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isaiah isaiah-perumalla

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isaiah-perumalla /
Last active Aug 29, 2015
Solution to Amazing-test problem from hackerearth


The essence of the problem is given a collection of n items of varying size, let Tsum be the sum of all item sizes. we need to able to answer the question, does a subset of items exist such that the sum of the items in the sumset is equal to K. where K is some integer that satisfies the constraints K <= Tmax and (total-K) <= Tmax . Where Tmax is the maximum time allowed.


Brute force approach would be iterate through all the subsets of the items.

Let S1, S2 ... S2n all subsets, we iterate through each subset to see if there exist a subset Sk such that the sum of Sk = K. This approach is guarenteed to be correct, however it is impossibly slow because we have to iterate 2n subsets. since the number of items in the collection can reach 100, in the worst case we would have to examine 2100 subsets this is clearly intractable.

using namespace std;
struct Elephant {
int iq, weight, index;
static bool iq_greater(const Elephant& e1, const Elephant& e2) {
return >=;
def water_collected(A):
sortedHeights = [(x,i) for i,x in enumerate(A)]
sortedHeights.sort(key=lambda v: v[0]*-1) #sort in decreasing order
leftIdx = rightIdx = sortedHeights[0][1]
waterLevel = 0
for i in xrange(1,len(A)):
x,idx = sortedHeights[i]
if idx > rightIdx:
water = (idx-rightIdx-1)*x
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
View x86mmodel.c
#include <pthread.h>
#include <stdio.h>
int x, y = 0;
int r0, r1;
void *core0 (void *arg)
x = 1;
asm volatile ("" ::: "memory"); // ensure GCC compiler will not reorder
View x86mfence.c
#include <pthread.h>
#include <stdio.h>
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
isaiah-perumalla /
Last active Dec 19, 2015
simple linear time algorithm for finding longest palindome substring in a give string
def longest_subpalindrome_slice(s):
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
isaiah-perumalla / change.fs
Last active Dec 20, 2015
make change algorithm. compute change with using minimum number of coins
View change.fs
[<Measure>] 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<p>) xs = List.filter (fun (coin, _) -> x >= coin) xs
let num_coins (coin:int<p>, qty:int) (amt:int<p>) = min qty (int(amt/coin))
let better ps bs =

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isaiah-perumalla / q-funcs.q
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)
2 1