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Problem statement =>
You are given queries. Each query consists of a single number N. You can perform any of the 2 operations on in each move:
1: If we take 2 integers a and b where N=a*b (a>1,b>1), then we can change N=max(a,b).
2: Decrease the value of N by 1.
Determine the minimum number of moves required to reduce the value of N to 0.
here is the link for better understanding.
https://www.hackerrank.com/challenges/down-to-zero-ii/problem
I know here are some overlapping sub-problems and we can use DP to ignore the computation of same sub-problems again and again.
Now, my question is how in this problem, same sub-problems have same solutions. Because we have to solve this from top to bottom and sub-problem have same solution if we solved them from bottom to top.
For example
N=4
1 possibility = 4->3->2->1->0
2 possibility = 4->2->1->0
Now in above two possibility, 2 is repeating and I can use DP, but how I store their values. I mean, in 1 possibility solution of 2 is different from 2nd possibility because in first one I've to traverse 4->3->2 here solution of 2 is 2 and in 2nd possibility we traverse 4->2 and solution of 2 here is 1 now these 2 same sub-problems have different values because of the solving from top to bottom. Now I'm totally confused here. Please someone help me out in this.
The solution for a number N should store the minimun steps required to make it 0
this is how the sol should look
int dp[1000001];
memset(dp,-1,sizeof(dp);
int sol(N){
if(N == 2){
return 2;
}
if(dp[n]!=-1){
return dp[n]'
}
int sol = 1+sol(min(move1,move2));
dp[n] = sol ;
return sol;
}
EDIT 2:
I think this is a solution for your problem. The solution is in JavaScript:
// ****************************************************************************
function findPaths(tree, depth = 0, path = [], paths = [-1, []]) {
const [node, children] = tree
path.push(node)
if (!children) {
// console.log(path, depth)
if (paths[0] === -1 || paths[0] > depth) {
paths[0] = depth
paths[1] = [paths.length]
} else if (paths[0] === depth) {
paths[1].push(paths.length)
}
paths.push([...path])
path.pop()
return
}
children.forEach((el) => {
findPaths(el, depth + 1, path, paths)
})
path.pop()
return paths
}
// ****************************************************************************
function downToZero(n) {
const tree = [n]
const divisors = []
for (let i = 2; i <= Math.sqrt(n); i++) {
if (n % i == 0) {
divisors.push(Math.max(i, n / i))
}
}
if (divisors.length) {
tree.push(divisors.map(downToZero))
} else if (n > 0) {
tree.push([downToZero(n - 1)])
}
return tree
}
// ****************************************************************************
function printPaths(paths) {
console.log('Total number of solutions:', paths.length - 2)
console.log('Total number of solutions with minimal moves:', paths[1].length)
console.log('Minimal moves:', paths[0])
paths[1].forEach((pathIndex) => {
let printPath = ''
paths[pathIndex].forEach((element) => {
printPath = `${printPath}${printPath === '' ? '' : '->'}${element}`
})
console.log(printPath)
})
console.log('')
}
// ****************************************************************************
// Test
printPaths(findPaths(downToZero(812849)))
printPaths(findPaths(downToZero(100)))
printPaths(findPaths(downToZero(19)))
printPaths(findPaths(downToZero(4)))
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) { }
I have a collection of arrays that include numbers from 1 to 10. The size of each array is 5. For example
[1,2,3,4,5]
[3,4,9,8,2]
[1,5,7,9,2]
[1,2,5,9,7]
...........
[3,8,1,6,9]
What algorithm should I use to find repeated triads in these arrays?
For example one of the results should be 1,2,5 since this triad is included in some of the arrays. I don't mind how many times some triad is repeated. I am looking the n most often (could be 3 or 4 or something else).
[1,2,3] is the same with [3,1,2] and each number is allowed only once. [3,3,4] is not valid.
This problem gets harder if we assume arrays of 10 or more numbers, so that each array could have a combination of triads. Just food for thought
[1,3,5,19,45,11,12,13,9,31]
[1,3,5,32,46,15,12,18,29,37]
result : (1,3,5) (1,3,12) (3,5,12) etc
I have completely reviewed my response :
**Bugs fixed** in : `array function computeRepettition(array $a);`
**Avoid** increment of repetition if triad was already found in pass-1
**Return** an array of arrays, and the number of repetition of each triad is set in '**numberOfRepetition**', the triad in self is the key of the array
**Support** number composed of 2 digits or more
**New** `array function iCombination(array $a);` reduce the number of probability for finding triad, since the order it is not important, and repetition of number is not allowed
**Update** of `array function detectRepetition(array $a);` detects all triad that can be found
<?php
define ("MIN_LENGTH_VECTOR" , 3 );
define ("KEY_SEPERATOR" , '-');
$src2D = array (
array(1,3,5,19,45,11,12,13,9, 100,31),
array(1,3,5,32,46,15,100, 12,18,29,37),
array(1222,32222,5,3222222,4622222,1522222,100, 12,182222,292222,372222));
var_dump (computeRepetition ($src2D));
function computeRepetition ($src2D) {
$repetition = array ();
for ($i=0 ; $i<count($src2D)-1 ; $i++) {
foreach ($repetition as &$rep) {
$rep['escape'] = TRUE;
}
for ($j=$i+1 ; $j<count($src2D) ; $j++) {
$t = buildTruth ($src2D[$i], $src2D[$j]);
$r = detectRepetition ($t);
if (is_null ($r)) continue;
$comb = iCombination ($r);
foreach ($comb as $cb) {
if (isset ($repetition[$cb]['escape']) && $repetition[$cb]['escape'] === TRUE) continue;
if (array_key_exists ($cb, $repetition)) {
$repetition[$cb]['numberOfRepetition']++;
} else {
$repetition[$cb]['numberOfRepetition'] = 2;
$repetition[$cb]['escape'] = FALSE;
}
}
}
}
return $repetition;
}
function detectRepetition ($t) {
$a = array ();
foreach ($t as $key => $value) {
if ($value === TRUE) {
$a[] = $key;
}
}
if (count($a) < MIN_LENGTH_VECTOR) return NULL;
return $a;
}
function iCombination ($array) {
$res = array ();
sort ($array, SORT_NUMERIC);
for ($i = 0 ; $i < count ($array) - 2 ; $i++) {
for ($k = $i + 1 ; $k < count ($array) - 1 ; $k++) {
for ($l = $k + 1 ; $l < count ($array) ; $l++) {
$res[] = $array[$i] . KEY_SEPERATOR . $array[$k] . KEY_SEPERATOR . $array[$l];
}
}
}
return $res;
}
function buildTruth ($vec1, $vec2) {
foreach ($vec1 as $v) {
$t[$v] = FALSE;
}
foreach ($vec2 as $v) {
if (isset ($t[$v]) && $t[$v] === FALSE ) $t[$v] = TRUE ;
}
return $t;
}
I'd go with brute force. Depending on how you define 'triad' (is [1,2,3] different than [3,2,1]? Is [3,3,3] admissible?) There are from C(10,3)=120 to 1000 possible triads, and each array generates C(5,3)=10 triads.
Prepare a (hash) table of counters for triads, iterate over the arrays updating the counters as you go, select triads with largest counts.
I'm working on a Texas Holdem game and i need to generate all possible k subsets from an Array of cards (represented as numbers in this example). This is how it looks so far:
public function getKSubsetsFromArray(arr:Array, k:int):Array {
var data:Array = new Array();
var result:Array = new Array();
combinations(arr, data, 0, arr.length - 1, 0, k, result, 0);
return result;
}
public function combinations(arr:Array, data:Array, start:int, end:int, index:int, r:int, resultArray:Array, resultIndex:int):int {
if (index == r) {
trace(resultIndex, data);
resultArray[resultIndex] = data;
return ++resultIndex;
}
for (var i:int = start; i<=end && end-i+1 >= r-index; i++) {
data[index] = arr[i];
resultIndex = combinations(arr, data, i + 1, end, index + 1, r, resultArray, resultIndex);
}
return resultIndex;
}
I am new to Actionscript, my idea is to have a function that takes an array of number and a parameter k, and returns an Array of arrays each of size k. However once i test the functions I get an array containing only the last combination nCk times. For example:
var testArray:Array = new Array(1, 2, 3, 4, 5);
trace(getKSubsetsFromArray(testArray, 3));
Returns:
0 1,2,3
1 1,2,4
2 1,2,5
3 1,3,4
4 1,3,5
5 1,4,5
6 2,3,4
7 2,3,5
8 2,4,5
9 3,4,5
The function output is
3,4,5,3,4,5,3,4,5,3,4,5,3,4,5,3,4,5,3,4,5,3,4,5,3,4,5,3,4,5
Of course it should print an array containing all the combinations listed before but it only prints the last one the right amount of times.
Thank your for your help.
The reason for the error is that when you are making array of arrays you are actually using the reference of the same array (data) so when the last combination is executed the contains of data array become 3,4,5 and each of index of resultArray points to data array so it prints out same values.
Solution :-
if (index == r) {
trace(resultIndex, data);
var result = new Array();
copy(result,data)
resultArray[resultIndex] = result;
return ++resultIndex;
}
Note :-
The above is pseudo code as i am not familiar with actionscript but you can implement copy function that copies values of data into result in actionscript syntax.
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);
}