17/q1 added

Day 18/q1: Maximum Difference Between Node and Ancestor/Tech-neophyte--c.md

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+## 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);
+    }
+};
+```

Day-17/Q1:Determine if two strings are close/Question.md

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+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"

Day-17/Q1:Determine if two strings are close/Tech-neophyte--c.md

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+## 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;
+    }
+};
+```