For the fire the source of energy is the "chemical" energy stored in the reacting compounds whereas for lightning the stored of energy is electrostatic in nature due to the separation of charges. Comparing power, [gas fire](https://physics.stackexchange.com/questions/194367/how-big-is-a-1kw-fire) - $4\times 10^3$ watts and [lightning strike](https://www.weather.gov/safety/lightning-power) - $3\times 10^8$ volts and $3\times 10^4$ amps $\Rightarrow \approx 10^{13}$ watts, you will note that the rate at which energy is dissipated in a lightning strike is much, much greater than that in a gas fire. Thus one would expect the lightning strike to last for a much shorter period of time than that for a chemical fire. --- @Peter-ReinstateMonica has made a valid comment that I only compared powers so here is an analysis resulting in a time for the event answer. A gas stove has a power rating of about $4\,\rm kW$ and suppose that it is run from a cylinder of propane which contains $10\,\rm kg$ of gas with a [calorific value](https://www.engineeringtoolbox.com/fuels-higher-calorific-values-d_169.html) of $50\,\rm MJ\,kg^{-1}$. So if run continuously the cylinder will last about **one day**. The paper [Measurement of the electrical properties of a thundercloud through muon imaging by the GRAPES-3 experiment](https://arxiv.org/abs/1903.09801) has some estimates on page 4 regarding thunderclouds. There is an estimate of $\ge 720 \,\rm GJ$ stored at a potential of approximately $1\,\rm GV$. Taking the figure for the current during a lightning strike to be $30\,\rm kA$ gives an estimated time for a complete discharge (which is unlikely) of the cloud to be $\mathbf 20\, \rm ms$.