The issue with the first circuit is that the three leds have a different forward current at the same forward voltage. The drop across the three leds will be the same. One led can't take 3 volts while it's parallel led only takes 2. Unequal voltages can't happen in parallel circuits.
And as each has a different current across them, one may dominate the amount of current available, and may go into thermal runaway conditions, killing it. Which kills the next one for the same reason. Etc. Highly depends on how much current can be pulled, i.e. the resistor value.
The third does not have this issue, as each is limited to the current by their individual resistor.
The voltage drop across each led and resistor will be equal. Current may vary.
simulate this circuit – Schematic created using CircuitLab
Notice that V-Total is the same as the Voltage source, 5 Volts. It will always be 5V. Notice that V-1 (R1 + D1), equals V-2 and V-3, but A-1 has a different current than A-2 and A-3. Yet V-4 and the nodes V-1/2/3 equal V-Total. And A-1 + A-2 + A-3 = A-4.
The very basics of series and parallel circuits.
Re: Your edit:
Practical application: I designed my original circuit with 18 LEDs in parallel, each with a 220 ohm resister. Now, I want to change the voltage to 9 volts (or 12 volts.) Will I need to switch out all of the original arrays or can I add one resistor in series.
Yes, you can. You need to resize R4 so that it takes up the same current at higher voltage. The formula stays the same. Basic Ohms law
R = V / I.
Since you know I, and the new V, just calculate for R. Assuming the values from above, we can change R4 and the Voltage Source but keep the same current of 15.45 mA.
R = (9 VSource - 1.909 VForward) / 0.01545 A = 459Ω
Of course you need to chose the next standard sized resistor. But we'll ignore that.
simulate this circuit
If you are just adding a resistor to an existing circuit like 2, then VForward would be the old VSource, i.e. 5V.