Lc problems

2540. Minimum Common Value/2540. Minimum Common Value.py

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+"""
+Given two integer arrays nums1 and nums2, sorted in non-decreasing order, return the minimum integer common to both arrays. If there is no common integer amongst nums1 and nums2, return -1.
+
+Note that an integer is said to be common to nums1 and nums2 if both arrays have at least one occurrence of that integer.
+
+
+
+Example 1:
+
+Input: nums1 = [1,2,3], nums2 = [2,4]
+Output: 2
+Explanation: The smallest element common to both arrays is 2, so we return 2.
+Example 2:
+
+Input: nums1 = [1,2,3,6], nums2 = [2,3,4,5]
+Output: 2
+Explanation: There are two common elements in the array 2 and 3 out of which 2 is the smallest, so 2 is returned.
+
+
+Constraints:
+
+1 <= nums1.length, nums2.length <= 105
+1 <= nums1[i], nums2[j] <= 109
+Both nums1 and nums2 are sorted in non-decreasing order.
+"""
+class Solution:
+    def getCommon(self, nums1: List[int], nums2: List[int]) -> int:
+        def binary_search(target,nums):
+            left,right = 0,len(nums)-1
+            while left <= right:
+                mid = left + (right- left)//2
+                if nums[mid] == target: return True
+                elif nums[mid] > target:
+                    right = mid - 1
+                else:
+                    left = mid + 1
+            return False
+        if len(nums1) > len(nums2):
+            for num in nums2:
+                if binary_search(num,nums1):
+                    return num
+            return -1
+        else:
+            for num in nums1:
+                if binary_search(num,nums2):
+                    return num
+            return -1

2540. Minimum Common Value/2540. Minimum Common Value_1.py

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+class Solution:
+    def getCommon(self, nums1: List[int], nums2: List[int]) -> int:
+        i, j = 0, 0
+        while i < len(nums1) and j < len(nums2):
+            if nums1[i] < nums2[j]:
+                i += 1
+            elif nums1[i] > nums2[j]:
+                j += 1
+            else:
+                return nums1[i]
+        return -1

647. Palindromic Substrings/647. Palindromic Substrings_2.py

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+class Solution:
+    def dfs(self, left, right,s):
+        final = 0
+        while left >= 0 and right < len(s) and s[left] == s[right]:
+            left -= 1
+            right += 1
+            final += 1
+        return final
+
+    def countSubstrings(self, s: str) -> int:
+        result = 0
+        for i in range(len(s)):
+            result += self.dfs(i,i,s)
+            result += self.dfs(i,i+1,s)
+        return result