Added tasks 34-51

g0001_0100/s0034_find_first_and_last_position_of_element_in_sorted_array/readme.md

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+34\. Find First and Last Position of Element in Sorted Array
+
+Medium
+
+Given an array of integers `nums` sorted in non-decreasing order, find the starting and ending position of a given `target` value.
+
+If `target` is not found in the array, return `[-1, -1]`.
+
+You must write an algorithm with `O(log n)` runtime complexity.
+
+**Example 1:**
+
+**Input:** nums = [5,7,7,8,8,10], target = 8
+
+**Output:** [3,4] 
+
+**Example 2:**
+
+**Input:** nums = [5,7,7,8,8,10], target = 6
+
+**Output:** [-1,-1] 
+
+**Example 3:**
+
+**Input:** nums = [], target = 0
+
+**Output:** [-1,-1] 
+
+**Constraints:**
+
+*   <code>0 <= nums.length <= 10<sup>5</sup></code>
+*   <code>-10<sup>9</sup> <= nums[i] <= 10<sup>9</sup></code>
+*   `nums` is a non-decreasing array.
+*   <code>-10<sup>9</sup> <= target <= 10<sup>9</sup></code>
+
+To solve this problem efficiently with a runtime complexity of O(log n), we can use binary search to find the starting and ending positions of the target value in the sorted array. Here are the steps:
+
+### Approach:
+
+1. **Binary Search for Starting Position:**
+   - Use binary search to find the starting position of the target value.
+   - When the target value is found, continue searching towards the left to find the leftmost occurrence.
+   - Update the left boundary to narrow down the search space.
+
+2. **Binary Search for Ending Position:**
+   - Use binary search to find the ending position of the target value.
+   - When the target value is found, continue searching towards the right to find the rightmost occurrence.
+   - Update the right boundary to narrow down the search space.
+
+3. **Handle Missing Target:**
+   - If the target value is not found, return [-1, -1].
+
+### Python Code:
+
+```python
+class Solution:
+    def searchRange(self, nums, target):
+        def search_boundary(nums, target, is_left):
+            left, right = 0, len(nums) - 1
+            boundary = -1
+
+            while left <= right:
+                mid = (left + right) // 2
+
+                if nums[mid] == target:
+                    boundary = mid
+                    if is_left:
+                        right = mid - 1
+                    else:
+                        left = mid + 1
+                elif nums[mid] < target:
+                    left = mid + 1
+                else:
+                    right = mid - 1
+
+            return boundary
+
+        left_boundary = search_boundary(nums, target, is_left=True)
+        right_boundary = search_boundary(nums, target, is_left=False)
+
+        return [left_boundary, right_boundary]
+
+# Example Usage:
+solution = Solution()
+
+# Example 1:
+nums1 = [5, 7, 7, 8, 8, 10]
+target1 = 8
+print(solution.searchRange(nums1, target1))  # Output: [3, 4]
+
+# Example 2:
+nums2 = [5, 7, 7, 8, 8, 10]
+target2 = 6
+print(solution.searchRange(nums2, target2))  # Output: [-1, -1]
+
+# Example 3:
+nums3 = []
+target3 = 0
+print(solution.searchRange(nums3, target3))  # Output: [-1, -1]
+```
+
+This code defines a `Solution` class with a `searchRange` method to find the starting and ending positions of the target value in the given array. The example usage demonstrates how to create an instance of the `Solution` class and call the `searchRange` method with different inputs.

g0001_0100/s0035_search_insert_position/readme.md

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+35\. Search Insert Position
+
+Easy
+
+Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
+
+You must write an algorithm with `O(log n)` runtime complexity.
+
+**Example 1:**
+
+**Input:** nums = [1,3,5,6], target = 5
+
+**Output:** 2 
+
+**Example 2:**
+
+**Input:** nums = [1,3,5,6], target = 2
+
+**Output:** 1 
+
+**Example 3:**
+
+**Input:** nums = [1,3,5,6], target = 7
+
+**Output:** 4 
+
+**Example 4:**
+
+**Input:** nums = [1,3,5,6], target = 0
+
+**Output:** 0 
+
+**Example 5:**
+
+**Input:** nums = [1], target = 0
+
+**Output:** 0 
+
+**Constraints:**
+
+*   <code>1 <= nums.length <= 10<sup>4</sup></code>
+*   <code>-10<sup>4</sup> <= nums[i] <= 10<sup>4</sup></code>
+*   `nums` contains **distinct** values sorted in **ascending** order.
+*   <code>-10<sup>4</sup> <= target <= 10<sup>4</sup></code>
+
+To solve this problem efficiently with a runtime complexity of O(log n), we can use binary search to find the insertion position of the target value in the sorted array. Here are the steps:
+
+### Approach:
+
+1. **Binary Search:**
+   - Implement binary search to find the insertion position of the target value.
+   - If the target value is found, return its index.
+   - If the target value is not found, binary search will determine the index where the target would be inserted to maintain the sorted order.
+
+### Python Code:
+
+```python
+class Solution:
+    def searchInsert(self, nums, target):
+        left, right = 0, len(nums) - 1
+
+        while left <= right:
+            mid = (left + right) // 2
+
+            if nums[mid] == target:
+                return mid
+            elif nums[mid] < target:
+                left = mid + 1
+            else:
+                right = mid - 1
+
+        # If the target is not found, return the insertion position (left).
+        return left
+
+# Example Usage:
+solution = Solution()
+
+# Example 1:
+nums1 = [1, 3, 5, 6]
+target1 = 5
+print(solution.searchInsert(nums1, target1))  # Output: 2
+
+# Example 2:
+nums2 = [1, 3, 5, 6]
+target2 = 2
+print(solution.searchInsert(nums2, target2))  # Output: 1
+
+# Example 3:
+nums3 = [1, 3, 5, 6]
+target3 = 7
+print(solution.searchInsert(nums3, target3))  # Output: 4
+
+# Example 4:
+nums4 = [1, 3, 5, 6]
+target4 = 0
+print(solution.searchInsert(nums4, target4))  # Output: 0
+
+# Example 5:
+nums5 = [1]
+target5 = 0
+print(solution.searchInsert(nums5, target5))  # Output: 0
+```
+
+This code defines a `Solution` class with a `searchInsert` method to find the insertion position of the target value in the given sorted array. The example usage demonstrates how to create an instance of the `Solution` class and call the `searchInsert` method with different inputs.

g0001_0100/s0039_combination_sum/readme.md

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+39\. Combination Sum
+
+Medium
+
+Given an array of **distinct** integers `candidates` and a target integer `target`, return _a list of all **unique combinations** of_ `candidates` _where the chosen numbers sum to_ `target`_._ You may return the combinations in **any order**.
+
+The **same** number may be chosen from `candidates` an **unlimited number of times**. Two combinations are unique if the frequency of at least one of the chosen numbers is different.
+
+It is **guaranteed** that the number of unique combinations that sum up to `target` is less than `150` combinations for the given input.
+
+**Example 1:**
+
+**Input:** candidates = [2,3,6,7], target = 7
+
+**Output:** [[2,2,3],[7]]
+
+**Explanation:**
+
+    2 and 3 are candidates, and 2 + 2 + 3 = 7. Note that 2 can be used multiple times.
+    7 is a candidate, and 7 = 7.
+    These are the only two combinations. 
+
+**Example 2:**
+
+**Input:** candidates = [2,3,5], target = 8
+
+**Output:** [[2,2,2,2],[2,3,3],[3,5]] 
+
+**Example 3:**
+
+**Input:** candidates = [2], target = 1
+
+**Output:** [] 
+
+**Example 4:**
+
+**Input:** candidates = [1], target = 1
+
+**Output:** [[1]] 
+
+**Example 5:**
+
+**Input:** candidates = [1], target = 2
+
+**Output:** [[1,1]] 
+
+**Constraints:**
+
+*   `1 <= candidates.length <= 30`
+*   `1 <= candidates[i] <= 200`
+*   All elements of `candidates` are **distinct**.
+*   `1 <= target <= 500`
+
+To solve the Combination Sum problem, you can use backtracking. Here are the steps you can follow:
+
+1. **Define the function**: Create a function, let's name it `combinationSum`, that takes `candidates`, `target`, and any other necessary parameters.
+
+2. **Sort the candidates**: Sort the `candidates` array. Sorting helps in pruning the search space during backtracking.
+
+3. **Initialize variables**: Initialize an empty list to store the result combinations. 
+
+4. **Define a helper function**: Create a recursive helper function, let's name it `backtrack`, that takes the current combination, the current index of the candidate being considered, and the remaining target.
+
+5. **Base case**: If the target is 0, add the current combination to the result list.
+
+6. **Iterate through candidates**: Start iterating from the current index to the end of the `candidates` array.
+
+7. **Check the sum condition**: For each candidate, check if adding it to the current combination keeps the sum less than or equal to the target. If so, recursively call the `backtrack` function with the updated combination, the current index, and the reduced target.
+
+8. **Backtrack**: After the recursive call returns, remove the last element from the current combination to backtrack.
+
+9. **Call the helper function**: Call the `backtrack` function initially with an empty combination, starting index 0, and the target.
+
+10. **Return the result**: Return the list of combinations obtained.
+
+Here's a Python code snippet implementing the above steps:
+
+```python
+class Solution:
+    def combinationSum(self, candidates, target):
+        def backtrack(start, target, path):
+            if target == 0:
+                result.append(path)
+                return
+            for i in range(start, len(candidates)):
+                if candidates[i] > target:
+                    break
+                backtrack(i, target - candidates[i], path + [candidates[i]])
+
+        result = []
+        candidates.sort()
+        backtrack(0, target, [])
+        return result
+
+# Test cases
+solution = Solution()
+print(solution.combinationSum([2,3,6,7], 7)) # Output: [[2, 2, 3], [7]]
+print(solution.combinationSum([2,3,5], 8))   # Output: [[2, 2, 2, 2], [2, 3, 3], [5, 3]]
+print(solution.combinationSum([2], 1))        # Output: []
+print(solution.combinationSum([1], 1))        # Output: [[1]]
+print(solution.combinationSum([1], 2))        # Output: [[1, 1]]
+```
+
+This implementation efficiently finds all unique combinations that sum up to the target.