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Tested Python solution for LeetCode 991 with 20 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 991 # by problem number
lcpy gen -s broken_calculator # by problem name
startValue on its display initially. In one operation, you can:
2, or1 from the number on display.startValue and target, return the minimum number of operations needed to display target on the calculator.
Input: startValue = 2, target = 3
Output: 2
Explanation: Use double operation and then decrement operation {2 -> 4 -> 3}.
Input: startValue = 5, target = 8
Output: 2
Explanation: Use decrement and then double {5 -> 4 -> 8}.
Input: startValue = 3, target = 10
Output: 3
Explanation: Use double, decrement and double {3 -> 6 -> 5 -> 10}.
class Solution:
# Time: O(log(target))
# Space: O(1)
def broken_calc(self, start_value: int, target: int) -> int:
ops = 0
while target > start_value:
if target % 2:
target += 1
else:
target //= 2
ops += 1
return ops + start_value - target
| Time | Space |
|---|---|
| O(log(target)) | O(1) |