Count Inversion in array

Problem Overview

Given an integer array nums, return the number of inversion in the array.

Inversion: Two number nums[i] and nums[j] forms and inversion if nums[i]>nums[j] and i<j.

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Sample Input Output

Input: nums = [0,5,2,3,1];
Output: 5

Javascript Solution

For every element i inversion count will be total numbers from i+1 to n count which are less than nums[i].

Brute Force Solution

We pick one number nums[i] and count the all number from i+1 to n that are less than nums[i].

Time Complexity: O(N*N)

Space Complexity: O(1)

Code

var inversionCount = function(nums) {
    let inversion = 0;

    for (let i = 0; i < nums.length; i++) {
        for (let j = i + 1; j < nums.length; j++) {
            if (nums[i] > 2 * nums[j]) {
                inversion++;
            }
        }
    }
    return inversion;
};

Using Devide and Conquer Algorithm

Iterate over the shortest array s
Keep accumulating (XOR ^ing) the elements of both s and t, after we have converted them to Numbers with CharCodeAt() so the XOR can work. Elements that appear both in t and s will give us zero ( X^X=0), yet that zero will not affect the total sum (X^0=X)
Return the converted back to character final sum, which is essentially the (desired) character not present in s

Original Solution Link GeorgeChryso;

const findTheDifference = function(s, t) {
    let sum=0 // initialize the accumulator

	// t.length > s.length because there will always be 1 extra letter on t (the one we want to return)
    for (let i = 0; i < s.length; i++) {
		sum^=                       // accumulate with XOR
		t[i].charCodeAt()^          // the converted to Number t[i]
        s[i].charCodeAt()           // along with a converted to Number s[i],

    }
    sum^=t[i].charCodeAt()          // add the remaining letter of t
   return String.fromCharCode(sum)  // return the converted to Character total

};

Quick example:
s='abcd' , t='dcbqa'

Same Algorithm Implementation Using ES6

/**
 * @param {number} x
 * @return {number}
 */
var findTheDifference = function (s, t) {
  return String.fromCharCode(
    [...s, ...t].reduce((acc, c) => {
      return acc ^ c.charCodeAt(0)
    }, 0)
  )
}