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30 Days of Code

Day 6: Let's Review

Solution

   Explanation: This is a simple problem. It asks to create 2 string one containing all the character at the even position and the other containing all the character at the odd position. 
  Just iterate through the string till the last character and keep finding the index mod 2 for each position. if the mod is 0 add the character at the position to the even string and if the mod is 1 add the character to the odd string. 
  Here index is the position of each character in the string. I have used substrings to to create the string in java code. There should be a similar way in all the other programming language.

Day 10: Binary Numbers

Solution

    Explanation: Take two integer variable. One storing the max and the second one storing the count of consecutive 1's. Now the input in the problem is a decimal number (number with base 10). 
  Here we need to think first how to convert the decimal to binary. and then count 1's. But doing this in two step will increase the execution time. So we have to do it in one step. If we go back to the school days we can get the answer.
  Execute a while loop till the number is greater then zero. Inside the loop number mod 2 result is the last digit of the binary. Then re assign the number value dividing the number by 2 (n=n/2). Now in the for loop the number value is changed so aging repeat the same step get the mod but this time time the result of mode is the second last digit of binary.
  As i mentioned the while loop should be running till the number is greater then 0 and the every time diving the number keep decreasing its value. This logic fetch the binary equivalent to decimal. But the core problem is still there.find the consecutive one.
  That should be easy now . Till the time mode consecutively returns 1 keep increasing the value of count variable by one. as soon as you get zero compare the count with the  max and if count is greater then max. Copy the count to max and initialize the max to zero again.
  even if the count is not greater then max just don copy the count to max but initialize it to zero. One thing to note. There will be situation where loop will terminate and the count is not compared with max. so every time before returning the result compare count and the max. 

Day 11: 2D Arrays

Solution

Day 18: Queues and Stacks

Solution

Day 20: Sorting

Solution

Day 22: Binary Search Trees

Solution

Day 23: BST Level-Order Traversal

Solution

Day 24: More Linked Lists

Solution

Day 25: Running Time and Complexity

Solution

Day 26: Nested Logic

Solution

Day 28: RegEx, Patterns, and Intro to Databases

Solution

Day 29: Bitwise AND

Solution

Algorithms

Warmup

Simple Array Sum

Java Solution

Scala Solution

Python Solution

   EXPLANATION :
   Sum all the elments of the Array.

Compare the Triplets

Java Solution

Scala Solution

Python Solution

  EXPLANATION :

Implementation

Strings

Sorting

Search

Graph Theory

Greedy

Dynamic Programming

Constructive Algorithms

Bit Manipulation

Recursion

Game Theory

NP Complete