# Leetcode 練習筆記
## [30 Days of LC JavaScript Challenge](https://leetcode.com/discuss/study-guide/3458761/Open-to-Registration!-30-Days-of-LC-JavaScript-Challenge/?utm_campaign=Banner9&utm_medium=Banner&utm_source=Banner&gio_link_id=xRxVYXjo)
## Day30
### Day
### Day
### Day
### Day
### Day
### Day
### Day
### Day
### Day
### Day
### Day
### Day
### Day
### Day
### Day
### Day
### Day
## Day12
### Day11
### Day10
>Input:
fn = function sum(a, b, c) { return a + b + c; }
inputs = [[1,2],[3]]
Output: 6
Explanation:
curriedSum(1, 2)(3) should return the same value as sum(1, 2, 3).

var curry = function(fn) {
    return function curried(...args) {
        if (args.length >= fn.length) {
            // 如果傳入的參數數量已達到原函式所需的參數數量，則直接執行原函式
            return fn(...args);
        } else {
            // 如果傳入的參數數量不足，則返回一個新的柯里化函式，等待更多的參數傳入
            return function(...moreArgs) {
                return curried(...args.concat(moreArgs));
            };
        }
    };
};

// 測試
function sum(a, b, c) {
    return a + b + c;
}

const curriedSum = curry(sum);
console.log(curriedSum()(1, 2, 3)); // 6
console.log(curriedSum(1)(2)(3)); // 6
console.log(curriedSum(1, 2)(3)); // 6
console.log(curriedSum(1)(2, 3)); // 6

### Day9 Memoize
>Input
"sum"
["call","call","getCallCount","call","getCallCount"]
[[2,2],[2,2],[],[1,2],[]]
Output
[4,4,1,3,2]
**Explanation**
const sum = (a, b) => a + b;
const memoizedSum = memoize(sum);
memoizedSum(2, 2); // Returns 4. sum() was called as (2, 2) was not seen before.
memoizedSum(2, 2); // Returns 4. However sum() was not called because the same inputs were seen before.
// Total call count: 1
memoizedSum(1, 2); // Returns 3. sum() was called as (1, 2) was not seen before.
// Total call count: 2

function memoize(fn) {
  const cache = {};//用一個物件來儲存 放入的input跟結果 
  return function (...args) {
    const key = JSON.stringify(args); 由於物件的key只能是string所以必須透過stringify來讓args這個array變成可以儲存的key
    if (key in cache) { 
      return cache[key];
    }
    const functionOutput = fn(...args);
    cache[key] = functionOutput;
    return functionOutput;
  };
}

### Day8 Allow One Function Call
>Input: fn = (a,b,c) => (a * b * c), calls = [[5,7,4],[2,3,6],[4,6,8]]
Output: [{"calls":1,"value":140}]
Explanation:
const onceFn = once(fn);
onceFn(5, 7, 4); // 140
onceFn(2, 3, 6); // undefined, fn was not called
onceFn(4, 6, 8); // undefined, fn was not called

var once = function(fn) {
     let hasBeenCalled = false;
    return function(...args){
 if (hasBeenCalled) {
      return undefined;
    } else {
      hasBeenCalled = true;
      return fn(...args);
    }
    }
};

### Day7 Function Composition
>Input: functions = [x => x + 1, x => x * x, x => 2 * x], x = 4
Output: 65
Explanation:
Evaluating from right to left ...
Starting with x = 4.
2 * (4) = 8
(8) * (8) = 64
(64) + 1 = 65

var compose = function(functions) {
	return function(x) {
        let result=x;
        for(let i=functions.length-1;i>=0;i--){
            result=functions[i](result);
        }
        return result;
    }
};

### Day6 Array Reduce Transformation
>Input:
nums = [1,2,3,4]
fn = function sum(accum, curr) { return accum + curr * curr; }
init = 100
Output: 130
Explanation:
initially, the value is init=100.
(100) + nums[0]^2 = 101
(101) + nums[1]^2 = 105
(105) + nums[2]^2 = 114
(114) + nums[3]^2 = 130
The final answer is 130.

var reduce = function(nums, fn, init) {
    let result=init;
    for(let i=0;i<nums.length;i++){
        result=fn(result,nums[i])
    }
    return result;
};

### Day5 Filter Elements from Array
>Example 1:
Input: arr = [0,10,20,30], fn = function greaterThan10(n) { return n > 10; }
Output: [20,30]
Explanation:
const newArray = filter(arr, fn); // [20, 30]
The function filters out values that are not greater than 10
>Example 2:
Input: arr = [1,2,3], fn = function firstIndex(n, i) { return i === 0; }
Output: [1]
Explanation:
fn can also accept the index of each element
In this case, the function removes elements not at index 0

var filter = function(arr, fn) {
  const result=[];
for(let i=0;i<arr.length;i++){
if (fn(arr[i],i)){
  result.push(arr[i])
}
}
return result;
};

### Day4 Apply Transform Over Each Element in Array
>Input: arr = [1,2,3], fn = function plusI(n, i) { return n + i; }
Output: [1,3,5]
Explanation: The function increases each value by the index it resides in.

var map = function(arr, fn) {
    return arr.map((item,i)=>fn(item,i));
};

### Day3 Counter II(closure)
>Input: init = 0, calls = ["increment","increment","decrement","reset","reset"]
Output: [1,2,1,0,0]
Explanation:
const counter = createCounter(0);
counter.increment(); // 1
counter.increment(); // 2
counter.decrement(); // 1
counter.reset(); // 0
counter.reset(); // 0

var createCounter = function(init) {
  let currentCount = init;

  function increment() {
    return ++currentCount;
  }

  function decrement() {
        currentCount -= 1;
       return currentCount;
  }

  function reset() {
      return (currentCount = init);
  }

  return { increment, decrement, reset };
};

### Day2 Counter(closure)
>Input:
n = -2
["call","call","call","call","call"]
Output: [-2,-1,0,1,2]
Explanation: counter() initially returns -2. Then increases after each sebsequent call.

var createCounter = function(n) {
    let input=n-1;
    return function() {
        input++;
        return input;
    };
};

### Day1 Create Hello World Function(closure)

var createHelloWorld = function() {
    return function(...args) {
        return "Hello World"
    }
};

1. question:
>Input: s = "codeleet", indices = [4,5,6,7,0,2,1,3]
Output: "leetcode"
Explanation: As shown, "codeleet" becomes "leetcode" after shuffling.

var restoreString = function(s, indices) {
  let shuffled = new Array(s.length);
  for (let i = 0; i < indices.length; i++) {
    shuffled[indices[i]] = s.charAt(i);
  }
  return shuffled.join('');
};
//let arr = new Array(); // -> [] 給arr長度但內容仍然是空的
//[ <8 empty items> ]
//charAt() 方法从一个字符串中返回指定的字符。

2. Deci-Binary Numbers
>Example 1:
Input: n = "32"
Output: 3
Explanation: 10 + 11 + 11 = 32

>Example 2:
Input: n = "82734"
Output: 8
82734-11111-11111-10111-10101-10100-10100-10100-10000
==>要做八次才能使得82734變成00000

>Example 3:
Input: n = "27346209830709182346"
Output: 9

（換言之就是找到最大的數字）

var minPartitions = function(n) {
    const nArr=n.split('');
    const numArr=nArr.map(s=>parseInt(s));
    return Math.max(...numArr);
};

3. Reduce
>const array1 = [1, 2, 3, 4];
// 0 + 1 + 2 + 3 + 4
const initialValue = 0;
const sumWithInitial = array1.reduce(
(accumulator, currentValue) => accumulator + currentValue,
initialValue
);
4. Question
>Input: nums = [10,4,8,3]
Output: [15,1,11,22]
Explanation: The array leftSum is [0,10,14,22] and the array rightSum is [15,11,3,0].
The array answer is [|0 - 15|,|10 - 11|,|14 - 3|,|22 - 0|] = [15,1,11,22].

var leftRigthDifference = function(nums) {
const n = nums.length;
const leftSum = Array(n).fill(0);
const rightSum = Array(n).fill(0);
for (let i = 1; i < n; i++) {
leftSum[i] = leftSum[i-1] + nums[i-1];
rightSum[n-i-1] = rightSum[n-i] + nums[n-i];
}
return nums.map((num, i) => Math.abs(leftSum[i] - rightSum[i]));
};