solution use sum of 3 for sum of two javascript code example
Example 1: two sum javascript
var twoSum = function(nums, target) {
//hash table
var hash = {};
for(let i=0; i<=nums.length; i++){
//current number
var currentNumber = nums[i];
//difference in the target and current number
var requiredNumber = target - currentNumber;
// find the difference number from hashTable
const index2 = hash[requiredNumber];
// if number found, return index
// it will return undefined if not found
if(index2 != undefined) {
return [index2, i]
} else {
// if not number found, we add the number into the hashTable
hash[currentNumber] = i;
}
}
};
Example 2: javascript sum of all the multiples of 3 or 5 below 1000 running total
(() => { // sum35 :: Int -> Int const sum35 = n => { // The sum of all positive multiples of // 3 or 5 below n. const f = sumMults(n); return f(3) + f(5) - f(15); }; // sumMults :: Int -> Int -> Int const sumMults = n => // Area under straight line // between first multiple and last. factor => { const n1 = quot(n - 1)(factor); return quot(factor * n1 * (n1 + 1))(2); }; // ------------------------- TEST -------------------------- // main :: IO () const main = () => fTable('Sums for n = 10^1 thru 10^8:')(str)(str)( sum35 )( enumFromTo(1)(8) .map(n => Math.pow(10, n)) ); // ------------------------ GENERIC ------------------------ // enumFromTo :: Int -> Int -> [Int] const enumFromTo = m => n => !isNaN(m) ? ( Array.from({ length: 1 + n - m }, (_, i) => m + i) ) : enumFromTo_(m)(n); // quot :: Int -> Int -> Int const quot = n => m => Math.floor(n / m); // ------------------------ DISPLAY ------------------------ // compose (<<<) :: (b -> c) -> (a -> b) -> a -> c const compose = (...fs) => // A function defined by the right-to-left // composition of all the functions in fs. fs.reduce( (f, g) => x => f(g(x)), x => x ); // fTable :: String -> (a -> String) -> (b -> String) // -> (a -> b) -> [a] -> String const fTable = s => // Heading -> x display function -> // fx display function -> // f -> values -> tabular string xShow => fxShow => f => xs => { const ys = xs.map(xShow), w = Math.max(...ys.map(length)); return s + '\n' + zipWith( a => b => a.padStart(w, ' ') + ' -> ' + b )(ys)( xs.map(x => fxShow(f(x))) ).join('\n'); }; // length :: [a] -> Int const length = xs => // Returns Infinity over objects without finite // length. This enables zip and zipWith to choose // the shorter argument when one is non-finite, // like cycle, repeat etc 'GeneratorFunction' !== xs.constructor.constructor.name ? ( xs.length ) : Infinity; // list :: StringOrArrayLike b => b -> [a] const list = xs => // xs itself, if it is an Array, // or an Array derived from xs. Array.isArray(xs) ? ( xs ) : Array.from(xs); // str :: a -> String const str = x => Array.isArray(x) && x.every( v => ('string' === typeof v) && (1 === v.length) ) ? ( x.join('') ) : x.toString(); // take :: Int -> [a] -> [a] // take :: Int -> String -> String const take = n => // The first n elements of a list, // string of characters, or stream. xs => 'GeneratorFunction' !== xs .constructor.constructor.name ? ( xs.slice(0, n) ) : [].concat.apply([], Array.from({ length: n }, () => { const x = xs.next(); return x.done ? [] : [x.value]; })); // zipWith :: (a -> b -> c) -> [a] -> [b] -> [c] const zipWith = f => // Use of `take` and `length` here allows zipping with non-finite lists // i.e. generators like cycle, repeat, iterate. xs => ys => { const n = Math.min(length(xs), length(ys)); return (([as, bs]) => Array.from({ length: n }, (_, i) => f(as[i])( bs[i] )))([xs, ys].map( compose(take(n), list) )); }; // --- return main();})();