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yes..HasMap will contain 26 letters key and each key will contain the linklist ,having name and phone number in each node, starting with key character in Alphabetically sorted.

Your Algorithm is absolutely correct.
The running code is:

#include<stdio.h>
#include<string.h>
int RegexMatch(char *S1, char* S2, int n1, int n2)
{
int ptr1 = 0; int ptr2 = 0;



while (ptr1 < n1 && ptr2 < n2)
{

if (S1[ptr1] == S2[ptr2])
{
ptr1++; ptr2++;
if (S2[ptr2] == '?')
ptr2++;
if (S2[ptr2] == '*')
{ ptr2++;
while(S1[ptr1-1] == S1[ptr1] && ptr1 < n1)
ptr1++;
}
}
else if (S2[ptr2+1] == '*')
{
ptr2 = ptr2+2;
}
else if (S2[ptr2+1] == '?')
{
ptr2 = ptr2+2;
}
else
{
return 0;
}
}
if (ptr1 < n1) return 0;
if (ptr2 < n2)
{
while(ptr2 < n2)
{
if (S2[ptr2] == '*' || S2[ptr2] == '?')
ptr2++;
else
{
if (S2[ptr2+1] == '?' || S2[ptr2+1] == '*')
ptr2 = ptr2 + 2;
else
return 0;
}
}
}
return 1;
}
int main()
{
char S1[]="aabcccd",S2[]="a*bc*d?e?";
int flag=RegexMatch(S1,S2,strlen(S1),strlen(S2));
if(!flag)
printf("Match Not Found\n");
else
printf("Match Found\n");
getch();
return 0;
}

Thanks Don!!! :)

Please give us some example for clarity and also tell us that in which case it will return true and in which case it will return false.

Here val1<va2 and
[1].treeNode* LowestCommonAncestor(treeNode *node,int val1,int val2);
[2]int hightCount(treeNode *node,int val);
[3]int findShortestLength(treeNode *node,int val1,int val2);
here we find the lowest common Ancestor and the find the hight of the left child found and then hight of right child found.After that we sum the hights.

treeNode* LowestCommonAncestor(treeNode *node,int val1,int val2)
{

if(!node)
return NULL ;
else if((val1<(node->data))&& (val2>(node->data)))
{
return node;
}
else if((val1>(node->data))&& (val2>(node->data)))
{
return(LowestCommonAncestor(node->right,val1,val2));
}
else if((val1<(node->data))&& (val2<(node->data)))
{

return (LowestCommonAncestor(node->left,val1,val2));
}
}
int hightCount(treeNode *node,int val)
{
int count;
if((!node)||(val==(node->data)))
return 0;
else if(val<(node->data))
{
count=(hightCount(node->left,val))+1;
}
else if(val>(node->data))
{
count=(hightCount(node->right,val))+1;
}
return count;
}
int findShortestLength(treeNode *node,int val1,int val2)
{
int leftLength=0,rightLength=0;
treeNode *LCA=LowestCommonAncestor(node,val1,val2);
leftLength=hightCount(LCA->left,val1)+1;
rightLength=hightCount(LCA->right,val2)+1;
return (leftLength+rightLength);
}