Blind 75 - Beats 100.00% [11/75]
Intuition\nImplements the bitwise addition of two integers a and b without using the + operator. It uses the concept of bitwise manipulation to perform addition
# Intuition\n*Implements the bitwise addition of two integers `a` and `b` without using the `+` operator. It uses the concept of bitwise manipulation to perform addition. Python use more bits to store integer so we need to shorten the bits, Anyhow I had also attached a Java solution.*\n# Approach\n\n1. **Initialization**:\n   - `bitShort` is initialized to `0xffffffff`, which is a 32-bit integer with all bits set to 1 except for the sign bit.\n\n2. **Bitwise Addition**:\n   - It enters a while loop that continues until `b` becomes zero or until all bits are consumed.\n   - Within the loop:\n     - It calculates the `carry` by performing a bitwise AND operation between `a` and `b`, and then left-shifts the result by 1.\n     - It updates `a` by performing a bitwise XOR operation between `a` and `b`.\n     - It updates `b` with the value of `carry`, which represents the carry for the next iteration.\n\n3. **Return**:\n   - After the loop, it returns `a` bitwise ANDed with `bitShort` if `b` is greater than 0, else it returns `a`.\n\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(1)\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nO(1)\nNow, let\'s analyze the time and space complexity:\n\n- **Time Complexity**: \n  - The code iterates until all bits of `b` are consumed. Since the maximum number of iterations is fixed (32 in this case for a 32-bit integer), the time complexity is O(1), indicating constant time complexity.\n\n- **Space Complexity**: \n  - The algorithm uses only a constant amount of extra space for variables (`bitShort`, `carry`, `a`, `b`, and the loop variable). Therefore, the space complexity is O(1), indicating constant space complexity.\n\nSo, summarizing:\n- Time Complexity: O(1)\n- Space Complexity: O(1)\n# Code\n```\nclass Solution:\n    def getSum(self, a: int, b: int) -> int:\n        \n        bitShort = 0xffffffff \n\n        while b & bitShort != 0:\n            carry = (a & b) << 1\n            a = a ^ b\n            b = carry\n\n        return a & bitShort if b > 0 else a\n```\n\n# Code - `Java Solution` `Beats: 100%` `runtime:0ms`\n\n\nThe straightforward solution, In Java integers, are by default  32-bit signed two\'s complement integers. So,  We can run the same logic. A simple explanation is as follows:\n\n1. The `getSum` method starts by entering a while loop that continues until `b` becomes zero.\n2. Inside the loop:\n   - It calculates the carry by performing a bitwise AND operation between `a` and `b`, then shifting the result one position to the left (`(a & b) << 1`).\n   - It updates `a` by performing a bitwise XOR operation between `a` and `b`. This step effectively calculates the sum of `a` and `b` without considering any carry.\n   - It updates `b` with the value of the carry calculated in the first step.\n3. Once `b` becomes zero, indicating that there\'s no more carry left, the method exits the loop and returns the final value of `a`, which represents the sum of the original `a` and `b` values.\n\n```\nclass Solution {\n    public int getSum(int a, int b) {\n\n        while (b != 0){\n            int carry = (a & b) << 1;\n            a = a ^ b;\n            b = carry;\n        } \n\n        return a;\n    }\n}\n```\n\n\n***Any doubts regarding the Python solution listed below,  Comment here, and I would be happy to help!!***\n\n> 1. Why `bitShort`?\n> 2. Why `carry`. Can\'t we initilaize `b = (a & b) << 1` before `a` \n> 3. Why `return a & bitShort if b > 0 else a` rather than `return a`?\n\n\n\n
['Java', 'Python3']