the previous LED array layout assumed that i am going to be having a twelve volt power source. It seems that when I use a 5 volt power source, the layout will be like this:

Solution 0: 1 x 18 array uses 18 LEDs exactly

+----|>|---/\/\/----+ R = 100 ohms

+----|>|---/\/\/----+ R = 100 ohms

+----|>|---/\/\/----+ R = 100 ohms

+----|>|---/\/\/----+ R = 100 ohms

+----|>|---/\/\/----+ R = 100 ohms

+----|>|---/\/\/----+ R = 100 ohms

+----|>|---/\/\/----+ R = 100 ohms

+----|>|---/\/\/----+ R = 100 ohms

+----|>|---/\/\/----+ R = 100 ohms

+----|>|---/\/\/----+ R = 100 ohms

+----|>|---/\/\/----+ R = 100 ohms

+----|>|---/\/\/----+ R = 100 ohms

+----|>|---/\/\/----+ R = 100 ohms

+----|>|---/\/\/----+ R = 100 ohms

+----|>|---/\/\/----+ R = 100 ohms

+----|>|---/\/\/----+ R = 100 ohms

+----|>|---/\/\/----+ R = 100 ohms

+----|>|---/\/\/----+ R = 100 ohms

does this mean that unlike the 12v array where leds were divided into groups of three on my terminal block (shown above) i am going to have to divide them to groups of one (if there is such a thing) and rewire my block?

it was my assumption also, that my block be rewired as shown:

is this correct?