how can we know this array is sorted in java

this is easy example

        int o = a[0] < a[1] ? 0 :1;
 for (int i = 1; i < a.length-1; i++) 
   if( 0!= (a[i] < a[i+1]?0:1) )   return "not sorted";
 return o ==0?"ascending" : "descending";

Project Euler problem 303

My effort to solve it

1. make 3digit generator and divide with n

2. 9999 is never solved in time.

3. find 9, 99, 999, 9999 in pattern.

4. finally solve it in 9s

find the ways to make specific number with dice

this problem is about hexanacci number

               long a = 0,b= 0,c= 0,d= 0,e= 0,f=1, t = 0;
  
  for (int i = 1; i <= n; i++) {
   t = a+b+c+d+e+f;   
   a=b;
   b=c;
   c=d;   
   d=e;
   e=f;
   f=t;
  }
  return t+"";

Hexanacci numbers: a(n+1) = a(n)+...+a(n-5) with a(0)=...=a(4)=0, a(5)=1. 

huffman coding ( encoding and decoding) algorithm in python

this is short code written by python.

Huffman encoding and decoding

I studied using this site and write code 

http://www.geeksforgeeks.org/greedy-algorithms-set-3-huffman-coding/

import random

dic = {}
class MinHeapNode:
    def setNode(self, left, right, freq, char):
        self.left = left
        self.right = right
        self.freq = freq
        self.char = char
        return self
def huffmanCoding(arr, freq):
    heapArr = []

    for i in range(len(arr)):
        heapArr.append(MinHeapNode().setNode(None, None , freq[i], arr[i]))
        print(heapArr[i].freq, heapArr[i].char)

    while(len(heapArr)>1):
       left = heapArr.pop(0)
       right = heapArr.pop(0)
       insertionSort(heapArr, MinHeapNode().setNode(left, right, left.freq+right.freq, left.char + right.char))

    rootheap= heapArr.pop()

    return rootheap


def printHeap(heap, str):
    if heap.left != None:
        printHeap(heap.left, str + '0')

    if heap.right !=None:
        printHeap(heap.right, str + '1' )

    if heap.left == None:
        print(heap.char, str)
        dic[heap.char] = str

def insertionSort(heapArr, heap):
    index = len(heapArr)
    for i in range(len(heapArr)):
        if heap.freq < heapArr[i].freq:
           index  = i
           break
    heapArr.insert(index, heap)

def makeNewString(arr, freq):

    freqcopy = freq[:]
    arrcopy = arr[:]
    strval = ''
    while len(freqcopy) >0:
        number = random.randrange(0,len(freqcopy))
        freqcopy[number] = freqcopy[number]-1        strval = strval + arrcopy[number]
        if freqcopy[number] == 0:
            freqcopy.pop(number)
            arrcopy.pop(number)

    return strval

def endcoding(str):
    strval = ''    for i in range(len(str)):
        strval = strval + dic[str[i]]

    return strval

def huffmandecode(rootheap, strvalencoded):

    index = 0;
    orgstr = '';

    while index < len(strvalencoded):
        str = getchar(rootheap, index, strvalencoded)
        index = index + len(dic[str])
        orgstr = orgstr + str
    return orgstr

def getchar(rootheap, index, strvalencoded):
    if rootheap.left == None:
        return rootheap.char

    number = int(strvalencoded[index])
    if number ==0:
        return getchar(rootheap.left, index+1, strvalencoded)

    if number ==1:
        return getchar(rootheap.right, index+1, strvalencoded)

arr = ['a', 'b', 'c', 'd', 'e', 'f']
freq = [5, 9, 12, 13, 16, 45]
rootheap = huffmanCoding(arr, freq)

printHeap(rootheap, '')

strval  = makeNewString(arr, freq)
print("org     ", strval)

strvalencoded = endcoding(strval)
print("encoded ",strvalencoded)

orgencoded = huffmandecode(rootheap, strvalencoded)
print("decoded ", orgencoded)

problem 293

This problem was some how easy.

Using DFS and Set

I could solve it.

python disjoint set (union find)

# make setdef MakeSet(number):
    return node().set(number)

# find rootdef find(x):
    if x.parent.number == x.number:
        return x
    else:
        return find(x.parent)

# merge two setsdef union(x, y):
    xRoot = find(x)
    yRoot = find(y)
    xRoot.parent = yRoot


class node():
    def set(self, number):
        self.parent = self        self.number = number
        return self
# make setsets = list(map(lambda x:MakeSet(x), range(10)))
for i in range(len(sets)):
    print(sets[i].number, sets[i].parent)

# one set is 0-4 super root is 4for i in range(0, 4):
    union(sets[i], sets[i+1])

# the otehr set is 5~9 super root is 9for i in range(5, len(sets)-1):
    union(sets[i], sets[i+1])

# test for all nodes finding root nodefor i in range(len(sets)):
    print(i ," parent -> ", find(sets[i]).number)

Problem 491

Examples:

Is the number 2547039 divisible by 11?

First find the difference between sum of its digits at odd and even places.

(Sum of digits at odd places) - (Sum of digits at even places)

= (9 + 0 + 4 + 2) - (3 + 7 + 5)

= 15 - 15 = 0

The number is 0, so the number 2547039 is divisible by 11

I approach this problem by using above condition.

1. recursive function

2. combination

my java code is below

package numberFrom400;

import java.util.Arrays;

import euler.util.EulerUtil;

public class number491 {
 
 public static long totalCase = 0;
 
 public static int total = 0;
 
 public static int sum = 90;
 
 public static void main(String args[]){
  
  long start = System.currentTimeMillis();
  int numbers[] = new int[10];
  searching(0, 0, numbers);
  
  System.out.println(totalCase);
  System.out.println(System.currentTimeMillis()-start);
 }
 
 
 public static void searching(int position, int count, int[]arr ){
  if(position>=arr.length){
   if(count == 10){
    // 여기서 시작한다. 왼쪽 섬
    int subsum = 0;
    for (int i = 0; i < arr.length; i++) {
     if(arr[i] != 0){
      int val = i *arr[i];
      subsum += val;
     }
    }
    
    if((subsum -(sum - subsum))%11 ==0){
//     System.out.println("this num is divisible");
//     System.out.println(Arrays.toString(arr));
//     System.out.println(Arrays.toString(reversse(arr)) + " re ");
     // 여기가 Divisible이 가능한 위치
     
     int twocount = 0;
     for (int i = 0; i < arr.length; i++) {
      if(arr[i]== 2){
       twocount ++;
      }
     }
     
     long leftCase = EulerUtil.getFactorial(10)/(long)Math.pow(2, twocount);
     if(arr[0] ==1  ){
      leftCase = (leftCase - EulerUtil.getFactorial(9)/(long)Math.pow(2, twocount));
     }else if (arr[0] ==2) {
      leftCase = (leftCase - EulerUtil.getFactorial(9)/(long)Math.pow(2, twocount-1));
     }
     
     int[] reverse = reversse(arr);
     
     // right case
     twocount = 0;
     for (int i = 0; i < reverse.length; i++) {
      if(reverse[i]== 2){
       twocount ++;
      }
     }
     
     long rightCase = EulerUtil.getFactorial(10)/(long)Math.pow(2, twocount);
     
     totalCase = totalCase + (leftCase * rightCase );
    }else{
    }
    
    total++;
    return;
   }else if(count > 5){
    return;
   }
   return; 
  }
  
  for (int i = 0; i < 3; i++) {
   
   arr[position] = i;
   searching(position+1, count+i, arr);
   arr[position] = 0;
  }
 }
 
 
 public static int[] reversse (int arr[]){
  int reverse[]= new int[arr.length];
  for (int i = 0; i < arr.length; i++) {
   reverse[i] = 2 - arr[i];
  }
  return reverse;
 }
 
}