isaiah-perumalla

#Problem

The essence of the problem is given a collection of n items of varying size, let T_sum 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 <= T_max and (total-K) <= T_max . Where T_max is the maximum time allowed.

#Solution

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

Let S₁, S₂ ... S_2ⁿ all subsets, we iterate through each subset to see if there exist a subset S_k such that the sum of S_k = 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.

isaiah-perumalla

 #include<iostream>
 #include<vector>
 #include<algorithm>
 using namespace std;
 struct Elephant { 
   int iq,  weight, index;
   static bool iq_greater(const Elephant& e1, const Elephant& e2) {
     return e1.iq >= e2.iq;
   }
};

isaiah-perumalla

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

isaiah-perumalla

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

isaiah-perumalla

#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 

isaiah-perumalla

#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

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

isaiah-perumalla

[<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 =

isaiah-perumalla

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:

isaiah-perumalla

//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