Black out some cells until each row and each column contains each symbol at most once. Cells that are horizontally or vertically adjacent are not allowed to both be blacked out and all remaining white cells have to be connected.
Suppose the uppermost 4 in the leftmost column will be coloured, then the 2 that is adjacent to it can not be coloured (because coloured squares are not allowed to be adjacent) and therefore the other 2 on this row will have to be coloured (because no two equal digits on the same row are allowed to remain white). Consequently the bottom 1 in the rightmost column would have to remain white and the topmost 1 would become coloured. Now the 4 in the rightmost column has to remain white since otherwise the bottom 1 will be disconnected from the rest of the white squares in the puzzle. However, now there are two 4's on row three that would have to remain white and that is not allowed. We can conclude that our assumption was wrong and instead it is the bottom 4 in the leftmost column that has to be coloured. The cells adjacent to this 4 we can mark as definitely white (with the right mouse button).
Because the 1 in the bottom left remains white, we can colour the 1 in the upper left corner, and now we see the entire second column has to remain white because coloured squares are not allowed to share a border and because no white parts are allowed to be disconnected from the others.
The 2 in the third column will be coloured because of his white left neighbour. His direct neighbours will be marked as definitely white and consequently the 1 in the upper right and the 3 at the bottom of the third column will be coloured. The 4 in the rightmost column also has to remain white in order for the white 1 above it not to become isolated, and this completes the puzzle.