 50 different buckets?
If the capacity of the tub is not very large ( not greater than 10^6 for an example), we can solve it using dynamic programming.
Approach:
Initialization: memo[X][Y] is an array to memorize the result. X = number of buckets, Y = maximum capacity of the tub. Initialize memo[][] with -1.
Code:
bool dp(int bucketNum, int curVolume){
if(curVolume > maxCap)return false; // pruning extra branches
if(curVolume>=minCap && curVolume<=maxCap){ // base case on success
return true;
}
int &ret = memo[bucketNum][curVolume];
if(ret != -1){ // this state has been visited earlier
return false;
}
ret = false;
for(int i = minC[bucketNum]; i < = maxC[bucketNum]; i++){
int newVolume = curVolume + i;
for(int j = bucketNum; j <= 3; j++){
ret|=dp(j,newVolume);
if(ret == true)return ret;
}
}
return ret;
}
Warning: Code not tested
Here's a naïve recursive solution in python that works just fine (although it doesn't find an optimal solution):
def match_helper(lower, upper, units, least_difference, fail = dict()):
if upper < lower + least_difference:
return None
if fail.get((lower,upper)):
return None
exact_match = [ u for u in units if u['lower'] >= lower and u['upper'] <= upper ]
if exact_match:
return [ exact_match[0] ]
for unit in units:
if unit['upper'] > upper:
continue
recursive_match = match_helper(lower - unit['lower'], upper - unit['upper'], units, least_difference)
if recursive_match:
return [unit] + recursive_match
else:
fail[(lower,upper)] = 1
return None
def match(lower, upper):
units = [
{ 'name': 'Bucket 1', 'lower': 5, 'upper': 10 },
{ 'name': 'Bucket 2', 'lower': 15, 'upper': 25 },
{ 'name': 'Bucket 3', 'lower': 10, 'upper': 50 }
]
least_difference = min([ u['upper'] - u['lower'] for u in units ])
return match_helper(
lower = lower,
upper = upper,
units = sorted(units, key = lambda u: u['upper']),
least_difference = min([ u['upper'] - u['lower'] for u in units ]),
)
result = match(100, 175)
if result:
lower = sum([ u['lower'] for u in result ])
upper = sum([ u['upper'] for u in result ])
names = [ u['name'] for u in result ]
print lower, "-", upper
print names
else:
print "No solution"
It prints "No solution" for 100-150, but for 100-175 it comes up with a solution of 5x bucket 1, 5x bucket 2.
Assuming you are saying that the "range" for each bucket is the amount of water that it may have when it reaches the tub, and all you care about is if they could possibly fill the tub...
Just take the "max" of each bucket and sum them. If that is in the range of what you consider the tub to be "filled" then it can.
Updated:
Given that buckets can be used multiple times, this seems to me like we're looking for solutions to a pair of equations.
Given buckets x, y and z we want to find a, b and c:
a*x.min + b*y.min + c*z.min >= bathtub.min
and
a*x.max + b*y.max + c*z.max <= bathtub.max
Re: http://en.wikipedia.org/wiki/Diophantine_equation
If bathtub.min and bathtub.max are both multiples of the greatest common divisor of a,b and c, then there are infinitely many solutions (i.e. we can fill the tub), otherwise there are no solutions (i.e. we can never fill the tub).
This can be solved with multiple applications of the change making problem.
Each Bucket.Min value is a currency denomination, and Bathtub.Min is the target value.
When you find a solution via a change-making algorithm, then apply one more constraint:
sum(each Bucket.Max in your solution) <= Bathtub.max
If this constraint is not met, throw out this solution and look for another. This will probably require a change to a standard change-making algorithm that allows you to try other solutions when one is found to not be suitable.
Initially, your target range is Bathtub.Range.
Each time you add an instance of a bucket to the solution, you reduce the target range for the remaining buckets.
For example, using your example buckets and tub:
Target Range = 100..150
Let's say we want to add a Bucket1 to the candidate solution. That then gives us
Target Range = 95..140
because if the rest of the buckets in the solution total < 95, then this Bucket1 might not be sufficient to fill the tub to 100, and if the rest of the buckets in the solution total > 140, then this Bucket1 might fill the tub over 150.
So, this gives you a quick way to check if a candidate solution is valid:
TargetRange = Bathtub.Range
foreach Bucket in CandidateSolution
TargetRange.Min -= Bucket.Min
TargetRange.Max -= Bucket.Max
if TargetRange.Min == 0 AND TargetRange.Max >= 0 then solution found
if TargetRange.Min < 0 or TargetRange.Max < 0 then solution is invalid
This still leaves the question - How do you come up with the set of candidate solutions?
Brute force would try all possible combinations of buckets.
Here is my solution for finding the optimal solution (least number of buckets). It compares the ratio of the maximums to the ratio of the minimums, to figure out the optimal number of buckets to fill the tub.
private static void BucketProblem()
{
Range bathTub = new Range(100, 175);
List<Range> buckets = new List<Range> {new Range(5, 10), new Range(15, 25), new Range(10, 50)};
Dictionary<Range, int> result;
bool canBeFilled = SolveBuckets(bathTub, buckets, out result);
}
private static bool BucketHelper(Range tub, List<Range> buckets, Dictionary<Range, int> results)
{
Range bucket;
int startBucket = -1;
int fills = -1;
for (int i = buckets.Count - 1; i >=0 ; i--)
{
bucket = buckets[i];
double maxRatio = (double)tub.Maximum / bucket.Maximum;
double minRatio = (double)tub.Minimum / bucket.Minimum;
if (maxRatio >= minRatio)
{
startBucket = i;
if (maxRatio - minRatio > 1)
fills = (int) minRatio + 1;
else
fills = (int) maxRatio;
break;
}
}
if (startBucket < 0)
return false;
bucket = buckets[startBucket];
tub.Maximum -= bucket.Maximum * fills;
tub.Minimum -= bucket.Minimum * fills;
results.Add(bucket, fills);
return tub.Maximum == 0 || tub.Minimum <= 0 || startBucket == 0 || BucketHelper(tub, buckets.GetRange(0, startBucket), results);
}
public static bool SolveBuckets(Range tub, List<Range> buckets, out Dictionary<Range, int> results)
{
results = new Dictionary<Range, int>();
buckets = buckets.OrderBy(b => b.Minimum).ToList();
return BucketHelper(new Range(tub.Minimum, tub.Maximum), buckets, results);
}
This is an interview question: "How to build a distributed algorithm to compute the balance of the parentheses ?"
Usually he balance algorithm scans a string form left to right and uses a stack to make sure that the number of open parentheses always >= the number of close parentheses and finally the number of open parentheses == the number of close parentheses.
How would you make it distributed ?
You can break the string into chunks and process each separately, assuming you can read and send to the other machines in parallel. You need two numbers for each string.
The minimum nesting depth achieved relative to the start of the string.
The total gain or loss in nesting depth across the whole string.
With these values, you can compute the values for the concatenation of many chunks as follows:
minNest = 0
totGain = 0
for p in chunkResults
minNest = min(minNest, totGain + p.minNest)
totGain += p.totGain
return new ChunkResult(minNest, totGain)
The parentheses are matched if the final values of totGain and minNest are zero.
I would apply the map-reduce algorithm in which the map function would compute a part of the string return either an empty string if parentheses are balanced or a string with the last parenthesis remaining.
Then the reduce function would concatenate the result of two returned strings by map function and compute it again returning the same result than map. At the end of all computations, you'd either obtain an empty string or a string containing the un-balanced parenthesis.
I'll try to have a more detailed explain on #jonderry's answer. Code first, in Scala
def parBalance(chars: Array[Char], chunkSize: Int): Boolean = {
require(chunkSize > 0, "chunkSize must be greater than 0")
def traverse(from: Int, until: Int): (Int, Int) = {
var count = 0
var stack = 0
var nest = 0
for (n <- from until until) {
val cur = chars(c)
if (cur == '(') {
count += 1
stack += 1
}
else if (cur == ')') {
count -= 1
if (stack > 0) stack -= 1
else nest -= 1
}
}
(nest, count)
}
def reduce(from: Int, until: Int): (Int, Int) = {
val m = (until + from) / 2
if (until - from <= chunkSize) {
traverse(from, until)
} else {
parallel(reduce(from, m), reduce(m, until)) match {
case ((minNestL, totGainL), (minNestR, totGainR)) => {
((minNestL min (minNestR + totGainL)), (totGainL + totGainR))
}
}
}
}
reduce(0, chars.length) == (0,0)
}
Given a string, if we remove balanced parentheses, what's left will be in a form )))(((, give n for number of ) and m for number of (, then m >= 0, n <= 0(for easier calculation). Here n is minNest and m+n is totGain. To make a true balanced string, we need m+n == 0 && n == 0.
In a parallel operation, how to we derive those for node from it's left and right? For totGain we just needs to add them up. When calculating n for node, it can just be n(left) if n(right) not contribute or n(right) + left.totGain whichever is smaller.
I have an array (of 9 elements, say) which I must treat as a (3 by 3) square.
For the sake of simplifying the question, this is a one-based array (ie, indexing starts at 1 instead of 0).
My goal is to determine valid adjacent squares relative to a starting point.
In other words, how it's stored in memory: 1 2 3 4 5 6 7 8 9
How I'm treating it:
7 8 9
4 5 6
1 2 3
I already know how to move up and down and test for going out of bounds (1 >= current_index <= 9)
edit: I know the above test is overly general but it's simple and works.
//row_size = 3, row_step is -1, 0 or 1 depending on if we're going left,
//staying put or going right respectively.
current_index += (row_size * row_step);
How do I test for an out of bounds condition when going left or right? Conceptually I know it involves determining if 3 (for example) is on the same row as 4 (or if 10 is even within the same square as 9, as an alternate example, given that multiple squares are in the same array back to back), but I can't figure out how to determine that. I imagine there's a modulo in there somewhere, but where?
Thanks very much,
Geoff
Addendum:
Here's the resulting code, altered for use with a zero-based array (I cleaned up the offset code present in the project) which walks adjacent squares.
bool IsSameSquare(int index0, int index1, int square_size) {
//Assert for square_size != 0 here
return (!((index0 < 0) || (index1 < 0))
&& ((index0 < square_size) && (index1 < square_size)))
&& (index0 / square_size == index1 / square_size);
}
bool IsSameRow(int index0, int index1, int row_size) {
//Assert for row_size != 0 here
return IsSameSquare(index0, index1, row_size * row_size)
&& (index0 / row_size == index1 / row_size);
}
bool IsSameColumn(int index0, int index1, int row_size) {
//Assert for row_size != 0 here
return IsSameSquare(index0, index1, row_size * row_size)
&& (index0 % row_size == index1 % row_size);
}
//for all possible adjacent positions
for (int row_step = -1; row_step < 2; ++row_step) {
//move up, down or stay put.
int row_adjusted_position = original_position + (row_size * row_step);
if (!IsSameSquare(original_position, row_adjusted_position, square_size)) {
continue;
}
for (int column_step = -1; column_step < 2; ++column_step) {
if ((row_step == 0) & (column_step == 0)) { continue; }
//hold on to the position that has had its' row position adjusted.
int new_position = row_adjusted_position;
if (column_step != 0) {
//move left or right
int column_adjusted_position = new_position + column_step;
//if we've gone out of bounds again for the column.
if (IsSameRow(column_adjusted_position, new_position, row_size)) {
new_position = column_adjusted_position;
} else {
continue;
}
} //if (column_step != 0)
//if we get here we know it's safe, do something with new_position
//...
} //for each column_step
} //for each row_step
This is easier if you used 0-based indexing. These rules work if you subtract 1 from all your indexes:
Two indexes are in the same square if (a/9) == (b/9) and a >= 0 and b >= 0.
Two indexes are in the same row if they are in the same square and (a/3) == (b/3).
Two indexes are in the same column if they are in the same square and (a%3) == (b%3).
There are several way to do this, I'm choosing a weird one just for fun. Use modulus.
Ase your rows are size 3 just use modulus of 3 and two simple rules.
If currPos mod 3 = 0 and (currPos+move) mod 3 = 1 then invalid
If currPos mod 3 = 1 and (currPos+move) mod 3 = 0 then invalid
this check for you jumping two a new row, you could also do one rule like this
if (currPos mod 3)-((currPos+move) mod 3)> 1 then invalid
Cheers
You should be using a multidimensional array for this.
If your array class doesn't support multidimensional stuff, you should write up a quick wrapper that does.