In this post, we're going to solve some JavaScript problems you may be asked to solve in a technical interview or meet in real live projects.

(1)Two Sum Problem

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 array = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

Solution 1

Let's assume the input is:

1. array=[1,3,10,11,14]
2. goal=13

//O(n^2) time complexity
const twoSum = (array, goal) => {
  let indexes = [];
  for (let i = 0; i < array.length; i++) {
    for (let j = i + 1; j < array.length; j++) {
      if (arry[i] + array[j] === goal) {
        indexes.push(i);
        indexes.push(j);
      }
    }
  }
  return indexes;
};

This will return an array of [1,2].
It works but if you check it out you'll notice it's running
a loop inside of a loop to figure out which two numbers add up to the goal.
That gives us a time complexity of O(n^2) which is pretty slow

A more optimized solution

1. array=[1,3,10,11,14]
2. goal=13

//0(n) time complexity
const twoSum = (array, goal) => {
  let mapOfNumbers = {};
  let twoIndexes = [];
  for (let i = 0; i < array.length; i++) {
    mapOfNumbers[array[i]] = i;
  }
  for (let i = 0; i < array.length; i++) {
    let target = goal - arr[i];
    if (mapOfNumbers[target] !== null && mapOfNumbers[target] !== i) {
      twoIndexes.push(i);
      twoIndexes.push(mapOfNumbers[target]);
    }
  }
  return twoIndexes;
};

(2) Search Insert position of K in a sorted array

Given a sorted array arr consisting of N distinct integers
and an integer K,the task is to find the index of K, if it's present in the array
arr.
Otherwise otherwise,find the index where K must be inserted to keep the array sorted
You may assume no duplicates in the array.

explanation

Input:
K=5,
const arr = [1, 3, 5, 6];
Output: 2
Explanation: Since 5 is found at
index 2 as arr[2]=5 the ouput is 5

Input: K=2,
const arr=[1,3,5,6]
Output:1
Explanation: Since 2 is not present
in the array but can be inserted at
index 1 to make the array sorted.

Solution

let findIndex = (arr, n, K) => {
  //Traverse the array
  for (let i = 0; i < n; i++) {
    //If K is found
    if (arr[i] == K) {
      return i;
    }
    //If current array element exceeds K
    else if (arr[i] > K) {
      return i;
    }
  }
  //if all elements are smaller than K
  return n;
};
//Driver code
let arr = [1, 3, 5, 6];
let n = arr.length;
let K = 2;
console.log(findIndex(arr, n, K));
//Output: 1

Time Complexity:O(N)

Auxiliary Space:O(1)

3)Check whether a triangle is valid or not if the sides are given.

Explanation

Input: a=3, b=8, c=10
Output: Valid


Input: a=3,b=4,c=1
Output: Invalid

Aproach:

A triangle is valid if the sum of its sides is greater than the third side.
If three sides are a,b, and c, then three conditions should be met.

1. a+b>c
2. a+c>a
3. a+c>b

Solution:

function checkValidity(a, b, c) {
  if (a + b > c && a + c > b && b + c > a) {
    return true;
  } else {
    return false;
  }
}

//Test code
console.log(checkValidity(3, 4, 1));