Problem 18

Problem Statement :Problem18

By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.

3
7 4
2 4 6
8 5 9 3

That is, 3 + 7 + 4 + 9 = 23.
Find the maximum total from top to bottom of the triangle below:

75
95 64
17 47 82
18 35 87 10
20 04 82 47 65
19 01 23 75 03 34
88 02 77 73 07 63 67
99 65 04 28 06 16 70 92
41 41 26 56 83 40 80 70 33
41 48 72 33 47 32 37 16 94 29
53 71 44 65 25 43 91 52 97 51 14
70 11 33 28 77 73 17 78 39 68 17 57
91 71 52 38 17 14 91 43 58 50 27 29 48
63 66 04 68 89 53 67 30 73 16 69 87 40 31
04 62 98 27 23 09 70 98 73 93 38 53 60 04 23

Approach:

• For solving this we generate all possible paths from the top to the bottom.
• At every point we have 2 directions to search and we explore every possible path through recursion.
• when the path terminates we put the result in  a global list.
• Next after generation of all possible paths we find the maximum from the global list.

Solution:

import java.util.ArrayList;
import java.util.List;

public class problem18 {
 static List<Integer> l = new ArrayList<Integer>();
 static int M[][] = { { 75, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 },
   { 95, 64, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 },
   { 17, 47, 82, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 },
   { 18, 35, 87, 10, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 },
   { 20, 04, 82, 47, 65, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 },
   { 19, 01, 23, 75, 03, 34, 0, 0, 0, 0, 0, 0, 0, 0, 0 },
   { 88, 02, 77, 73, 07, 63, 67, 0, 0, 0, 0, 0, 0, 0, 0 },
   { 99, 65, 04, 28, 06, 16, 70, 92, 0, 0, 0, 0, 0, 0, 0 },
   { 41, 41, 26, 56, 83, 40, 80, 70, 33, 0, 0, 0, 0, 0, 0 },
   { 41, 48, 72, 33, 47, 32, 37, 16, 94, 29, 0, 0, 0, 0, 0 },
   { 53, 71, 44, 65, 25, 43, 91, 52, 97, 51, 14, 0, 0, 0, 0 },
   { 70, 11, 33, 28, 77, 73, 17, 78, 39, 68, 17, 57, 0, 0, 0 },
   { 91, 71, 52, 38, 17, 14, 91, 43, 58, 50, 27, 29, 48, 0, 0 },
   { 63, 66, 04, 68, 89, 53, 67, 30, 73, 16, 69, 87, 40, 31, 0 },
   { 4, 62, 98, 27, 23, 9, 70, 98, 73, 93, 38, 53, 60, 4, 23 } },
   n = 15;

 public static void getSum(int csum, int x, int y) {
  if (x == n - 1) {
   csum = csum + M[x][y];
   l.add(csum);
   return;
  }
  csum = csum + M[x][y];
  int bsum, rsum, lsum;
  getSum(csum, x + 1, y);
  getSum(csum, x + 1, y + 1);
 }

 public static void main(String args[]) {
  long t1 = System.currentTimeMillis();
  getSum(0, 0, 0);
  int max = l.get(0);
  for (int i = 1; i < l.size(); i++) {
   if (l.get(i) >= max) {
    max = l.get(i);
   }
  }
  System.out.println(max);
  long t2 = System.currentTimeMillis();
  System.out.println("TIME: " + (t2 - t1) + " millisec");
 }
}

Problem 17

Problem Statement :Problem17

If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.
If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used?

NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20 letters. The use of "and" when writing out numbers is in compliance with British usage.

Algorithm:

• The algorithm for this problem is brute.
• Numbers with length 2 have been broken into their corresponding units and tens places
• Similarly number with 3 digits have been broken into their corresponding hundreds,units and tens places.
• Then we use the same function to reuse the broken numbers and extract the corresponding number name.

Solution:

public class problem17_1 {
 public static String numberTransfer(int number) {
  String s = null;

  if (number <= 19 && number >= 1) {
   switch (number) {
   case 1:
    s = "one";
    break;
   case 2:
    s = "two";
    break;
   case 3:
    s = "three";
    break;
   case 4:
    s = "four";
    break;
   case 5:
    s = "five";
    break;
   case 6:
    s = "six";
    break;
   case 7:
    s = "seven";
    break;
   case 8:
    s = "eight";
    break;
   case 9:
    s = "nine";
    break;
   case 10:
    s = "ten";
    break;
   case 11:
    s = "eleven";
    break;
   case 12:
    s = "twelve";
    break;
   case 13:
    s = "thirteen";
    break;
   case 14:
    s = "fourteen";
    break;
   case 15:
    s = "fifteen";
    break;
   case 16:
    s = "sixteen";
    break;
   case 17:
    s = "seventeen";
    break;
   case 18:
    s = "eighteen";
    break;
   case 19:
    s = "nineteen";
    break;
   }
  } else if (number == 20) {
   s = "twenty";
  } else if (number >= 20 && number <= 29) {
   number -= number / 10 * 10;
   s = "twenty-" + numberTransfer(number);
  } else if (number == 30) {
   s = "thirty";
  } else if (number >= 31 && number <= 39) {
   number -= number / 10 * 10;
   s = "thirty-" + numberTransfer(number);
  } else if (number == 40) {
   s = "forty";
  } else if (number >= 41 && number <= 49) {
   number -= number / 10 * 10;
   s = "forty-" + numberTransfer(number);
  } else if (number == 50) {
   s = "fifty";
  } else if (number >= 51 && number <= 59) {
   number -= number / 10 * 10;
   s = "fifty-" + numberTransfer(number);
  } else if (number == 60) {
   s = "sixty";
  } else if (number >= 61 && number <= 69) {
   number -= number / 10 * 10;
   s = "sixty-" + numberTransfer(number);
  } else if (number == 70) {
   s = "seventy";
  } else if (number >= 71 && number <= 79) {
   number -= number / 10 * 10;
   s = "seventy-" + numberTransfer(number);
  } else if (number == 80) {
   s = "eighty";
  } else if (number >= 81 && number <= 89) {
   number -= number / 10 * 10;
   s = "eighty-" + numberTransfer(number);
  } else if (number == 90) {
   s = "ninety";
  } else if (number >= 91 && number <= 99) {
   number -= number / 10 * 10;
   s = "ninety-" + numberTransfer(number);
  } else if (number == 100 || number == 200 || number == 300
    || number == 400 || number == 500 || number == 600
    || number == 700 || number == 800 || number == 900) {
   number /= 100;
   s = numberTransfer(number) + " hundred";
  } else if (number >= 101 && number <= 999) {
   int h = number / 100;
   number = number - h * 100;
   s = numberTransfer(h) + " hundred and " + numberTransfer(number);
  } else if (number == 1000) {
   s = "one thousand";
  } else if (number >= 1000 && number <= 9999) {
   int t = number / 1000;
   number = number - t * 1000;
   s = numberTransfer(t) + " thoustand " + numberTransfer(number);
  } else {
   s = "ERROR";
  }

  return s;
 }

 public static void main(String[] args) {
  int result = 0;
  String s;

  for (int i = 1; i <= 1000; i++) {
   s = numberTransfer(i);
   for (int j = 0; j < s.length(); j++) {
    if (s.charAt(j) != ' ' && s.charAt(j) != '-') {
     result++;
    }
   }
  }
  System.out.println(result);
 }
}

After-Thought:

The problem doesnt involve any aspect of numbers as such just breaking them and using a large number of special cases to generate the answer.

Problem 16

Problem Statement :Problem16

2¹⁵ = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
What is the sum of the digits of the number 2¹⁰⁰⁰?

Algorithm:

• The solution use of string multiplication which internally uses string addition.
• For optimization purposes we use 2^39 for the generation of 2^1000 and the remainder 2^25 is taken care in the end.
• In the end  just  sum up the answer that is obtained.

Solution:

import java.util.ArrayList;

public class problem16 {
 // to add two numbers using strings a,b are two numbers
 public static String stringAddition(String a, String b) {
  String sum = "";
  int carry = 0;
  if (a.length() > b.length()) {
   String temp = a;
   a = b;
   b = temp;
  }
  for (int i = 0; i < a.length(); i++) {
   int s = a.charAt(a.length() - 1 - i) + b.charAt(b.length() - 1 - i)
     - 96 + carry;
   if (s > 9) {
    carry = (s - s % 10) / 10;
    s = s % 10;
   } else {
    carry = 0;
   }
   sum = s + sum;
  }

  for (int i = 0; i < b.length() - a.length(); i++) {
   int s = b.charAt(b.length() - a.length() - 1 - i) + carry - 48;

   if (s > 9) {
    carry = (s - s % 10) / 10;
    s = s % 10;
   } else {
    carry = 0;
   }
   sum = s + sum;

  }
  if (carry != 0)
   return carry + sum;
  else
   return sum;
 }

 public static String stringMultiplication(String s1, String s2) {
  String zero = "", num = "";
  ArrayList<String> list = new ArrayList<String>();
  for (int i = s1.length() - 1; i >= 0; i--) {
   int carry = 0;
   int c1 = s1.charAt(i) - '0';
   num = "";
   for (int j = s2.length() - 1; j >= 0; j--) {
    int c2 = s2.charAt(j) - '0';
    int prod = c1 * c2 + carry;
    int add = prod % 10;
    carry = prod / 10;
    num = add + num;
   }
   if (carry != 0)
    num = carry + num;
   list.add(num + zero);
   zero = zero + "0";
  }
  num = "0";
  for (int i = 0; i < list.size(); i++)
   num = stringAddition(num, list.get(i));
  return num;
 }
 
 public static void main(String[] args) {
  String num1="549755813888";//2^39 
  String num2="33554432";//2^25
  String ans="1";
  for(int i=0;i<1000/39;i++)//main power stuff
   ans=stringMultiplication(ans, num1);
  
  ans=stringMultiplication(ans, num2);//remainder stuff
  int sum=0;
  for(int i=0;i<ans.length();i++)
   sum=sum+ans.charAt(i)-'0';
  System.out.println(ans);
  System.out.println(sum);
 }
}

Problem 15

Problem Statement :Problem15

Starting in the top left corner of a 22 grid, there are 6 routes (without backtracking) to the bottom right corner.

How many routes are there through a 2020 grid?

Algorithm:

• We need 40 steps to reach to the bottom right corner
• We have a choice of 2 directions Down and Right
• Fill any 20 steps with D the remaining will be filled with R
• The straight of answer is 40C20 ie (40!/(20!*20!))

Probelm 14

Problem Statement :Problem14

The following iterative sequence is defined for the set of positive integers:

n n/2 (n is even)
n 3n + 1 (n is odd)

Using the rule above and starting with 13, we generate the following sequence:

13 40 20 10 5 16 8 4 2 1

It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.
Which starting number, under one million, produces the longest chain?

NOTE: Once the chain starts the terms are allowed to go above one million.

Algorithm:

• The approach to this problem is simple and straightforward.
• Iterate over all the numbers less than 10^6 
• For every number check the chain length and find the biggest chain length.

Solution:

public class problem14 {
  public static void main(String[] args) {
  int r = 0,bign = 0, temp = 0;
  long n = 0;
  for (int i = 1; i < 1000000; i++) {
   n = i;
   temp = 0;
   while (n != 1) {
    if (n % 2 == 0)
     n = n / 2;
    else
     n = 3 * n + 1;
    temp++;
   }

   if (temp > bign) {
    bign = temp;
    r = i;
   }
  }
  System.out.println(r);
 }
}

Problem 13

Problem Statement :Problem13

Work out the first ten digits of the sum of the following one-hundred 50-digit numbers.

37107287533902102798797998220837590246510135740250
46376937677490009712648124896970078050417018260538
74324986199524741059474233309513058123726617309629
91942213363574161572522430563301811072406154908250
23067588207539346171171980310421047513778063246676
89261670696623633820136378418383684178734361726757
28112879812849979408065481931592621691275889832738
44274228917432520321923589422876796487670272189318
47451445736001306439091167216856844588711603153276
70386486105843025439939619828917593665686757934951
62176457141856560629502157223196586755079324193331
64906352462741904929101432445813822663347944758178
92575867718337217661963751590579239728245598838407
58203565325359399008402633568948830189458628227828
80181199384826282014278194139940567587151170094390
35398664372827112653829987240784473053190104293586
86515506006295864861532075273371959191420517255829
71693888707715466499115593487603532921714970056938
54370070576826684624621495650076471787294438377604
53282654108756828443191190634694037855217779295145
36123272525000296071075082563815656710885258350721
45876576172410976447339110607218265236877223636045
17423706905851860660448207621209813287860733969412
81142660418086830619328460811191061556940512689692
51934325451728388641918047049293215058642563049483
62467221648435076201727918039944693004732956340691
15732444386908125794514089057706229429197107928209
55037687525678773091862540744969844508330393682126
18336384825330154686196124348767681297534375946515
80386287592878490201521685554828717201219257766954
78182833757993103614740356856449095527097864797581
16726320100436897842553539920931837441497806860984
48403098129077791799088218795327364475675590848030
87086987551392711854517078544161852424320693150332
59959406895756536782107074926966537676326235447210
69793950679652694742597709739166693763042633987085
41052684708299085211399427365734116182760315001271
65378607361501080857009149939512557028198746004375
35829035317434717326932123578154982629742552737307
94953759765105305946966067683156574377167401875275
88902802571733229619176668713819931811048770190271
25267680276078003013678680992525463401061632866526
36270218540497705585629946580636237993140746255962
24074486908231174977792365466257246923322810917141
91430288197103288597806669760892938638285025333403
34413065578016127815921815005561868836468420090470
23053081172816430487623791969842487255036638784583
11487696932154902810424020138335124462181441773470
63783299490636259666498587618221225225512486764533
67720186971698544312419572409913959008952310058822
95548255300263520781532296796249481641953868218774
76085327132285723110424803456124867697064507995236
37774242535411291684276865538926205024910326572967
23701913275725675285653248258265463092207058596522
29798860272258331913126375147341994889534765745501
18495701454879288984856827726077713721403798879715
38298203783031473527721580348144513491373226651381
34829543829199918180278916522431027392251122869539
40957953066405232632538044100059654939159879593635
29746152185502371307642255121183693803580388584903
41698116222072977186158236678424689157993532961922
62467957194401269043877107275048102390895523597457
23189706772547915061505504953922979530901129967519
86188088225875314529584099251203829009407770775672
11306739708304724483816533873502340845647058077308
82959174767140363198008187129011875491310547126581
97623331044818386269515456334926366572897563400500
42846280183517070527831839425882145521227251250327
55121603546981200581762165212827652751691296897789
32238195734329339946437501907836945765883352399886
75506164965184775180738168837861091527357929701337
62177842752192623401942399639168044983993173312731
32924185707147349566916674687634660915035914677504
99518671430235219628894890102423325116913619626622
73267460800591547471830798392868535206946944540724
76841822524674417161514036427982273348055556214818
97142617910342598647204516893989422179826088076852
87783646182799346313767754307809363333018982642090
10848802521674670883215120185883543223812876952786
71329612474782464538636993009049310363619763878039
62184073572399794223406235393808339651327408011116
66627891981488087797941876876144230030984490851411
60661826293682836764744779239180335110989069790714
85786944089552990653640447425576083659976645795096
66024396409905389607120198219976047599490197230297
64913982680032973156037120041377903785566085089252
16730939319872750275468906903707539413042652315011
94809377245048795150954100921645863754710598436791
78639167021187492431995700641917969777599028300699
15368713711936614952811305876380278410754449733078
40789923115535562561142322423255033685442488917353
44889911501440648020369068063960672322193204149535
41503128880339536053299340368006977710650566631954
81234880673210146739058568557934581403627822703280
82616570773948327592232845941706525094512325230608
22918802058777319719839450180888072429661980811197
77158542502016545090413245809786882778948721859617
72107838435069186155435662884062257473692284509516
208496039i80134001723930671666823555245252804609722
53503534226472524250874054075591789781264330331690

Algorithm:

• The main highlight of the solution is the add method
• It can add two strings directly and returns the answer as a string
• Extraxt the first ten digits of the answer.

Solution:

import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;

public class problem13 {
 // to add two numbers using strings a,b are two numbers
 public static String add(String a, String b) {
  String sum = "";
  int carry = 0;
  if (a.length() > b.length()) {
   String temp = a;
   a = b;
   b = temp;
  }
  for (int i = 0; i < a.length(); i++) {
   int s = a.charAt(a.length() - 1 - i) + b.charAt(b.length() - 1 - i)
     - 96 + carry;
   if (s > 9) {
    carry = (s - s % 10) / 10;
    s = s % 10;
   } else {
    carry = 0;
   }
   sum = s + sum;
  }

  for (int i = 0; i < b.length() - a.length(); i++) {
   int s = b.charAt(b.length() - a.length() - 1 - i) + carry - 48;

   if (s > 9) {
    carry = (s - s % 10) / 10;
    s = s % 10;
   } else {
    carry = 0;
   }
   sum = s + sum;

  }
  if (carry != 0)
   return carry + sum;
  else
   return sum;
 }

 public static void main(String args[]) throws IOException {
  List<String> l = new ArrayList<String>();
  BufferedReader reader = new BufferedReader(new FileReader(new File("/home/salvo/input13")));
  String line = null, sum = "0";
  while ((line = reader.readLine()) != null)
   sum = add(sum, line);
  
  reader.close();
  System.out.println(sum);
 }
}

Problem 12

Problem Statement :Problem12

The sequence of triangle numbers is generated by adding the natural numbers. So the 7^th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

Let us list the factors of the first seven triangle numbers:

 1: 1
 3: 1,3
 6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28

We can see that 28 is the first triangle number to have over five divisors.
What is the value of the first triangle number to have over five hundred divisors?

Algorithm:

• We  solve the problem on a basic assumption that if one factor of the number is present within root of the number the we need not iterate over all the factors till n/2.
• count can be incremented twice just because of the above reason.
• The value of  n = i * (i+1) / 2 is evident for i in the  

Solution:

public class problem12 {
 public static void main(String args[]) {
  for (int i = 10;; i++) {
   int n = i * (i + 1) / 2;
   int count = 2;
   for (int k = 2; k * k <= n; k++) {
    if (n % k == 0)
     count += 2;
   }
   if (count >= 500) {
    System.out.println(n);
    break;
   }
  }
 }
}

Problem10

Problem Statement :Problem10

Algorithm:
Well I solved this in a straight forward manner running over all primes less than the limit.
We can also use Eratosthenes Sieve or the Segmented Sieve to speed up the algo. 
Solution:

public class problem10 {
 public static boolean isPrime(double num) {
  for (int i = 2; i <= Math.sqrt(num); i++) {
   if (num % i == 0)
    return false;
  }
  return true;
 }
 public static void main(String args[])
 {
  long sum=2;
  for(long i=3;;i=i+2)
  {  
   if(i>2000000)
    break;
   if(isPrime(i))
    sum+=i;
  }
 System.out.println(sum);
 }
}

Problem 9

Problem Statement :Problem9

Algorithm:
We  solve the two given equations in the problem ie

• a² + b² = c²
• a + b + c = 1000

This leads to value of  b = (1000000 - 2000 * a) /  (2000 - 2 * a).

The remaining part is olved by iterating over 1000 values of a using the above value of b and checking for c.

Solution:

public class problem9 {
 public static void main(String args[]) {
  for (int a = 999; a >= 1; a--) {
   double b = (1000000 - 2000 * a) / (double) (2000 - 2 * a);
   if ((a * a + b * b) == (1000 - a - b) * (1000 - a - b) && a != b
     && (b - (int) b) == 0 && b > 0)
    System.out.println(a * b * (1000 - a - b));
  }
 }
}

Problem 7

Problem Statement :Problem7

Algorithm:

• Because we can not run a direct loop we keep adding primes to a list and check whether the size of the list exceeds 10001.
• As soon as it does we break and print the prime.

Solution:

import java.util.ArrayList;
import java.util.List;

public class problem7 {
 public static boolean isPrime(double num) {
  for (int i = 2; i <= Math.sqrt(num); i++) {
   if (num % i == 0)
    return false;
  }
  return true;
 }
 
 public static List<Long> primes = new ArrayList<Long>();
 
 public static void main(String args[])
 {
  primes.add((long)2);
  for(long i=3;;i=i+2)
  {
   if(isPrime(i))
    primes.add(i);
   if(primes.size()==10001)
    break;
  }
 System.out.println(primes.get(10000));
 }
 
}

Problem 6

Problem Statement :Problem6

Algorithm:
Well i will say this ones is pretty direct as Problem 1 on my blog.You dont need any code for this

• Sum of k^2 is N*(N+1)*(2*N+1)/6 
• Sum of k whole square is (N*(N+1)/2)^2
• Subtract the second value from the first for N=100 for the answer 

Problem 4

Problem Statement :Problem4

Algorithm:
We brute over all possible palindromes from 999*999 to 100*100. Then we try to find all possible palindromes and find the largest among them.

Solution:

public class problem4 {

 public static boolean isPalind(int n) {
  String num = n + "";
  for (int j = 0; j < num.length(); j++) {
   if (num.charAt(j) != num.charAt(num.length() - 1 - j))
    return false;
  }
  return true;
 }

 public static void main(String args[]) {
  int max=Integer.MIN_VALUE; 
  for (int i = 999; i >= 100; i--) {
   for (int j = i; j >= 100; j--) {
    int prod = i * j;
    if (isPalind(prod)) {
     if(max<prod)
      max=prod;
    }
   }
  }
  System.out.println(max);
 }
}

Problem 3

Problem Statement : Problem 3

Algorithm:
We try to iterate over all the factors of the number.
The algorithm runs over all prime factors of a given number that are less than N^.5.

Solution:

public class problem3 {
 public static boolean isPrime(double num) {
  for (int i = 2; i <= Math.sqrt(num); i++) {
   if (num % i == 0)
    return false;
  }
  return true;
 }

 public static void main(String rgp[]) {
  long num = (long) 600851475143.0;
  
  for (int i = (int)Math.pow(num,.5);  i>=3; i--) {
   if (num % i == 0) {
    if (isPrime(i)) {
     System.out.println(i);
     break;
    }
   }
  }
 }
}

Problem 2

Problem Statement :Problem 2


Algorithm:
Generating the Fibonacci series can be read here. 

Solution:
A linear time algorithm can is written here:

public class problem2 {
 public static void main(String args[]) {
  int a = 1, b = 2;
  int sum = 2;
  while (true) {
   int c = a + b;
   a = b;
   b = c;
   if (c > 4000000)
    break;

   if (c % 2 == 0)
    sum = sum + c;
  }
  System.out.println(sum);
 }
}

Problem 1

Problem Statement : Problem 1


Algorithm:
This problem is pretty simple and requires basic set theory.

Explanation about the principle can be read here.

• Calculate sum of all multiples of 3 less than 1000
• Calculate sum of all multiples of 5 less than 1000
• Calculate sum of all multiples of 15 less than 1000

Solution:
A linear time running algorithm to calculate this is written below:

public class problem1 {
 public static void main(String args[]) {
  int sum = 0;
  for (int i = 1; i < 1000; i++) {
   if (i % 3 == 0) {
    sum += i;
    continue;
   }
   if (i % 5 == 0)
    sum += i;
  }
  System.out.println(sum);
 }
}