Problem
The infinite spiral starts at the top-left cell. The first four rows and columns are shown below. Continue the same pattern in expanding square layers. Coordinates start at 1.
1 2 9 10
4 3 8 11
5 6 7 12
16 15 14 13Return one cell value without constructing the entire grid.
Implement number_spiral(row, column).
Input
row: The row counted from the top, starting at 1.
column: The column counted from the left, starting at 1.
Output
Return the number at that position.
Constraints
1 ≤ row, column ≤ 1,000,000,000. The value fits a signed 64-bit integer.
Examples
Input:
row = 3
column = 2
Expected output:
expected = 6Input:
long long row = 3;
long long column = 2;
Expected output:
long long expected = 6;Input:
long row = 3;
long column = 2;
Expected output:
long expected = 6;Explanation
In the displayed third row, the second value is 6.
Input:
row = 1
column = 4
Expected output:
expected = 10Input:
long long row = 1;
long long column = 4;
Expected output:
long long expected = 10;Input:
long row = 1;
long column = 4;
Expected output:
long expected = 10;Explanation
The fourth square layer begins at 10 in the top row.
How to approach it
Try on your own first, then open one hint at a time.
Hint 1
The spiral is built in square layers, so a cell belongs to a layer determined by the larger coordinate.
Hint 2
Use the parity of that layer to decide whether values increase along a row or a column.
Hint 3
Pay attention to whether the layer reaches from the top-left corner or the opposite corner first.
Expected complexity
O(1) time, O(1) extra space
References
This challenge is adapted for CodeUtility. Follow the statement and constraints above.