|
| 1 | +def binary_search(array, target): |
| 2 | + left = 0 |
| 3 | + right = len(array) - 1 |
| 4 | + while left <= right: |
| 5 | + mid = (left + right) // 2 |
| 6 | + if array[mid] == target: |
| 7 | + return mid |
| 8 | + if array[mid] < target: |
| 9 | + left = mid + 1 |
| 10 | + else: |
| 11 | + right = mid - 1 |
| 12 | + |
| 13 | + |
| 14 | + |
| 15 | +################################################################## |
| 16 | + |
| 17 | + |
| 18 | + |
| 19 | +def single_non_duplicate(nums): |
| 20 | + |
| 21 | + # Initialize left and right pointers |
| 22 | + left = 0 |
| 23 | + right = len(nums) - 1 |
| 24 | + |
| 25 | + while left != right: |
| 26 | + |
| 27 | + # Calculate the midpoint of the array |
| 28 | + middle = left + (right - left) // 2 |
| 29 | + |
| 30 | + # If mid-point is odd, decrement to make it even |
| 31 | + if middle % 2 != 0: |
| 32 | + middle -= 1 |
| 33 | + |
| 34 | + # If middle and middle + 1 are the same, then single element after mid-point |
| 35 | + if nums[middle] == nums[middle + 1]: |
| 36 | + left = middle + 2 |
| 37 | + |
| 38 | + # Otherwise search for single element before mid-point |
| 39 | + else: |
| 40 | + right = middle |
| 41 | + |
| 42 | + # Return single element value |
| 43 | + return nums[left] |
| 44 | + |
| 45 | + |
| 46 | + |
| 47 | +# Time Complexity = O(logn) |
| 48 | +# Space Complexity = O(1) |
| 49 | + |
| 50 | + |
| 51 | + |
| 52 | +################################################################## |
| 53 | + |
| 54 | + |
| 55 | + |
| 56 | +# Driver code |
| 57 | +def main(): |
| 58 | + |
| 59 | + nums = [[1, 2, 2, 3, 3, 4, 4], [1, 1, 2, 2, 3, 4, 4, 5, 5], [1, 1, 2, 3, 3], [1, 1, 2], [0, 2, 2, 3, 3, 4, 4, 5, 5]] |
| 60 | + |
| 61 | + for i in range(len(nums)): |
| 62 | + print(str(i + 1) + ".\tThe Array = ", nums[i] , sep = "") |
| 63 | + print("\tSingle Element Found: ", single_non_duplicate(nums[i]), sep = "") |
| 64 | + print("-" * 100) |
| 65 | + |
| 66 | +if __name__ == '__main__': |
| 67 | + main() |
| 68 | + |
| 69 | + |
| 70 | + |
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