Refactored the project

Author: jenilChristo

src/Main.java

@@ -0,0 +1,6 @@
+public class Main {
+
+    public static void main(String[] args) {
+        System.out.println("Hello World!");
+    }
+}

src/google/AddToInteger.java

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+package google;
+
+import java.util.*;
+import java.lang.*;
+
+//This problem was recently asked by Google.
+//
+//Given a list of numbers and a number k, return whether any two numbers from the list add up to k.
+//
+//For example, given [10, 15, 3, 7] and k of 17, return true since 10 + 7 is 17.
+//
+//Bonus: Can you do this in one pass?
+public class AddToInteger {
+    public static  void main(String[] args){
+        AddToInteger instance = new AddToInteger();
+        int sum = 13;
+        int[] array = {10, 15, 3, 7};
+        System.out.println(instance.willAddToInteger(sum, array));
+    }
+    public boolean willAddToInteger(int sum, int[] array){
+        HashMap<Integer,Boolean> temp = new HashMap<Integer,Boolean>();
+        int i = 0;
+        boolean result = false;
+
+        while(i < array.length){
+            try{
+                int currentElement = array[i];
+                int pair = sum - currentElement;
+
+                if(temp.containsKey(pair)){
+                    result = true;
+                }else{
+                    temp.put(currentElement, true);
+                }
+                i++;
+            }catch (Exception e){
+                System.out.println(e);
+            }
+
+        }
+
+        return result;
+    }
+}

src/google/BinarySerialize.java

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+package google;
+
+/*
+This problem was asked by Google.
+
+Given the root to a binary tree, implement serialize(root), which serializes the tree into a string, and deserialize(s), which deserializes the string back into the tree.
+
+For example, given the following Node class
+
+class Node:
+    def __init__(self, val, left=None, right=None):
+        self.val = val
+        self.left = left
+        self.right = right
+The following test should pass:
+
+node = Node('root', Node('left', Node('left.left')), Node('right'))
+assert deserialize(serialize(node)).left.left.val == 'left.left'
+
+ */
+public class BinarySerialize {
+    public static  void main(String[] args){
+
+    }
+
+}

src/google/Fibonacci.java

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+package com.company.com.company.google;
+
+public class Fibonacci {
+    public Fibonacci(){
+
+    }
+    public static void main(String[] args){
+        Fibonacci instance = new Fibonacci();
+        System.out.println(instance.getFibonacci(100));
+    }
+
+    public int getFibonacci(int n){
+        int result;
+        int memo[] = new int[n+1]; // memoize array
+        memo[0] = 1; //handle base case
+        memo[1] =1;
+        //back-track memoize array in O(n)
+        for(int i=2; i <= n ;i++){
+            memo[i] = memo[i-1] + memo[i-2];
+        }
+        return memo[n];
+    }
+}

src/google/FirstMissingPositiveInt.java

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+package google;
+
+/*
+This problem was asked by Stripe.
+
+Given an array of integers, find the first missing positive integer in linear time and constant space. In other words, find the lowest positive integer that does not exist in the array. The array can contain duplicates and negative numbers as well.
+
+For example, the input [3, 4, -1, 1] should give 2. The input [1, 2, 0] should give 3.
+
+You can modify the input array in-place.
+ */
+public class FirstMissingPositiveInt {
+    public static void main(String[] args){
+        int[] arr = {1, 2, 0};
+        FirstMissingPositiveInt instance = new FirstMissingPositiveInt();
+        System.out.println(instance.getFirstMissing(arr));
+     }
+     public int getFirstMissing(int[] arr){
+        int n = arr.length;
+        int min= arr[0] ,i=0;
+
+        for(int j=0;j<n;j++){
+
+        }
+
+        while(i<n){
+            if(arr[i] < min && arr[i] > 0 ){
+                min = arr[i];
+            }
+            i++;
+        }
+        return min;
+     }
+}

src/uber/MultiplyAllExceptIndex.java

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+package uber;/*This problem was asked by Uber.
+
+Given an array of integers, return a new array such that each element at index i of the new array
+is the product of all the numbers in the original array except the one at i.
+
+For example, if our input was [1, 2, 3, 4, 5], the expected output would be [120, 60, 40, 30, 24].
+If our input was [3, 2, 1], the expected output would be [2, 3, 6].
+
+Follow-up: what if you can't use division?
+*/
+
+import java.util.Arrays;
+
+public class MultiplyAllExceptIndex {
+    public static void main(String[] args){
+        MultiplyAllExceptIndex instance = new MultiplyAllExceptIndex();
+        int[] arr = {1, 2, 3, 4, 5};
+        System.out.println(instance.getProductArray((arr)));
+    }
+    //O(n) solution
+    public String getProductArray(int[] arr){
+        int n = arr.length;
+        int[] left = new int [n];
+        int[] right = new int [n];
+        int[] result = new int[n];
+        left[0] = 1;
+        right[n-1] = 1;
+
+        for(int i=1;i<n;i++){
+            left[i] = left[i-1]*arr[i-1];
+        }
+        for(int j= n-2;j>=0;j--){
+            right[j] = right[j+1]*arr[j+1];
+        }
+        for(int k= 0; k< n; k++){
+            result[k] = left[k]*right[k];
+        }
+
+        return Arrays.toString(result);
+    }
+}