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# isaiah isaiah-perumalla

Last active Nov 26, 2019
some python functions
View functions.py
 def calc_avg(items): if not items: return 0 total = sum(items) return float(total)/len(items) def std_deviation(items): if not items: return 0 mean = calc_avg(items)
Last active Nov 6, 2019
some q functions
View q-funcs.q
 //function return all subset of a list sset: {[xs] (enlist ()) {x, (y,) each x }/ xs} //or using the map each right /: sset: {[xs] (enlist ()) {x, y ,/: x }/ xs} //q)sset (1; 2; 3;`a) () ,1 ,2 2 1
Last active Nov 12, 2017
View keybase.md

### Keybase proof

I hereby claim:

• I am isaiah-perumalla on github.
• I am isaiahperumalla (https://keybase.io/isaiahperumalla) on keybase.
• I have a public key whose fingerprint is E099 3317 C358 C9E7 6CD6 12C8 B55B AF0E 50E1 63B1

To claim this, I am signing this object:

Last active Dec 20, 2015
make change algorithm. compute change with using minimum number of coins
View change.fs
 [] type p let count ps = let rec countcs acc = function |[] -> acc |(_, qty)::xs -> (countcs (acc+qty) xs) in countcs 0 ps let lessThan (x:int

) xs = List.filter (fun (coin, _) -> x >= coin) xs let num_coins (coin:int

, qty:int) (amt:int

) = min qty (int(amt/coin)) let better ps bs =

Last active Dec 19, 2015
simple linear time algorithm for finding longest palindome substring in a give string
View subpalindrome.py
 def longest_subpalindrome_slice(s): longest=(0,0) start_idx=0 #index of palindrone ending at i repeated = True #loop invariant we know longest palindrome in s[0:i] #and we know longest palidrom ending at i for i in xrange(1,len(s)): if(start_idx != 0 and s[start_idx-1] == s[i]) : start_idx = start_idx-1 repeated = False
Last active Dec 18, 2015
View x86mfence.c
 #include #include int x, y = 0; int r0, r1; void *core0 (void *arg) { x = 1; asm volatile ("mfence" ::: "memory"); // ensure compiler or hardware will not reorder, store buffer is drained
Created Jun 18, 2013
x86 memory model
View x86mmodel.c
 #include #include int x, y = 0; int r0, r1; void *core0 (void *arg) { x = 1; asm volatile ("" ::: "memory"); // ensure GCC compiler will not reorder
Created Aug 18, 2015
solution to codility challenge https://codility.com/programmers/challenges/magnesium2014
View longest_path.py
 def solution(N, A, B, C): edges = [(C[i],A[i],B[i]) for i in xrange(len(A))] edges.sort(key=lambda e: e[0]) maxLenAt = [0]*N costAt = [0]*N def updatePath(u,costu,lenu, v,costv, lenv, cost): if costu < cost and lenu+1 > lenv: maxLenAt[v] = lenu+1 costAt[v] = cost
Last active Aug 29, 2015