# Missing Numbers¶

## JavaScript¶

### Solution 1¶

/**
* Finds numbers that are present in ys but missing in xs.
*
* - T.C: O(n).
* - S.C: O(n).
*
* @param {number[]} xs
* @param {number[]} ys
* @returns {number[]} The array of the missing numbers (the
*   difference).
*/
function missingNums(arr, brr) {
var freqXs = {},
freqYs = {},
i,
n;

for (i = 0; n = xs[i], i < xs.length; ++i)
freqXs[n] = freqXs[n] + 1 || 1;

for (i = 0; n = ys[i], i < ys.length; ++i)
freqYs[n] = freqYs[n] + 1 || 1;

return Object.keys(freqYs).reduce((missing, key) => {
if (freqYs[key] === freqXs[key])
return missing;

missing.push(Number(key));

return missing;
}, []);
}


Solution using frequency counters. There are three loops involved but neither is nested, which makes for time complexity $$O(n)$$.

### Solution 2 with helper¶

/**
* Returns a hash map of the frequencies of the values in xs.
*
* - T.C: O(n).
* - S.C: O(n).
*
* @param {number[]} xs
* @returns {{ [key: string]: number }}
*/
function countFreq(xs) {
return xs.reduce(function reducer(freqs, x) {
freqs[x] = freqs[x] + 1 || 1;
return freqs;
}, {});
}

/**
* Finds numbers that are present in ys but missing in xs.
*
* - T.C: O(n).
* - S.C: O(n).
*
* @param {number[]} xs
* @param {number[]} ys
* @returns {number[]} The array of the missing numbers (the
*   difference).
*/
function missingNums(xs, ys) {
var freqXs = countFreq(xs);
var freqYs = countFreq(ys);

return Object.keys(freqYs).reduce((missing, key) => {
if (freqYs[key] === freqXs[key]) return missing;

return [...missing, Number(key)];
}, []);
}


Essentially the same as solution 1, but extracting part of the logic into a helper function.

### Solution 3 with sets, spread, sort and splice¶

/**
* Compares two numbers for ascending sorting.
*
* - T.C: O(1).
* - S.C: O(1).
*
* @param {number} a
* @param {number} b
* @returns {number}
*/
function sortAsc(a, b) {
return a - b;
}

/**
* Finds numbers that are present in ys but missing in xs.
*
* - T.C: O(n²).
* - S.C: O(1).
*
* @param {number[]} xs
* @param {number[]} ys
* @returns {number[]} The array of the missing numbers (the
*   difference).
*/
function missingNums(xs, ys) {
for (const n of xs) {
if (ys.includes(n)) {
const idx = ys.indexOf(n);
ys.splice(idx, 1);
}
}

//
// • Use sets to remove duplicates.
// • Convert back to array with spread syntax.
//
return [...new Set(ys)].sort(sortAsc);
}


For each number in xs, try to find it in ys and remove it from ys if present. In the end, only numbers in ys that are missing in xs will remain in ys.

Also use some ECMAScript set and spread stuff to return the final result.