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Commitbaef259

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Added tasks 2630-2637
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2630\. Memoize II
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Hard
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Given a function`fn`, return a**memoized** version of that function.
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A**memoized**function is a function that will never be called twice with the same inputs. Instead it will return a cached value.
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`fn` can be any function and there are no constraints on what type of values it accepts. Inputs are considered identical if they are`===` to each other.
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**Example 1:**
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**Input:**
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getInputs = () =>[[2,2],[2,2],[1,2]]
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fn = function (a, b) { return a + b; }
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**Output:**[{"val":4,"calls":1},{"val":4,"calls":1},{"val":3,"calls":2}]
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**Explanation:**
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const inputs = getInputs();
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const memoized = memoize(fn);
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for (const arr of inputs) {
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memoized(...arr);
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}
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For the inputs of (2, 2): 2 + 2 = 4, and it required a call to fn().
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For the inputs of (2, 2): 2 + 2 = 4, but those inputs were seen before so no call to fn() was required.
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For the inputs of (1, 2): 1 + 2 = 3, and it required another call to fn() for a total of 2.
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**Example 2:**
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**Input:**
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getInputs = () =>[[{},{}],[{},{}],[{},{}]]
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fn = function (a, b) { return ({...a, ...b}); }
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**Output:**[{"val":{},"calls":1},{"val":{},"calls":2},{"val":{},"calls":3}]
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**Explanation:** Merging two empty objects will always result in an empty object. It may seem like there should only be 1 call to fn() because of cache-hits, however none of those objects are === to each other.
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**Example 3:**
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**Input:**
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getInputs = () => { const o = {}; return[[o,o],[o,o],[o,o]]; }
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fn = function (a, b) { return ({...a, ...b}); }
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**Output:**[{"val":{},"calls":1},{"val":{},"calls":1},{"val":{},"calls":1}]
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**Explanation:** Merging two empty objects will always result in an empty object. The 2nd and 3rd third function calls result in a cache-hit. This is because every object passed in is identical.
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**Constraints:**
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* <code>1 <= inputs.length <= 10<sup>5</sup></code>
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* <code>0 <= inputs.flat().length <= 10<sup>5</sup></code>
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*`inputs[i][j] != NaN`
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// #Hard #2023_08_31_Time_264_ms_(98.86%)_Space_115.9_MB_(61.71%)
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typeFn=(...params:any)=>any
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functionmemoize(fn:Fn):Fn{
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constcache=newMap();
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returnfunction(...args){
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letcurrentCache;
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if(cache.has(args.length)){
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currentCache=cache.get(args.length);
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}
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else{
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currentCache=newMap();
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cache.set(args.length,currentCache);
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}
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for(leti=0,len=args.length;i<=len;i++){
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constarg=args[i];
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constisEnd=i>=len-1;
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if(currentCache.has(arg)){
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if(isEnd){
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returncurrentCache.get(arg);
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}
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else{
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currentCache=currentCache.get(arg);
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}
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}
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elseif(isEnd){
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break;
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}else{
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constnewSubCache=newMap();
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currentCache.set(arg,newSubCache);
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currentCache=newSubCache;
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}
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}
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letvalue=fn(...args);
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currentCache.set(args[args.length-1],value);
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returnvalue;
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}
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}
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/*
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* let callCount = 0;
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* const memoizedFn = memoize(function (a, b) {
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* callCount += 1;
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* return a + b;
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* })
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* memoizedFn(2, 3) // 5
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* memoizedFn(2, 3) // 5
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* console.log(callCount) // 1
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*/
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export{memoize}
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2631\. Group By
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Medium
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Write code that enhances all arrays such that you can call the`array.groupBy(fn)` method on any array and it will return a**grouped** version of the array.
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A**grouped** array is an object where each key is the output of`fn(arr[i])` and each value is an array containing all items in the original array with that key.
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The provided callback`fn` will accept an item in the array and return a string key.
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The order of each value list should be the order the items appear in the array. Any order of keys is acceptable.
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Please solve it without lodash's`_.groupBy` function.
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**Example 1:**
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**Input:**
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array = [
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{"id":"1"},
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{"id":"1"},
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{"id":"2"}
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],
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fn = function (item) {
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return item.id;
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}
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**Output:**
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{
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"1": [{"id": "1"}, {"id": "1"}],
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"2": [{"id": "2"}]
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}
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**Explanation:**
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Output is from array.groupBy(fn).
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The selector function gets the "id" out of each item in the array.
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There are two objects with an "id" of 1. Both of those objects are put in the first array.
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There is one object with an "id" of 2. That object is put in the second array.
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**Example 2:**
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**Input:**
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array = [
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[1, 2, 3],
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[1, 3, 5],
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[1, 5, 9]
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]
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fn = function (list) {
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return String(list[0]);
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}
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**Output:**
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{
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"1": [[1, 2, 3], [1, 3, 5], [1, 5, 9]]
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}
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**Explanation:**
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The array can be of any type. In this case, the selector function defines the key as being the first element in the array.
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All the arrays have 1 as their first element so they are grouped together.
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{
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"1": [[1, 2, 3], [1, 3, 5], [1, 5, 9]]
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}
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**Example 3:**
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**Input:**
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array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
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fn = function (n) {
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return String(n > 5);
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}
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**Output:**
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{
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"true": [6, 7, 8, 9, 10],
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"false": [1, 2, 3, 4, 5]
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}
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**Explanation:** The selector function splits the array by whether each number is greater than 5.
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**Constraints:**
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* <code>0 <= array.length <= 10<sup>5</sup></code>
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*`fn returns a string`
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// #Medium #2023_08_31_Time_101_ms_(99.50%)_Space_63.8_MB_(87.11%)
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declare global{
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interfaceArray<T>{
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groupBy(fn:(item:T)=>string):Record<string,T[]>
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}
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}
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Array.prototype.groupBy=function<T>(fn:(item:T)=>string){//NOSONAR
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constreturnObject:Record<string,T[]>={};
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for(constitemofthis){
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constkey=fn(item);
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if(keyinreturnObject){
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returnObject[key].push(item);
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}else{
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returnObject[key]=[item];
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}
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}
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returnreturnObject;
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};
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/*
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* [1,2,3].groupBy(String) // {"1":[1],"2":[2],"3":[3]}
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*/
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export{}
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2634\. Filter Elements from Array
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Easy
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Given an integer array`arr` and a filtering function`fn`, return a filtered array`filteredArr`.
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The`fn` function takes one or two arguments:
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*`arr[i]` - number from the`arr`
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*`i` - index of`arr[i]`
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`filteredArr` should only contain the elements from the`arr` for which the expression`fn(arr[i], i)` evaluates to a**truthy** value. A**truthy** value is a value where`Boolean(value)` returns`true`.
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Please solve it without the built-in Array.filter method.
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**Example 1:**
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**Input:** arr =[0,10,20,30], fn = function greaterThan10(n) { return n > 10; }
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**Output:**[20,30]
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**Explanation:**
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const newArray = filter(arr, fn); //[20, 30]
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The function filters out values that are not greater than 10
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**Example 2:**
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**Input:** arr =[1,2,3], fn = function firstIndex(n, i) { return i === 0; }
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**Output:**[1]
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**Explanation:**
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fn can also accept the index of each element
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In this case, the function removes elements not at index 0
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**Example 3:**
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**Input:** arr =[-2,-1,0,1,2], fn = function plusOne(n) { return n + 1 }
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**Output:**[-2,0,1,2]
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**Explanation:** Falsey values such as 0 should be filtered out
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**Constraints:**
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*`0 <= arr.length <= 1000`
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* <code>-10<sup>9</sup> <= arr[i] <= 10<sup>9</sup></code>
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// #Easy #2023_08_31_Time_44_ms_(98.04%)_Space_42.7_MB_(69.67%)
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functionfilter(arr:number[],fn:(n:number,i:number)=>boolean):number[]{
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constfilteredArr:number[]=[]
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for(leti=0;i<arr.length;i++){
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if(fn(arr[i],i))filteredArr.push(arr[i])
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}
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returnfilteredArr
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}
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export{filter}
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2635\. Apply Transform Over Each Element in Array
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Easy
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Given an integer array`arr` and a mapping function`fn`, return a new array with a transformation applied to each element.
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The returned array should be created such that`returnedArray[i] = fn(arr[i], i)`.
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Please solve it without the built-in`Array.map` method.
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**Example 1:**
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**Input:** arr =[1,2,3], fn = function plusone(n) { return n + 1; }
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**Output:**[2,3,4]
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**Explanation:**
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const newArray = map(arr, plusone); //[2,3,4]
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The function increases each value in the array by one.
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**Example 2:**
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**Input:** arr =[1,2,3], fn = function plusI(n, i) { return n + i; }
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**Output:**[1,3,5]
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**Explanation:** The function increases each value by the index it resides in.
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**Example 3:**
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**Input:** arr =[10,20,30], fn = function constant() { return 42; }
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**Output:**[42,42,42]
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**Explanation:** The function always returns 42.
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**Constraints:**
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*`0 <= arr.length <= 1000`
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* <code>-10<sup>9</sup> <= arr[i] <= 10<sup>9</sup></code>
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*`fn returns a number`
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// #Easy #2023_08_31_Time_43_ms_(98.46%)_Space_42.2_MB_(92.83%)
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functionmap(arr:number[],fn:(n:number,i:number)=>number):number[]{
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constres:number[]=[]
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for(leti=0;i<arr.length;i++){
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res.push(fn(arr[i],i))
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}
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returnres
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}
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export{map}

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