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
1from 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 <= startValue, target <= 109
class Solution:
def brokenCalc(self, startValue: int, target: int) -> int:
ans = 0
while startValue < target:
if target & 1:
target += 1
else:
target >>= 1
ans += 1
ans += startValue - target
return ansclass Solution {
public int brokenCalc(int startValue, int target) {
int ans = 0;
while (startValue < target) {
if ((target & 1) == 1) {
target++;
} else {
target >>= 1;
}
ans += 1;
}
ans += startValue - target;
return ans;
}
}class Solution {
public:
int brokenCalc(int startValue, int target) {
int ans = 0;
while (startValue < target) {
if (target & 1) {
target++;
} else {
target >>= 1;
}
++ans;
}
ans += startValue - target;
return ans;
}
};func brokenCalc(startValue int, target int) (ans int) {
for startValue < target {
if target&1 == 1 {
target++
} else {
target >>= 1
}
ans++
}
ans += startValue - target
return
}