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clingoramclingoram
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Add problem no.1390 in LeetCode,and no.17 in codewars.
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Lines changed: 131 additions & 62 deletions

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‎javascript/LeetCode/Array/1390.js‎

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/**
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* 1390. Four Divisors
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*
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* 參數為數值陣列，找出元素能夠被整除4次的為何？並將能整除該元素的數字加總回傳
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* @param {number[]} nums
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* @return {number}
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*/
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var sumFourDivisors = function(nums) {
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// 能整除元素的除了最小的1之外，還有它自己，所以固定整除的有2個
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// solution 1.此方法可用，但若用在大資料，會tle
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// let ans = 0;
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// for(let i = 0;i < nums.length;++i) {
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// let arr = divisors(nums[i]);
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// if(arr.length === 4){
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// ans += arr.reduce((a,b) => a + b,0);
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// }
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// }
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// return ans;
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// /**
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// * 每個元素能被整除的數字有哪些
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// * @param {number} e
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// * @returns {number[]}
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// */
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// function divisors(e){
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// let divisor = [];
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// for(let i = 1;i <= e;++i) {
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// if(e % i === 0){
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// divisor.push(i);
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// }
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// }
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// return divisor;
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// }
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// solution 2.
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let ans = 0;
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for(const a of nums){
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let divisorsCount = 0;
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let sum = 0;
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for(let i = 1;i * i <= a;++i) {
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if(a % i === 0){
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divisorsCount++;
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sum += i;
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if (i * i !== a) {
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divisorsCount++;
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sum += a / i;
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}
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}
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}
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if(divisorsCount === 4){
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ans += sum;
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}
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}
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return ans;
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};
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let nums = [21,4,7];
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/***
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* ans: 32
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*
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* 21 has 4 divisors: 1, 3, 7, 21
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* 4 has 3 divisors: 1, 2, 4
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* 7 has 2 divisors: 1, 7
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* The answer is the sum of divisors of 21 only.
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*/
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// let nums = [21,21];
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// 64 (32 + 32)
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console.log(sumFourDivisors(nums));

‎javascript/codewar/array/17.js‎

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/**
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* 3kyu - How many are smaller than me II?
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*
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* 回傳arr[i]的右邊有幾個是小於自己的
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*
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* @param {number[]} arr
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* @returns {number[]}
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*/
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function smaller(arr) {
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// 這方法ok，但不適用於large test cases
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// let ans = [];
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// for(let i = 0;i < arr.length;++i) {
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// let count = 0;
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// for(let j = 0;j < arr.length;j++) {
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// // if(arr[i] === arr[j]){
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// // continue;
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// // }
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// if(arr[i] > arr[j]){
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// count++;
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// }
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// }
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// ans[i] = count;
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// }
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// return ans;
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return arr.map((current, i) => {
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let count = 0;
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// 比較當前元素右邊所有的元素
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for (let j = 0; j < arr.length; j++) {
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if (arr[j] < current) {
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count++;
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}
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}
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return count;
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});
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// binary search
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// const result = new Array(arr.length).fill(0);
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// const sortedArray = [];
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// // 從右往左處理每個元素
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// for (let i = arr.length - 1; i >= 0; i--) {
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// const current = arr[i];
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// // binary search
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// let left = 0;
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// let right = sortedArray.length;
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// while (left < right) {
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// const mid = Math.floor((left + right) / 2);
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// if (sortedArray[mid] < current) {
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// left = mid + 1;
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// } else {
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// right = mid;
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// }
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// }
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// result[i] = left;
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// sortedArray.splice(left, 0, current);
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// }
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// return result;
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}
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console.log(assert.deepEqual(smaller([5, 4, 7, 9, 2, 4, 1, 4, 5, 6]), [5, 2, 6, 6, 1, 1, 0, 0, 0, 0]));
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console.log(assert.deepEqual(smaller([5, 4, 3, 2, 1]), [4, 3, 2, 1, 0]))

‎javascript/index.js‎

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*/
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// console.log(specialTriplets(nums));
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/**
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* 3kyu - How many are smaller than me II?
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*
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* 回傳arr[i]的右邊有幾個是小於自己的
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*
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* @param {number[]} arr
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* @returns {number[]}
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*/
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function smaller(arr) {
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// 這方法ok，但不適用於large test cases
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// let ans = [];
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// for(let i = 0;i < arr.length;++i) {
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// let count = 0;
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// for(let j = 0;j < arr.length;j++) {
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// // if(arr[i] === arr[j]){
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// // continue;
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// // }
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// if(arr[i] > arr[j]){
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// count++;
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// }
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// }
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// ans[i] = count;
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// }
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// return ans;
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return arr.map((current, i) => {
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let count = 0;
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// 比較當前元素右邊所有的元素
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for (let j = 0; j < arr.length; j++) {
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if (arr[j] < current) {
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count++;
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}
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}
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return count;
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});
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// binary search
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// const result = new Array(arr.length).fill(0);
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// const sortedArray = [];
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// // 從右往左處理每個元素
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// for (let i = arr.length - 1; i >= 0; i--) {
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// const current = arr[i];
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// // binary search
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// let left = 0;
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// let right = sortedArray.length;
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// while (left < right) {
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// const mid = Math.floor((left + right) / 2);
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// if (sortedArray[mid] < current) {
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// left = mid + 1;
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// } else {
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// right = mid;
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// }
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// }
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// result[i] = left;
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// sortedArray.splice(left, 0, current);
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// }
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// return result;
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}
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console.log(assert.deepEqual(smaller([5, 4, 7, 9, 2, 4, 1, 4, 5, 6]), [5, 2, 6, 6, 1, 1, 0, 0, 0, 0]));
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console.log(assert.deepEqual(smaller([5, 4, 3, 2, 1]), [4, 3, 2, 1, 0]))

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