leetcode patterns

2D to 1D mapping

In binary search, we deal with 1D values, l to r. However, in this problem, values are retreived by 2D index (row, col)
Since, the matrix elements are sorted, treat it as on single sorted array and perform binary search
Thus, we'll convert 1D mid index to find the value in matrix using % and / as appropriate

values           1D index           2D index (row,col) mapping
[ 1, 3, 5, 7]    [ 0, 1,  2,  3]    [  ,  ,      ,  ]
[10,11,16,20] -> [ 4, 5,  6,  7] -> [  ,  , (1,2),  ]
[23,30,34,60]    [ 8, 9, 10, 11]    [  ,  ,      ,  ]

- Let's say mid is at pos 6
- 1D index [6] is converted to 2D index [1,2] to get value 16

int mid = l + (r - l) / 2;
int midVal = matrix[mid / cols][mid % cols];

BacktoBackSWE 🎥

36. Valid Sudoku

use HashSets to keep track of numbers in a row, col, and box
For box coords, box row = row/3 and box col = col/3

while/switch

Using many while loops one after other instead of using only one

929. Unique Email Addresses

use switch to find + and find @ and append substring after it and break (use i = email.len instead of break)