diff options
Diffstat (limited to 'min_or_remaining_element.py')
-rw-r--r-- | min_or_remaining_element.py | 40 |
1 files changed, 40 insertions, 0 deletions
diff --git a/min_or_remaining_element.py b/min_or_remaining_element.py new file mode 100644 index 0000000..7e37a35 --- /dev/null +++ b/min_or_remaining_element.py @@ -0,0 +1,40 @@ +# 3022. Minimize OR of Remaining Elements Using Operations + +""" +given a 0 indexed integer array 'nums' and an integer 'k'. in one operation, +you can pick any index 'i' of 'nums' such that '0 <= i < nums.len - 1' and +replace 'nums[i]' and 'nums[i + 1]' with a single occurence of 'nums[i] & +nums[i + 1]' where & represents the bitwise and operation. return the minimum +possible value of the bitwise or of the remaining elements of 'nums' after +applying at most 'k' operations. +""" + + +class Solution(object): + def minOrAfterOperations(self, nums, k): + """ + :type nums: List[int] + :type k: int + :rtype: int + """ + mask = 0 + for bit in range(29, -1, -1): + mask |= 1 << bit + ops_needed, and_res = 0, 0 + for i in nums: + if and_res != 0: + and_res &= i + ops_needed += 1 + elif i & mask != 0: + and_res = i & mask + if and_res != 0: + ops_needed += 1 + if ops_needed > k: + mask -= 1 << bit + return (1 << 30) - mask - 1 + + +if __name__ == "__main__": + obj = Solution() + print(obj.minOrAfterOperations(nums=[3, 5, 3, 2, 7], k=2)) + print(obj.minOrAfterOperations(nums=[7, 3, 15, 14, 2, 8], k=4)) |