Excel Sheet Column Number

Related to question Excel Sheet Column Title

Given a column title as appear in an Excel sheet, return its corresponding column number.

For example:

A -> 1
B -> 2
C -> 3
...
Z -> 26
AA -> 27
AB -> 28

/**
 * @param {string} s
 * @return {number}
 */
var titleToNumber = function(s) {
  var a = s.split('').reverse();
  s = a.join('');
  var ans = a.reduce(function(pre, item, index) {
    return pre + Math.pow(26, index) * (s.charCodeAt(index) - 64);
  }, 0);
  return ans;
};

Excel Sheet Column Title

Given a positive integer, return its corresponding column title as appear in an Excel sheet.

For example:

1 -> A
2 -> B
3 -> C
...
26 -> Z
27 -> AA
28 -> AB

/**
 * @param {number} n
 * @return {string}
 */
var convertToTitle = function(n) {
  var ans = '';
  while (n) {
    if (n % 26 === 0) {
      ans += String.fromCharCode(26 + 64);
      n = (n / 26) - 1;
    }
    else {
      ans += String.fromCharCode(n % 26 + 64);
      n = ~~(n / 26);
    }
  }

  return ans.split('').reverse().join('');
};

Factorial Trailing Zeroes

Given an integer n, return the number of trailing zeroes in n!.

Note: Your solution should be in logarithmic time complexity.

/**
 * @param {number} n
 * @return {number}
 */
var trailingZeroes = function(n) {
  var ans = 0;
  while (n) {
  	ans += ~~(n / 5);
  	n /= 5;
  	n = ~~n;
  }

  return ans;
};

Find All Anagrams in a String

Given a string s and a non-empty string p, find all the start indices of p’s anagrams in s.

Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.

The order of output does not matter.

Example 1:

Input:
s: "cbaebabacd" p: "abc"

Output:
[0, 6]

Explanation:
The substring with start index = 0 is "cba", which is an anagram of "abc".
The substring with start index = 6 is "bac", which is an anagram of "abc".

Example 2:

Input:
s: "abab" p: "ab"

Output:
[0, 1, 2]

Explanation:
The substring with start index = 0 is "ab", which is an anagram of "ab".
The substring with start index = 1 is "ba", which is an anagram of "ab".
The substring with start index = 2 is "ab", which is an anagram of "ab".

"use strict";

/**
 * @param {string} s
 * @param {string} p
 * @return {number[]}
 */
var findAnagrams = function(s, p) {
  let len = p.length;
  let hash = {};
  let ans = {};
  let ret = [];

  for (let i = 0, l = p.length; i < l; i++) {
    let index = p.charCodeAt(i) - 97;
    ans[index] = ~~ans[index] + 1;
  }

  for (let i = 0, l = s.length; i < l; i++) {
    let index = s.charCodeAt(i) - 97;
    hash[index] = ~~hash[index] + 1;

    if (i >= len) { // remove
      let index = s.charCodeAt(i - len) - 97;
      hash[index] = ~~hash[index] - 1;
    }

    if (i + 1 >= len) { // can compare
      help() && ret.push(i - len + 1);
    }
  }

  function help() {
    for (let i = 0; i < 26; i++) {
      if (~~hash[i] !== ~~ans[i])
        return false;
    }

    return true;
  }

  return ret;
};

Find All Duplicates in an Array

Given an array of integers, $1 ≤ a[i] ≤ n$ (n = size of array), some elements appear twice and others appear once.

Find all the elements that appear twice in this array.

Could you do it without extra space and in $O(n)$ runtime?

Example:

Input:
[4,3,2,7,8,2,3,1]

Output:
[2,3]

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var findDuplicates = function(nums) {
  let len = nums.length;
  let ans = [];

  for (let i = 0; i < len; i++) {
    let item = nums[i];
    let val = Math.abs(item) - 1;

    if (nums[val] < 0)
      ans.push(val + 1);
    else
      nums[val] *= -1;
  }

  return ans;
};