99% beats. Full Explanation :)
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \nThis function takes two
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThis function takes two integer arguments a and b, and returns their sum using bitwise operations. Here\'s how it works:\n\nThe function enters a while loop that runs as long as b is not equal to zero. This loop performs the addition of a and b bit by bit, by calculating the sum of each pair of corresponding bits, along with any carry from the previous pair.\n\nInside the loop, the function first calculates the carry by performing a bitwise AND operation between a and b, and storing the result in the variable carry. This operation sets each bit in carry that has a corresponding 1 bit in both a and b, which represents the carry generated from adding the previous bit pairs.\n\nThe function then calculates the sum by performing a bitwise XOR operation between a and b, and storing the result in a. This operation sets each bit in a that has a corresponding 1 bit in either a or b, but not in both, which represents the sum of the current bit pairs.\n\nThe function updates b by shifting the carry left by one position and storing the result in b. This operation prepares the carry for the next iteration of the loop, by shifting it one position to the left to match the position of the next bit pair.\n\nFinally, the function checks if both b and carry are equal to zero, and if so, it breaks out of the loop. This is an optimization that allows the function to exit early if no more carry bits are left to be added, which can improve performance for some input values.\n\nThe function returns the final value of a, which represents the sum of the input integers a and b.\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(log n)\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nO(1)\n# Code\n```\n/**\n * @param {number} a\n * @param {number} b\n * @return {number}\n */\nvar getSum = function(a, b) {\n    while (b !== 0) {\n        const carry = a & b;\n        a ^= b;\n        b = carry << 1;\n        if (b === 0 && carry === 0) break;\n    }\n    return a;\n};\n```
['JavaScript']