The source free RLC series/parallel circuit
This answer uses the example of a series capacitor, inductor and resistor and assumes the capacitor is previously charged to some voltage level.

If the over-damping is massive i.e. a massive series resistor then it dominates the inductance and the circuit becomes a resistor discharging a capacitor hence, the discharge time can be very long if R is very large.

As R reduces, the discharge time reduces. But as you approach critical damping (Q = 0.5) from the over-damped case, the decay changes from being a simple exponential situation to a decaying ripple where the frequency of the ripple is a decaying sinewave having a frequency of \$\omega_n\sqrt{1-\zeta^2}\$, where \$\zeta = \frac{1}{2Q}\$ and \$\omega_n\$ is the natural resonant frequency.

At Q = 0.6 (slightly less than critical damping), the decaying sinewave is very limited in its extent. The overshoot of the first cycle is about 0.9% and, to most folk would hardly count. As the sinewave decays, the first undershoot is about 0.008% so it's pretty much over as soon as it began.

The ratio of the magnitude of the 1st positive peak to the first negative peak is approximately 0.9/0.008 = ~112.

However, as Q rises more and more, the overshoot and undershoot become more dominant. For instance at Q = 1, the first overshoot peak is about 16% and the first undershoot peak is about 2.7%. You can see that these are closer in magnitude than was the case when Q was only 0.6. Ratio is about 6.

As Q rises even more you eventually reach the situation where the overshoot and undershoot continue for a very long time. This means that energy is kept in the reactive components for a very long time.

Conclusion: The energy is dissipated more quickly in a series RLC circuit at around critical damping. Either side of critical damping it takes longer and, at extremes (R = infinity, R = 0), it takes an infinite length of time.