Skip to content

Commit

Permalink
Merge pull request #417 from Tech-neophyte/main
Browse files Browse the repository at this point in the history
 17/q1 added
  • Loading branch information
karishma-2020 authored Jan 17, 2024
2 parents 0578d22 + fa2d45f commit 3bd6179
Show file tree
Hide file tree
Showing 3 changed files with 100 additions and 0 deletions.
Original file line number Diff line number Diff line change
@@ -0,0 +1,33 @@
## Approach:
<br /> 1. Using dfs find the max value node and min value node in each branch and calculate their diffence.
<br /> 2. Return the max difference.
## Code:
```
/**
* 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:
int maxAncestorDiff(TreeNode* root) {
return dfs(root,root->val,root->val);
}
int dfs(TreeNode* root,int mini,int maxx){
if (root==NULL){
return maxx-mini;
}
mini = std::min(mini, root->val);
maxx = std::max(maxx, root->val);
int left_diff = dfs(root->left, mini, maxx);
int right_diff = dfs(root->right, mini, maxx);
return std::max(left_diff, right_diff);
}
};
```
27 changes: 27 additions & 0 deletions Day-17/Q1:Determine if two strings are close/Question.md
Original file line number Diff line number Diff line change
@@ -0,0 +1,27 @@
Two strings are considered close if you can attain one from the other using the following operations:
Operation 1: Swap any two existing characters.
For example, abcde -> aecdb
Operation 2: Transform every occurrence of one existing character into another existing character, and do the same with the other character.
For example, aacabb -> bbcbaa (all a's turn into b's, and all b's turn into a's)
You can use the operations on either string as many times as necessary.
Given two strings, word1 and word2, return true if word1 and word2 are close, and false otherwise.

### Example 1:
Input: word1 = "abc", word2 = "bca"
Output: true
Explanation: You can attain word2 from word1 in 2 operations.
Apply Operation 1: "abc" -> "acb"
Apply Operation 1: "acb" -> "bca"

### Example 2:
Input: word1 = "a", word2 = "aa"
Output: false
Explanation: It is impossible to attain word2 from word1, or vice versa, in any number of operations.

### Example 3:
Input: word1 = "cabbba", word2 = "abbccc"
Output: true
Explanation: You can attain word2 from word1 in 3 operations.
Apply Operation 1: "cabbba" -> "caabbb"
Apply Operation 2: "caabbb" -> "baaccc"
Apply Operation 2: "baaccc" -> "abbccc"
40 changes: 40 additions & 0 deletions Day-17/Q1:Determine if two strings are close/Tech-neophyte--c.md
Original file line number Diff line number Diff line change
@@ -0,0 +1,40 @@
## Approach:
<br />1. Get the freq of each unique character in word1 and word2.
<br />2. Check if all the char present in word1 are present in word2. ( word2 has no unique char).
<br />3. Sort the frequencies and check if the sorted frequencies match.
## C++ code:
```
class Solution {
public:
bool closeStrings(string word1, string word2) {
if (word1.size() != word2.size()) {
return false;
}
std::unordered_map<char, int> freq1;
for (char c : word1) {
freq1[c]++;
}
std::unordered_map<char, int> freq2;
for (char c : word2) {
freq2[c]++;
}
for (char c : word1) {
if (freq2.find(c) == freq2.end()) {
return false;
}
}
std::vector<int> freqVec1, freqVec2;
for (const auto& pair : freq1) {
freqVec1.push_back(pair.second);
}
for (const auto& pair : freq2) {
freqVec2.push_back(pair.second);
}
std::sort(freqVec1.begin(), freqVec1.end());
std::sort(freqVec2.begin(), freqVec2.end());
return freqVec1 == freqVec2;
}
};
```

0 comments on commit 3bd6179

Please sign in to comment.