day 18-q2 py, day21-q2, day 22-q1, 23-q1, 19-q3, 21-q3

Day-19/q3-Letter Combinations of a Phone Number/Tech-neophyte--p.md

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+## Approach:
+<br /> 1. Initialize a Mapping: Create a dictionary that maps each digit from 2 to 9 to their corresponding letters on a telephone's buttons.
+<br /> 2. Base Case: If the input string digits is empty, return an empty list.
+<br /> 3. Iteratively make Combinations: Start with an empty combination in a list and iteratively build the combinations by processing each digit in the input string.
+For each existing combination, append each corresponding letter for the current digit, building new combinations.
+<br /> 4. Result: Return the generated combinations as the final result.
+## Python code:
+```
+class Solution:
+    def letterCombinations(self, digits: str) -> List[str]:
+        if not digits:
+            return []
+
+        phone_map = {
+            '2': 'abc',
+            '3': 'def',
+            '4': 'ghi',
+            '5': 'jkl',
+            '6': 'mno',
+            '7': 'pqrs',
+            '8': 'tuv',
+            '9': 'wxyz'
+        }
+        combinations=[""]
+        for i in digits:
+            new_comb=[]
+            for j  in combinations:
+                for l in phone_map[i]:
+                    new_comb.append(j+l)
+            combinations=new_comb
+        return combinations
+
+
+```

Day-21/q2 : Container With Most Water/Tech-neophyte--c.md

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+## Approach:
+<br />1. Initiate two pointers, one from start (left) and one from end (right).
+<br />2. While left<right find the max area possible by inc left and dec right wherever necessary.
+## cpp code: 
+```
+class Solution {
+public:
+    int maxArea(vector<int>& height) {
+        int left = 0;
+        int right = height.size() - 1;
+        int maxArea = 0;
+
+        while (left < right) {
+            int currentArea = min(height[left], height[right]) * (right - left);
+            maxArea = max(maxArea, currentArea);
+
+            if (height[left] < height[right]) {
+                left++;
+            } else {
+                right--;
+            }
+        }
+
+        return maxArea;
+    }
+};
+```

Day-21/q3: Unique Paths/Tech-neophyte--pc.md

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+## Python code:
+```
+class Solution:
+    def uniquePaths(self, m: int, n: int) -> int:
+        row = [1]*n
+
+        for i in range(m-1):
+            newRow = [1]*n
+
+            for j in range(n-2, -1, -1):
+                newRow[j] = newRow[j+1] + row[j]
+            row = newRow 
+
+        return row[0]
+```
+## cpp code: Using dp
+```
+class Solution {
+public:
+    int uniquePaths(int m, int n) {
+
+        vector<vector<int>> dp(m, vector<int>(n, 0));
+
+
+        for (int i = 0; i < m; ++i) {
+            dp[i][0] = 1;
+        }
+        for (int j = 0; j < n; ++j) {
+            dp[0][j] = 1;
+        }
+
+
+        for (int i = 1; i < m; ++i) {
+            for (int j = 1; j < n; ++j) {
+                dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
+            }
+        }
+
+
+        return dp[m - 1][n - 1];
+    }
+};
+```
+## Cpp code: Using combinations:
+```
+class Solution {
+public:
+    int nCr(int n, int r) {
+        if (r > n) return 0;
+        if (r == 0 || n == r) return 1;
+        double res = 0;
+        for (int i = 0; i < r; i++) {
+            res += log(n-i) - log(i+1);
+        }
+        return (int)round(exp(res));
+    }
+
+    int uniquePaths(int m, int n) {
+        int t = (m-1) + (n-1);
+
+           return nCr(t,m-1);
+
+    }
+};
+```  

Day-22/q1: Minimum Number of Operations to Make Array Empty/Tech-neophyte--c.md

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+## Approach:
+
+<br /> 1. Create a hashmap object to count the occurrences of each element in the given array nums. Initialize a variable ans = 0 to keep track of the minimum number of operations required.
+<br /> 2. Check if frequency is equal to 1. If yes, return -1, as it is not possible to perform the required operations on a single element.
+Else increment the answer ans by the ceiling division of c by 3.
+<br />3. After iterating through all counts in the Counter, return the final value of ans as the minimum number of operations required to empty the array.
+```
+class Solution {
+public:
+    int minOperations(vector<int>& nums) 
+    {
+        unordered_map<int, int> counter;
+        int ans = 0;
+        for (int i=0;i<nums.size();i++)
+            counter[nums[i]]++;
+        for (auto it : counter)
+        {
+            if (it.second == 1)
+                return -1;
+            ans = ans + ceil( (double)(it.second) / 3) ;
+        }
+        return ans;
+    }
+};
+```