latex adjustment

Author: Zanger67

markdowns/_919. Complete Binary Tree Inserter.md

@@ -31,23 +31,37 @@
 > ```
 > 
 > $$\left \lfloor{\log_{2}(1)}\right \rfloor = lvl = 0$$
+> 
 > $$\left \lfloor{\log_{2}(2)}\right \rfloor = lvl = 1$$
+> 
 > $$\left \lfloor{\log_{2}(3)}\right \rfloor = lvl = 1$$
+> 
 > $$\left \lfloor{\log_{2}(4)}\right \rfloor = lvl = 2$$
+> 
 > $$\left \lfloor{\log_{2}(5)}\right \rfloor = lvl = 2$$
+> 
 > $$\left \lfloor{\log_{2}(6)}\right \rfloor = lvl = 2$$
+> 
 > $$\left \lfloor{\log_{2}(7)}\right \rfloor = lvl = 2$$
 > 
+> 
 > Their remainders go as follows:
 > 
 > $$Remainder(1) = 0 = 0b0$$
+> 
 > $$Remainder(2) = 0 = 0b0$$
+> 
 > $$Remainder(3) = 1 = 0b1$$
+> 
 > $$Remainder(4) = 0 = 0b00$$
+> 
 > $$Remainder(5) = 1 = 0b01$$
+> 
 > $$Remainder(6) = 2 = 0b10$$
+> 
 > $$Remainder(7) = 3 = 0b11$$
 > 
+> 
 > Ignoring the `root`, if we take the binary string and read it from 
 > left to right (not including the `0b` binary indicator), we can 
 > find the path to the proper index of the new node where `0` indicates

my-submissions/m919.md

@@ -9,23 +9,37 @@
 ```
 
 $$\left \lfloor{\log_{2}(1)}\right \rfloor = lvl = 0$$
+
 $$\left \lfloor{\log_{2}(2)}\right \rfloor = lvl = 1$$
+
 $$\left \lfloor{\log_{2}(3)}\right \rfloor = lvl = 1$$
+
 $$\left \lfloor{\log_{2}(4)}\right \rfloor = lvl = 2$$
+
 $$\left \lfloor{\log_{2}(5)}\right \rfloor = lvl = 2$$
+
 $$\left \lfloor{\log_{2}(6)}\right \rfloor = lvl = 2$$
+
 $$\left \lfloor{\log_{2}(7)}\right \rfloor = lvl = 2$$
 
+
 Their remainders go as follows:
 
 $$Remainder(1) = 0 = 0b0$$
+
 $$Remainder(2) = 0 = 0b0$$
+
 $$Remainder(3) = 1 = 0b1$$
+
 $$Remainder(4) = 0 = 0b00$$
+
 $$Remainder(5) = 1 = 0b01$$
+
 $$Remainder(6) = 2 = 0b10$$
+
 $$Remainder(7) = 3 = 0b11$$
 
+
 Ignoring the `root`, if we take the binary string and read it from 
 left to right (not including the `0b` binary indicator), we can 
 find the path to the proper index of the new node where `0` indicates