That's a good one, I almost fell for the seemingly logical solution without giving it a proper thought, but in the end I got it. 😁
green and red are children of first, red has, with a 100, the higher index than green with its 2 and therefore is on top of green. black and blue are children of second, blue with 50 has the higher index than black with 0 and therefore is in the front.
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That's a good one, I almost fell for the seemingly logical solution without giving it a proper thought, but in the end I got it. 😁
green and red are children of first, red has, with a 100, the higher index than green with its 2 and therefore is on top of green.
black and blue are children of second, blue with 50 has the higher index than black with 0 and therefore is in the front.