LinkedIn Zip #490 Answer & Analysis

The solve begins at R6C4 (1). From there, the path goes up to R5C4 for 2, then bends right to R5C5 and down to R6C5 for 3. That opening keeps the lower-middle area flexible while setting up the long sweep around the outside.

After 3 at R6C5, the path runs around the lower-right corner: R6C6, R5C6, R4C6, R4C5, and then into R4C4 for 4. This is a natural way to keep the outer rim of the board active while staying fully connected.

From 4 at R4C4, the path goes up to R3C4 and R2C4, then right to R2C5 for 5 and down to R3C5 for 6. After that it continues across the upper-right corridor through R3C6, R2C6, R1C6, and R1C5 before arriving at R1C4 for 7. This is the part of the solution that links the center to the top row without crossing any cell twice.

From 7 at R1C4, the path goes left through 8 at R1C3, then continues across the top edge to R1C2 and R1C1. From there it drops down the left column through R2C1, R3C1, R4C1, R5C1, and R6C1, then steps right through R6C2 for 9, R6C3 for 10, and R5C3 for 11. This edge-heavy section is what gives the puzzle its full-grid coverage.

The last section threads the center-left corridor: R5C3 to R5C2, R4C2, R3C2 for 12, R2C2 for 13, then R2C3, R3C3, and R4C3 for 14. This final bend is important because it uses the remaining open cells without revisiting any earlier part of the path.