Add twice linear

Author: dwqs

codewars/201808/twice-linear.md

@@ -0,0 +1,54 @@
+
+## Description
+
+Consider a sequence u where u is defined as follows:
+
+* The number u(0) = 1 is the first one in u.
+* For each x in u, then y = 2 * x + 1 and z = 3 * x + 1 must be in u too.
+There are no other numbers in u.
+* Ex: u = [1, 3, 4, 7, 9, 10, 13, 15, 19, 21, 22, 27, ...]
+
+1 gives 3 and 4, then 3 gives 7 and 10, 4 gives 9 and 13, then 7 gives 15 and 22 and so on...
+
+Task: Given parameter n the function dbl_linear (or dblLinear...) returns the element u(n) of the ordered (with <) sequence u.
+
+Example: dbl_linear(10) should return 22
+
+Note: Focus attention on efficiency
+
+**Kata's link**: [Twice linear](https://www.codewars.com/kata/twice-linear)
+
+## Best Practices
+
+**First:**
+```js
+function dblLinear(n) {
+ var ai = 0, bi = 0, eq = 0;
+ var sequence = [1];
+ while (ai + bi < n + eq) {
+ var y = 2 * sequence[ai] + 1;
+ var z = 3 * sequence[bi] + 1;
+ if (y < z) { sequence.push(y); ai++; }
+ else if (y > z) { sequence.push(z); bi++; }
+ else { sequence.push(y); ai++; bi++; eq++; }
+ }
+ return sequence.pop();
+}
+```
+
+**Second:**
+```js
+function dblLinear(n) {
+ 
+ var u = [1], pt2 = 0, pt3 = 0; //two pointer
+ 
+ for(var i = 1;i<=n;i++){
+ u[i] = Math.min(2* u[pt2] + 1, 3*u[pt3] + 1);
+ if(u[i] == 2 * u[pt2] + 1) pt2++;
+ if(u[i] == 3 * u[pt3] + 1) pt3++;
+ }
+
+ return u[n];
+ 
+}
+```

readme.md

@@ -6,6 +6,9 @@ Some best practices for practice questions on [codewars](http://www.codewars.com
 Here is my codewars homepage: https://www.codewars.com/users/dwqs.
 
 ## Lists
+### 2018-08
+* [Twice linear](codewars/201808/twice-linear.md)
+
 ### 2018-07
 * [parseInt reloaded](codewars/201807/parseint-reloaded.md)
 * [Return substring instance count](codewars/201807/substring-instance-count.md)