Algorithm problems

Some useful steps to develop reusable algorithm :

1. Model the problem
2. Find a algorithm to solve the problem.
3. Analyze algorithm asking yourself, is it fast enough?, Does it fit in memory?
4. If not, figure why is that?
5. Find a way to address the problem.
6. Iterate until satisfied.

Problem #1. Find maximum repeating element and count.

 var countFreq = ary => {

    let tempHashmap = {}, maxFreqCount = 0, maxFreqElem = 0;
    for(let elem of ary){
        !tempHashmap[elem]? tempHashmap[elem] = 1 : tempHashmap[elem] += 1;
        if(tempHashmap[elem] > maxFreqCount){
            maxFreqCount = tempHashmap[elem];
            maxFreqElem = elem;
        }
    }
    return { maxFreqCount, maxFreqElem };
};

countFreq([1, 1, 2, 1 ]) results { maxFreCount : 3, maxFreqElem = 1 }

Problem #2, common elements in sorted array

var commonElementsInSortedArray = (array1, array2) => {
 let resultArray = [], pointer1 = 0, pointer2 = 0;
 while(pointer1 < array1.length && pointer2 < array2.length){
  if(array1[pointer1] == array2[pointer2]){
   resultArray.push(array1[pointer1]);
   pointer1++;
   pointer2++;
        }else if(array1[pointer1] > array1[pointer2]){
   pointer2++;
        }else{
   pointer1++;
        }
 }
 return resultArray;
};

commonElementsInSortedArray([1, 3, 5], [1, 5]); //results [1, 5]

Problem #3, 

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?

Soln : const isSumExists = (nums, sum) => { 
 let tempHash = {};
    for(let n of nums){
  let requireNum = sum - n;;
  if(tempHash[requireNum] === n){
   return true;
        }
  tempHash[n] = requireNum;
    }
return false;
};

Problem 4 from leetcode [Easy].

Given an array of integers, return indices of the two numbers such that they 

add up to a specific target.
You may assume that each input would have exactly one solution, and you may not 

use the same element twice.
Example:
Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

Ans:

var twoSum = function(nums, target) {
    let tempHash = {};
    for(let  [index, element ] of nums.entries()){
  let neededNumber = target - element;
  if(tempHash[neededNumber] > -1){ 
            return [tempHash[neededNumber], index]; 
        }
        tempHash[element] = index;
    }
    return [];
};

//twoSum([3,2,4], 6);
twoSum([2, 7, 11, 15], 9);

Feedback : 

Runtime: 60 ms, faster than 71.96% of JavaScript online submissions for Two Sum.
Memory Usage: 37.4 MB, less than 7.44% of JavaScript online submissions for Two Sum.

Problem #5, Daily Coding Problem: Problem #2 [Hard].

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?

Soln :

var elemProductAry = function(srcAry){
    let productOfAll = 1;
    for(const n of srcAry){
        productOfAll *= n;
    }

    return srcAry.map(function(n){
        return productOfAll / n;
    });
}

elemProductAry([1, 2, 3, 4, 5]);
//elemProductAry([3, 2, 1]);

Problem #5, Daily Coding Problem: Problem #3 [Medium].
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'

Reference :

1. https://www.dailycodingproblem.com/blog/tips-to-ace-your-coding-interview/