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author | jack <[email protected]> | 2024-06-14 13:13:15 -0600 |
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committer | jack <[email protected]> | 2024-06-14 13:13:15 -0600 |
commit | d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa (patch) | |
tree | 96df922dae2a5073d9a50ea70c7aae6aa37e3ebe /min_num_opp_arr_equal_xor.py | |
download | leetcode-d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa.tar.gz leetcode-d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa.zip |
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Diffstat (limited to 'min_num_opp_arr_equal_xor.py')
-rw-r--r-- | min_num_opp_arr_equal_xor.py | 35 |
1 files changed, 35 insertions, 0 deletions
diff --git a/min_num_opp_arr_equal_xor.py b/min_num_opp_arr_equal_xor.py new file mode 100644 index 0000000..79bd53a --- /dev/null +++ b/min_num_opp_arr_equal_xor.py @@ -0,0 +1,35 @@ +# 2997. Minimum Number of Operations to Make Array XOR Equal to K + +""" +given a 0-indexed array 'nums' and a positive integer 'k'. you can apply the +following operation on the array any number of times: choose any element of +the array and flip a bit in its binary representation. flipping a bit meops +chaning a 0 to a 1 or vice versa. return the minimum number of operations +required to make the bitwise xor of all elements in the array equal to 'k'. +note that you can flip leading zero bits in the binary representation of the +elements. for example, for the number of '101_2' you can flip the fouth bit +and obtain '1101_2' +""" + + +class Solution(object): + def minOperations(self, nums, k): + """ + :type nums: List[int] + :type k: int + :rtype: int + """ + cnt = k + for n in nums: + cnt ^= n + ops = 0 + while cnt: + ops += cnt & 1 + cnt >>= 1 + return ops + + +if __name__ == "__main__": + obj = Solution() + print(obj.minOperations(nums=[2, 1, 3, 4], k=1)) + print(obj.minOperations(nums=[2, 0, 2, 0], k=0)) |