10 Algos 🚀 to Boost Your JavaScript Skills 🦄

1) Find the missing number in the array

Input: [1, 2, 3, 4, 6, 7, 8, 9, 10]
Output: 5

const find_missing = function(input) {
  let n = input.length + 1;

  let sum = 0;
  for (let i in input) {
    sum += input[i];
  }

  return Math.floor((n * (n + 1)) / 2) - sum;
};

Hint: Arithmetic series sum formula:

2) Reverse Integer

Input: num = 123
Output: 321
Input: num = -123
Output: -321

const reverse = function(num) {
    let result = 0;
    while (num !== 0) {
      result = result * 10 + num % 10;
      num = Math.trunc(num / 10);
    }

    if (result > 2**31 || result < -(2**31)) return 0;
    return result;
};

3) Array Permutations

Input: [1,2,3]
Output: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]

const permute = function(nums) {
    let results = [];

    let go = (current) => {
      if (current.length === nums.length){
        results.push(current);
        return;
      }
      nums.forEach(n => {
        if (!current.includes(n)){
          go([...current, n]);
        }
      });
    }
    go([]);
    return results;
};

4) Permutation in String

Input: s1 = "ab", s2 = "eidbao"
Output: true
Input: s1 = "aa", s2 = "eidbao"
Output: false

const checkPermutation = function(s1, s2) {
  const len1 = s1.length, len2 = s2.length;
  if (len1 > len2) return false;

  const count = Array(26).fill(0);
  for (let i = 0; i < len1; i++) {
      count[s1.charCodeAt(i)-97]++;
      count[s2.charCodeAt(i)-97]--;
  }
  if (!count.some(e => e !== 0)) return true;

  for (let i = len1; i < len2; i++) {
      count[s2.charCodeAt(i)-97]--;
      count[s2.charCodeAt(i-len1)-97]++;
      if (!count.some(e => e !== 0)) return true;
  }
  return false;
};

5) Longest Valid Parentheses

Input: "(()"
Output: 2
Input: ")()())"
Output: 4

const longestValidParentheses = function(S) {
  let stack = [-1], ans = 0;
  for (let i = 0; i < S.length; i++)
    if (S[i] === '(') stack.push(i)
    else if (stack.length === 1) stack[0] = i
    else stack.pop(), ans = Math.max(ans, i - stack[stack.length-1])
  return ans
};

6) 4Sum

const fourSum = function(nums, target) {
  let result = [];
  let length = nums.length;
  if (length < 4) return result; 
  nums = nums.sort((a, b) => a - b );

  for (let i = 0; i < length - 3; i++) {
    if (nums[i] === nums[i - 1]) continue;
    for (let j = i + 1; j < length - 2; j++) {
      if (j > i + 1 && nums[j] === nums[j - 1]) continue;

      let k = j + 1;
      let l = length - 1;

      while (k < l) {
        const sum = nums[i] + nums[j] + nums[k] + nums[l];

        if (sum === target) {
          result.push([nums[i], nums[j], nums[k], nums[l]])
        }

        if (sum <= target) {
          k += 1;
          while (nums[k] === nums[k - 1]) {
            k += 1;
          }
        }

        if (sum >= target) {
          l -= 1;
          while (nums[l] === nums[l + 1]) {
            l -= 1;
          }
        }
      }
    }
  }

  return result;
};

7) Multiply Strings

Input: num1 = "2", num2 = "3"
Output: "6"

const multiply = function(num1, num2) {
    if (num1 == 0 || num2 == 0) return '0';
    const result = [];

    for (let a = num1.length - 1; a >= 0; a--) {
        for (let b = num2.length - 1; b >= 0; b--) {
            const p1 = a + b;
            const p2 = a + b + 1;
            const sum = (result[p2] ?? 0) + num1[a] * num2[b];

            result[p1] = (result[p1] ?? 0) + Math.floor(sum / 10);
            result[p2] = sum % 10;
        }
    }
    result[0] == 0 && result.shift();
    return result.join('');
};

8) Shortest Palindrome

Input: s = "aacecaaa"
Output: "aaacecaaa"
Input: s = "abcd"
Output: "dcbabcd"

const shortestPalindrome = function(s) {
  let index = 0;
  for (let i = s.length - 1; i >= 0; i--) {
    if (s[i] === s[index]) index++;
  }
  if (index === s.length) return s;
  let remainingRev = s.substring(index, s.length);
  console.log(remainingRev);
  remainingRev = reverse(remainingRev);

  return remainingRev + shortestPalindrome(s.substring(0, index)) + s.substring(index);
};

function reverse(string) {
  let myString = '';
  for (let i = string.length - 1; i >= 0; i--) {
    myString = myString + string[i];
  }
  return myString;
};

9) Integer to English Words

Input: num = 123
Output: "One Hundred Twenty Three"
Input: num = 1234567
Output: "One Million Two Hundred Thirty Four Thousand Five Hundred Sixty Seven"

const numberToWords = function(num) {
  let result = toHundreds(num % 1000);
  const bigNumbers = ["Thousand", "Million", "Billion"];
  for (let i = 0; i < 3; ++i) {
    num = Math.trunc(num / 1000);
    result = num % 1000 !== 0 ? [toHundreds(num % 1000), bigNumbers[i], result].filter(Boolean).join(" ") : result;
  }
  return result.length === 0 ? "Zero" : result;
}

function toHundreds(num) {
  const numbers = ["", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten",
    "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"];
  const tens = ["", "", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"];
  const result = Array(3).fill("");
  let a = Math.trunc(num / 100), b = num % 100, c = num % 10;
  result[0] = a > 0 && `${numbers[a]} Hundred`;
  result[1] = b < 20 ? numbers[b] : tens[Math.trunc(b / 10)]
  result[2] = b >= 20 && `${numbers[c]}`;
  return result.filter(Boolean).join(" ");
}

10) Ransom Note

Input: ransomNote = "aa", magazine = "ab"
Output: false
Input: ransomNote = "aa", magazine = "aab"
Output: true

const canConstruct = function(ransomNote, magazine) {
  if (ransomNote.length > magazine.length) return false;
  let magMap = new Map();

  for(let char of magazine) {
    magMap.set(char, (magMap.get(char) || 0 ) + 1);
  }

   for(let note of ransomNote) {
    let counter = magMap.get(note);
    if (!counter) return false;

    magMap.set(note, --counter);
   }
  return true;
};

After doing some research, I came out with these 10 algorithms, each of them have special tricks that not only will increase our JavaScript skills, but it will also help us to improve our critical thinking ability 🙂