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Commit ded3bd4

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O(nlogm) time with binary search. M being longer and n being shorter.
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"""
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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.
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Note that an integer is said to be common to nums1 and nums2 if both arrays have at least one occurrence of that integer.
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Example 1:
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Input: nums1 = [1,2,3], nums2 = [2,4]
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Output: 2
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Explanation: The smallest element common to both arrays is 2, so we return 2.
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Example 2:
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Input: nums1 = [1,2,3,6], nums2 = [2,3,4,5]
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Output: 2
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Explanation: There are two common elements in the array 2 and 3 out of which 2 is the smallest, so 2 is returned.
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Constraints:
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1 <= nums1.length, nums2.length <= 105
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1 <= nums1[i], nums2[j] <= 109
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Both nums1 and nums2 are sorted in non-decreasing order.
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"""
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class Solution:
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def getCommon(self, nums1: List[int], nums2: List[int]) -> int:
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def binary_search(target,nums):
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left,right = 0,len(nums)-1
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while left <= right:
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mid = left + (right- left)//2
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if nums[mid] == target: return True
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elif nums[mid] > target:
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right = mid - 1
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else:
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left = mid + 1
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return False
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if len(nums1) > len(nums2):
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for num in nums2:
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if binary_search(num,nums1):
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return num
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return -1
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else:
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for num in nums1:
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if binary_search(num,nums2):
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return num
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return -1

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