By Long Luo

# Intuition

1. If $\textit{startValue} \gt \textit{target}$, so return $\textit{startValue} \gt \textit{target}$;
2. Considering such cases: $2 \to 3$, $2 \to 5$, $1 \to 128$, $1 \to 100$;
3. $\textit{startValue} = \textit{startValue} \times 2^i$, when $\textit{startValue}$ grows bigger, the last $\textit{startValue}$ is closer to $\textit{target} / 2$ that we only need double $\textit{startValue}$ once.

So think reversly:

1. if target is even, the minimum operand is convert $startValue$ equal to $\textit{target} / 2 + 1$;
2. if target is odd, the minimum operand is convert $startValue$ equal to $(\textit{target} + 1) / 2 + 2$.

Let’s coding.

## Analysis

• Time Complexity: $O(logn)$.
• Space Complexity: $O(1)$.

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