## igor grudenic

BAN USER<pre lang="" line="1" title="CodeMonkey10275" class="run-this">Simple brute force would be to keep all the previous solutions and then iterate over

them to find the minimal one that multiplied by 2 or 5 gives the number greater

then the current element. History search should start at the latest number and continue by decreasing solutions until the first one found and multiplied by 5 is lesser then current element. Complexity is linear space, ant time is certainly better than O(n^2) but one should prove that limited history search is better than O(n) (and it is).

void GetElementsSimple(int n){

list<int> solutions;

int curSolution;

int bestNext;

solutions.push_back(1);

while(n){

curSolution=solutions.back();

cout<<solutions.back()<<" ";

bestNext=curSolution*2;

for(list<int>::reverse_iterator i=solutions.rbegin();

i!=solutions.rend()&& (*i)*5>curSolution;

i++){

if(((*i)*5>curSolution)&&((*i)*5<bestNext)) bestNext=(*i)*5;

if(((*i)*2>curSolution)&&((*i)*2<bestNext)) bestNext=(*i)*2;

}

solutions.push_back(bestNext);

n--;

}

}

Additionally one may try to exercise O(n) solution that takes a numeber 2^i*5^j

and tries to create 2 exchanges. First is to convert a subset of 2^i to 2^(i-n)*5^n. The second is to convert 5^j to 2^m*5^(j-m). Conversion that gets smaller solution must be used.

Optimal conversion for all parameters i and j can be precomputed. This leads to provable O(n) time after precomputation.

</pre><pre title="CodeMonkey10275" input="yes">

</pre>

<pre lang="" line="1" title="CodeMonkey59148" class="run-this">char* LongestPrefix(char **strings,int n){

int i,prefixLength,zeroLength;

char *result;

zeroLength=strlen(strings[0]);

for(prefixLength=0;prefixLength<zeroLength;prefixLength++){

for(i=1;i<n;i++){

if((strings[i][prefixLength]==0)||

(strings[i][prefixLength]!=strings[0][prefixLength]))

break;

}

if(i<n) break;

};

result=(char *)malloc(sizeof(*result)*(prefixLength+1));

strncpy(result,strings[0],prefixLength);

result[prefixLength]=0;

return result;

}</pre><pre title="CodeMonkey59148" input="yes">

</pre>

<pre lang="" line="1" title="CodeMonkey85370" class="run-this">I think this one can be somewhere between linear and n^2. Algorithm goes recursively and it looks into 2 pointers t1 and t2. Suppose t1->data < t2->data. If that holds then children of t1 and t2 point to data that belong to following subsets:

t1->left [-MAXINT,t1->data]

t1->right [t1->data+1,t2->data] U <t2->data,MAXINT]

t2->left [-MAXINT,t1->data] U <t1->data,t2->data]

t2->right <t2->data,MAXINT>

So basically we have to make following recursive calls

...

function(t1->left,t2->left) but limited to print numbers from [-MAXINT,t1->data]

cout<<t1->data;

function(t1->right,t2->left) but limited to print numbers from <a,t2->data]

cout<<t2->data;

function(t1->right,t2->right) but limited to print numbers from <t2->data,MAXINT]

What is the complexity of the solution? Well, for pair of nodes on the top level we create 3 recursive calls on topLevel+1. This leads to 3^level complexity. Since level is log2(n) we have 3^(log2(n))=n^(log2(3))=n^1.5849.

This can be improved by preventing recursive calls on subtrees than do not meet limits requirements, but analysis of this would be much more complicated then I am willing to do at the moment :-).

The entire code for the routine is here:

static void printSorted2TreesInternalNew(

TreeNode *t1,

TreeNode *t2,

int min,

int max){

if((t1==NULL)&&(t2==NULL)) return;

if(t1==NULL)

swapTreePointers(t1,t2);

if(t2==NULL){

printSorted2TreesInternalNew(t1->left_,t2,min,max);

std::cout<<t1->data_<<" ";

printSorted2TreesInternalNew(t1->right_,t2,min,max);

return;

}

if(t2->data_<t1->data_)

swapTreePointers(t1,t2);

printSorted2TreesInternalNew( t1->left_,

t2->left_,

min,

std::min(max,t1->data_));

if((t1->data_>min)&&(t1->data_<=max))

std::cout<<t1->data_<<" ";

printSorted2TreesInternalNew( t1->right_,

t2->left_,

std::max(min,t1->data_),

std::min(max,t2->data_));

if((t2->data_>=min)&&(t2->data_<=max))

std::cout<<t2->data_<<" ";

printSorted2TreesInternalNew( t1->right_,

t2->right_,

std::max(min,t2->data_),

max);

}

Of course you have to call it with min and max set to int range.

</pre><pre title="CodeMonkey85370" input="yes">

</pre>

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There is a recursive way to find all the solutions:

TrieNode is a node in a prefix tree. Node has a wordMark field that denotes whether set of

- igor grudenic December 08, 2011characters formed by visiting all the parent nodes in a tree constitute a valid word.

Method call newNode=oldNode.moveForward(c) returns new trie node reachable from oldNode

using character c.

Complexity of finding one solution is by O(N*C) (N-sentence length, C-lenght of the longest

dictionary word). Complexity of moveForward(c) is O(1).

Number of solutions may vary greatly depending on a sentence length and size of the

dictionary. For instance if one uses Morse alphabet the number of valid sentences may easily

become exponential.