There is a broken calculator that has the integer startValue on its display initially. In one operation, you can:
multiply the number on display by 2, or
subtract 1 from the number on display.Given two integers startValue and target, return __the minimum number of operations needed to display __target__ on the calculator__.
Example 1: Input: startValue = 2, target = 3 Output: 2 Explanation: Use double operation and then decrement operation {2 -> 4 -> 3}. Example 2: Input: startValue = 5, target = 8 Output: 2 Explanation: Use decrement and then double {5 -> 4 -> 8}. Example 3: Input: startValue = 3, target = 10 Output: 3 Explanation: Use double, decrement and double {3 -> 6 -> 5 -> 10}.
Constraints:
1 <= x, y <= 109
- code TLE bfs
class Solution:
def brokenCalc(self, startValue: int, target: int) -> int:
q = deque([startValue])
visited = set()
step = 0
while q:
for _ in range(len(q)):
cur = q.popleft()
if cur == target: return step
if cur - 1 not in visited:
q.append(cur - 1)
visited.add(cur - 1)
if cur * 2 not in visited:
q.append(cur * 2)
visited.add(cur * 2)
step += 1
- code
class Solution:
def brokenCalc(self, startValue: int, target: int) -> int:
ans = 0
while target > startValue:
ans += 1
if target % 2: target += 1
else: target //= 2
return ans + startValue - target