rupalbarman

/*  The code allows rotation of an 1D array (right and left)
    using XOR operator
    
    ie. Shifts elements of an Array to left or right directions,
    wraps the elements circularly.
*/

#include <stdio.h>

int arr[]={1,2,3,4,5,6,7,8,9,10};

rupalbarman

//SIMPLE INTERMEDIATE CODE GENERATOR (3 ADDRESS CODE)
//Author:   Rupal Barman
//Modified: 5/11/16
// Does not consider the operator precedence, ie. generates the 3 address code from left to right directly.
// INPUT example:   t = q * p + g ;   (don't forget to end the statement with a ';' or any character, even a space would work)
// OUTPUT example:  v=r*j;  w=v+h                                                                                                                                                                                                                                                                                                                         

#include <iostream>
#include <stdio.h>
#include <string.h>

rupalbarman

'''
	Binary Searc Tree Simple

	Rupal Barman
	[email protected]

	This is a demonstration of implementing BST in python.
	Since, it's basic. Feel free to add in more features.
'''
class Node(object):

rupalbarman

/*
	Binary Search Tree GoLang
	Rupal barman
	[email protected]
*/
	
package main

import (
	"fmt"

rupalbarman

# get adjancy matrix

n= int(input()) #number of vertices
am=[]
for i in range(n):
  am.append([int(x) for x in input().split()])
print(am) #show matrix
print()

#conver to adjancy list

rupalbarman

def get_pair_with_sum(a, need):
  s= set()
  for i in a:
    if (need- i) in s:
      print(need-i, i)
    else:
      s.add(i)

a=[4,2,4,2]
b=[3,6,2,5]

rupalbarman

# bin packing algo (estimation)
# since it is Np- Hard 
# for better methods (first fit and best fit) use self balancing trees
# this one is Next fit

def bins_needed(item_weights, bin_capacity):
  remain= bin_capacity
  bins=0
  for i in item_weights:
    if i> bin_capacity:

rupalbarman

#generate all subsets of a sequence
f = lambda x: [[y for j, y in enumerate(set(x)) if (i >> j) & 1] for i in range(2**len(set(x)))]
print(f([1,2,3]))

#print((2>>1)&1)  #1010-> 0101

# max() using lambda
maxx= lambda x,y: y if y>x else x
print(maxx(23,12))

rupalbarman

'''
Alice is taking a cryptography class and finding anagrams to be very useful. We consider two strings to be anagrams of each other if the first string's letters can be rearranged to form the second string. In other words, both strings must contain the same exact letters in the same exact frequency For example, bacdc and dcbac are anagrams, but bacdc and dcbad are not.

Alice decides on an encryption scheme involving two large strings where encryption is dependent on the minimum number of character deletions required to make the two strings anagrams. Can you help her find this number?

Given two strings, and , that may or may not be of the same length, determine the minimum number of character deletions required to make and anagrams. Any characters can be deleted from either of the strings. 
'''
from collections import Counter
c1 = Counter('baaaaccd')
c2 = Counter('abbaa')

rupalbarman

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

int main() {
  int a[]= {1, 2, 90, 43, 23};
  int b[]= {1, 90, 2, 43, 99};
  
  vector<int> v(10);