**Description**\nGiven a positive integer `n`, return the minimum number of step to reduce `n` to 0. \nIn one step, you can add or subtract a power of two (2^i, i >= 0).\n\n**Example 1**\n```\nInput: 14\nOutput: 2\nExplanation: 14(01110) -> 16(10000) -> 0\n```\n\n**Example 2**\n```\nInput: 27\nOutput: 3\nExplanation: 27(011011) -> 31(011111) -> 32(100000) -> 0\n```\n\n**Example 3**\n```\nInput: 1\nOutput: 1\n```\n\n\n**My Best Attempt**\n```\nint minStep(long n) {\n    int one = 0, zero = 0;\n    while (n) {\n        if (n&1) {\n            one++;\n        } else {\n            zero++;\n        }\n        n >>= 1;\n    }\n    return min(one, zero+2);\n}\n```\nThis pass 9 test cases out of 15.\n\nFor me, this question is tough. My thought is something like dfs + memo but couldn\'t implement it. Any comment is welcomed!\n\n**Solution**\nThanks to @leetcode_dafu\n```\nint minStep(long n) {\n    int one = __builtin_popcountl(n);\n    if (one <= 1) return one;\n    return 1 + min(minStep(n+(n&(-n))), minStep(n-(n&(-n))));\n}\n```

Description\nGiven a positive integer n, return the minimum number of step to reduce n to 0. \nIn one step, you can add or subtract a power of two (2^i, i >= 0)