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This is not my puzzle, but I really wanted to share it. You can solve it interactively here.

In this maze you jump horizontally or vertically to a new square (not necessarily adjacent) with the constraint that the symbol you land on must be different from either of the two previous symbols. Starting from the diamond at top left, can you find the shortest route to the diamond at bottom right? Good luck!

enter image description here

Transcription (D = diamond, C = club, S = spade, H = heart)

D D C D D
C S C S D
D S C H D
H S H H H
D S D D D

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    $\begingroup$ Thanks for the useful edit. $\endgroup$ – Dmitry Kamenetsky Nov 26 '20 at 6:18
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I can solve it in

11 moves.

The path is:

R1C1 (Diamond) - R2C1 (Club) - R2C4 (Spades) - R3C4 (Heart) - R3C3 (Club) -
R5C3 (Diamond) - R4C3 (Heart) - R1C3 (Club) - R1C2 (Diamond) - R4C2 (Spades) -
R4C5 (Heart) - R5C5 (Diamond).

I'm pretty sure this is optimal and unique (no other path with the same move count exists), because

Thinking backwards from the destination, six moves out of it are unavoidable (up to R5C3), and to achieve the same move count, the four tiles visited between R1C1 and R5C3 cannot be diamonds. Then the first and the last (out of the four tiles) must be Clovers, and the two in the middle must be Spades and Heart in some order. This can only be achieved by going through the exact sequence above.

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