Add Day 6

Author: MadhavBahl

Day1/README.md

@@ -1,3 +1,5 @@
+![Fizz Buzz](./cover.png)
+
 # Day 1 -- The Fizz Buzz Challenge
 
 **Brief History** - Fizz Buzz is a group word game for children to teach them about division. Players take turns to count incrementally, replacing any number divisible by three with the word "fizz", and any number divisible by five with the word "buzz" and any number divisible by both three and five is replaced with the word "fizzbuzz"
@@ -7,7 +9,7 @@
 
 **Question**- Write a program that prints the numbers from 1 to n and for multiples of '3' print "Fizz" instead of the number, for the multiples of '5' print "Buzz", and for the numbers which are divisible by both 3 and 5, print FizzBuzz.
 
-![Fizz Buzz](./cover.png)
+![ques](./ques.png)
 
 ## JavaScript Implementation
 

Day2/README.md

@@ -1,9 +1,13 @@
+![cover](./cover.png)
+
 # Day 2 -- String Reversal and Palindrome
 
 ## Part A -- String Reversal
 
 **Question** - Given a string, write a program to return a new string with reversed order of characters.
 
+![ques](./ques.png)
+
 ## JavaScript Implementation
 
 ### [Solution 1](./JavaScript/sol1.js)

Day3/README.md

@@ -1,3 +1,5 @@
+![Hamming](./cover.png)
+
 # Day 3 -- The Hamming Distance Problem
 
 The Hamming distance between two strings of equal length is the number of positions at which the corresponding symbols are different.
@@ -8,7 +10,7 @@ Read more about Hamming Distance [here…](https://en.wikipedia.org/wiki/Hamming
 
 **Question**- Given 2 strings, we will find the number of positions at which the corresponding characters are different.
 
-![Hamming](./cover.png)
+![ques](./ques.png)
 
 **Example**
 

day1/ques.png

@@ -1,11 +1,13 @@
-# Day 4 -- Number of Vowels and Max Chars
-
 ![cover](./cover.png)
 
+# Day 4 -- Number of Vowels and Max Chars
+
 ## Part A -- Number of Vowels
 
 **Question** - Given a string, Write a program that prints the number of vowels in it.
 
+![ques](./ques.png)
+
 ## JavaScript Implementation
 
 ### [Solution 1](./JavaScript/partA_sol1.js)

day2/ques.png

@@ -1,10 +1,10 @@
+![Patterns](./cover.png)
+
 # Day 5 -- Patterns
 
 Pattern programs like printing a pyramid, inverted pyramid etc. are some very famous problems and a good way to test your logic and knowledge of nested loops. Hence, on the day 5 of Daily Codes, let’s do some pattern based programs 😀
 
-![Patterns](./cover.png)
-
-Here's the questions for today 😁
+Here's the questions for today 
 
 ### Pattern 1
 
@@ -117,6 +117,8 @@ input = 5
  *                 *
 ```
 
+![ques](./ques.png)
+
 ## Pattern 1
 
 ```

day3/ques.png

@@ -0,0 +1,23 @@
+/**
+ * @author MadhavBahlMD
+ * @date 27/12/2018
+ */
+
+import java.util.Scanner;
+
+public class SentenceCap1 {
+    public static void main(String[] args) {
+        Scanner input = new Scanner(System.in);
+        System.out.print("Enter the sentence: ");
+        String str = input.nextLine();
+
+        String[] words = str.split("\\s+");
+
+        for (int i=0; i<words.length; i++) {
+            words[i] = Character.toUpperCase(words[i].charAt(0)) + words[i].substring(1, words[i].length());
+        }
+
+        // Join the array
+        System.out.print(String.join(" ", words));
+    }
+}

day4/README.md

@@ -0,0 +1,29 @@
+/**
+ * @author MadhavBahlMD
+ * @date 27/12/2018
+ */
+ 
+import java.util.Scanner;
+
+public class SentenceCap2 {
+    public static void main(String[] args) {
+        Scanner input = new Scanner (System.in);
+        System.out.print("Enter the sentence: ");
+        String sentence = input.nextLine();
+        String capitalized = "";
+
+        for (int i=0; i<sentence.length(); i++) {
+            if (i==0)   capitalized += Character.toUpperCase(sentence.charAt(i));
+            else {
+                if (sentence.charAt(i-1) == ' ') {
+                    capitalized  += Character.toUpperCase(sentence.charAt(i));
+                } else {
+                    capitalized += sentence.charAt(i);
+                }
+            }
+        }
+
+        // Print the results
+        System.out.println("Capitalized String is: " + capitalized);
+    }
+}

day4/ques.png

@@ -0,0 +1,29 @@
+/**
+ * @author MadhavBahlMD
+ * @date 27/12/2018
+ */
+
+import java.util.Scanner;
+import java.lang.*;
+
+public class WordRev {
+    public static void main(String[] args) {
+        Scanner input = new Scanner (System.in);
+        System.out.print("Enter the sentence: ");
+        String sentence = input.nextLine();
+
+        String[] words = sentence.split("\\s+");
+        String reversed;
+
+        for (int i=0; i<words.length; i++) {
+            reversed = "";
+            for (int j=0; j<words[i].length(); j++) {
+                reversed = words[i].charAt(j) + reversed;
+            }
+            words[i] = reversed;
+        }
+
+        String wordsReversed = String.join(" ", words);
+        System.out.println("String with reversed words: " + wordsReversed);
+    }
+}

day5/README.md

@@ -0,0 +1,31 @@
+/**
+ * @author MadhavBahlMD
+ * @date 27/12/2018
+ * METHOD -- Check the lengths of both strings, sort them and then check whether they are same
+ */
+
+function anagram (str1, str2) {
+    let len1 = str1.length,
+        len2 = str2.length;
+    
+    // Compare lengths
+    if (len1 !== len2) {
+        console.log ('Invalid Input');
+        return -1;
+    }
+
+    // sort the strings
+    let sortedStr1 = str1.split('').sort().join(''),
+        sortedStr2 = str2.split('').sort().join('');
+
+    // Compare both strings
+    if (sortedStr1 === sortedStr2) {
+        console.log(`"${str1}" and "${str2}" are Anagrams`);
+        return 1;
+    } else {
+        console.log(`"${str1}" and "${str2}" are not Anagrams`);
+        return 0;
+    }
+}
+
+anagram ('LISTEN', 'SILENT');

day5/ques.png

@@ -0,0 +1,42 @@
+/**
+ * @author MadhavBahlMD
+ * @date 27/12/2018
+ * METHOD -- Prepare 2 objects which stores frequency of every character in both strings, compare those 2 objects  (dictionaries in python)
+ */
+
+function anagram (str1, str2) {
+    let len1 = str1.length,
+        len2 = str2.length;
+    
+    // Compare lengths
+    if (len1 !== len2) {
+        console.log ('Invalid Input');
+        return -1;
+    }
+
+    // Make  frequency objects
+    let countObj1 = {},
+        countObj2 = {};
+
+    for (let i=0; i<len1; i++) {
+        countObj1[str1[i]] = countObj1[str1[i]] + 1 || 1;
+    }
+
+    for (let i=0; i<len2; i++) {
+        countObj2[str2[i]] = countObj2[str2[i]] + 1 || 1;
+    }
+
+    // compare frequency objects
+    // Please note that there is no direct way of comparing 2 objects.
+    // We can either use some librries like Lowdash, or we can check the equality of each key value pair in objects, which is indeed a tedious task, but still, lets do it :)
+    for (let key in countObj1) {
+        if (countObj1[key] !== countObj2[key]) {
+            console.log(`"${str1}" and "${str2}" are not Anagrams`);
+            return 0;
+        }
+    }
+
+    console.log(`"${str1}" and "${str2}" are Anagrams`);
+}   
+
+anagram ('LISTEN', 'MILENT');

day6/Java/SentenceCap1.java

@@ -0,0 +1,49 @@
+/**
+ * @author MadhavBahlMD
+ * @date 27/12/2018
+ * A simple method which first compares the lengths of strings and then iterates through the characters of any string and check whether it exists in the other one as well and does same for the other string
+ * Please note that this is not at all an efficient method. Do not use this.
+ */
+
+function anagram (str1, str2) {
+    let len1 = str1.length,
+        len2 = str2.length;
+
+    // Lengths of both strings must be same
+    if (len1 !== len2) {
+        console.log ('Invalid Input');
+        return -1;
+    }
+
+    // check characters of string 1 are there in string 2
+    let flag = 1;
+    for (let char of str1) {
+        if (str2.indexOf(char) < 0) {
+            flag = 0;
+            break;
+        }
+    }
+
+    if (flag !== 1) {
+        console.log (`${str1} and ${str2} are not Anagrams`);
+        return 0;
+    }
+
+    for (let char of str2) {
+        if (str1.indexOf(char) < 0) {
+            flag = 0;
+            break;
+        }
+    }
+
+    if (flag !== 1) {
+        console.log (`${str1} and ${str2} are not Anagrams`);
+        return 0;
+    } 
+    else {
+        console.log (`${str1} and ${str2} are Anagrams`);
+        return 1;
+    }
+}
+
+anagram ('LISTEN', 'SILENT');

day6/Java/SentenceCap2.java

@@ -0,0 +1 @@
+    

day6/Java/WordRev.java

@@ -0,0 +1,49 @@
+/**
+ * @author MadhavBahlMD
+ * @date 27/12/2018
+ */
+
+// Without inbuilt reverse or split, a straightforward method
+
+function wordRev (line) {
+    // iterate through the string and add each word in a new array (splitting it on white space)
+    let words = [], count = 0, word = '';
+
+    for (let i=0; i<line.length; i++) {
+        if (line[i] !== ' ') {
+            word += line[i];
+        } else {
+            words[count] = word;
+            word = '';
+            count++;
+        }
+    }
+    // Add the last word as well to the words array as well
+    words[count] = word;
+    count++;
+
+    // Reverse the words
+    let temp;
+    for (let i=0; i<count; i++) {
+        temp = '';
+        for (let j=words[i].length-1; j>=0; j--) {
+            temp += words[i][j];
+        }
+        words[i] = temp;
+    }
+
+    // join the elements (Ok, let's not use the traditional join() method -_-)
+    let reversed = '';
+    for (let i=0; i<count; i++) {
+        if (i!=count-1) {
+            reversed += words[i] + ' ';
+        } else {
+            reversed += words[i];
+        }
+    }
+
+    // print the result
+    return reversed;
+}
+
+console.log(wordRev ('hello world greetings'));