# Daily Coding Challenge #64

This is a series of Daily Coding Challenge. Each day I show a few solutions written in C++. The questions are from coding practice/contest sites such as HackerRank, LeetCode, Codeforces, Atcoder and etc.

/*
Goat Latin

A sentence S is given, composed of words separated by spaces. Each word consists of lowercase and uppercase letters only.

We would like to convert the sentence to "Goat Latin" (a made-up language similar to Pig Latin.)

The rules of Goat Latin are as follows:

If a word begins with a vowel (a, e, i, o, or u), append "ma" to the end of the word.
For example, the word 'apple' becomes 'applema'.

If a word begins with a consonant (i.e. not a vowel), remove the first letter and append it to the end, then add "ma".
For example, the word "goat" becomes "oatgma".

Add one letter 'a' to the end of each word per its word index in the sentence, starting with 1.
For example, the first word gets "a" added to the end, the second word gets "aa" added to the end and so on.
Return the final sentence representing the conversion from S to Goat Latin.

Example 1:

Input: "I speak Goat Latin"
Output: "Imaa peaksmaaa oatGmaaaa atinLmaaaaa"
Example 2:

Input: "The quick brown fox jumped over the lazy dog"
Output: "heTmaa uickqmaaa rownbmaaaa oxfmaaaaa umpedjmaaaaaa overmaaaaaaa hetmaaaaaaaa azylmaaaaaaaaa ogdmaaaaaaaaaa"

Notes:

S contains only uppercase, lowercase and spaces. Exactly one space between each word.
1 <= S.length <= 150.
*/

class Solution {
public:
string toGoatLatin(string S) {
string ans,s,a="a";
istringstream ss(S);
while(ss>>s){
char f=s;
ans+=' ';
if(f=='a'||f=='e'||f=='i'||f=='o'||f=='u'||f=='A'||f=='E'||f=='I'||f=='O'||f=='U'){
// If a word begins with a vowel (a, e, i, o, or u), append "ma" to the end of the word.
// For example, the word 'apple' becomes 'applema'.
ans+=s+"ma";
}else{
// If a word begins with a consonant (i.e. not a vowel),
// remove the first letter and append it to the end, then add "ma".
ans+=s.substr(1)+f+"ma";
}
// Add one letter 'a' to the end of each word per its word index in the sentence, starting with 1.
ans+=a;
a+="a";
}
return ans.substr(1);
}
};


/*
Same Tree

Given two binary trees, write a function to check if they are the same or not.

Two binary trees are considered the same if they are structurally identical and the nodes have the same value.

Example 1:

Input:     1         1
/ \       / \
2   3     2   3

[1,2,3],   [1,2,3]

Output: true
Example 2:

Input:     1         1
/           \
2             2

[1,2],     [1,null,2]

Output: false
Example 3:

Input:     1         1
/ \       / \
2   1     1   2

[1,2,1],   [1,1,2]

Output: false
*/

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
bool isSameTree(TreeNode* p, TreeNode* q) {
// if both null, that is same
if(p==NULL&&q==NULL) return true;
// if either one is null, that is not same
if(p==NULL||q==NULL) return false;
// if either one has different value, that is not same
if(p->val!=q->val) return false;
// traverse left and right branches
return isSameTree(p->left,q->left) && isSameTree(p->right, q->right);
}
};


The source code is available in corresponding repo below. Star and watch for timely updates!

## wingkwong / atcoder

### Discussion   