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how to get sum of odd, even using reduce method, i have done as show in below code but returning undefined , #js-beginner
//code below
nums= [1,2,3,4,5,6,7,8,9]
function getOddEvenSum(numbers){
let{even,odd} = numbers.reduce((acc, cuu) => cuu%2 === 0?acc.even + cuu:acc.odd+cuu,{even:0, odd:0})
return {even, odd}
}
console.log(getOddEvenSum(nums)
//output i am getting below
{even:undefined, odd:undefined}
The value that you return from your reduce callback will be the value of acc upon the next invocation/iteration of your array of numbers. Currently, your acc starts off as an object, but as you're only returning a number from your first iteration, all subsequent iterations will use a number as acc, which don't have .even or .odd properties. You could instead return a new object with updated even/odd properties so that acc remains as an object through all iterations:
const nums = [1,2,3,4,5,6,7,8,9];
function getOddEven(numbers){
return numbers.reduce((acc, cuu) => cuu % 2 === 0
? {odd: acc.odd, even: acc.even + cuu}
: {even: acc.even, odd: acc.odd+cuu},
{even:0, odd:0});
}
console.log(getOddEven(nums));
You can use Array.prototype.reduce like this:
const nums = [1, 2, 3, 4, 5, 6, 7, 8, 9];
const [odds, evens] = nums.reduce(
([odds, evens], cur) =>
cur % 2 === 0 ? [odds, evens + cur] : [odds + cur, evens],
[0, 0]
);
console.log(odds);
console.log(evens);
This is not how the syntax of reduce works. One possible implementation:
function getOddEven(nums) {
return nums.reduce(
({odd, even}, num) => num % 2 === 0 ?
{odd, even: even + num} :
{odd: odd + num, even},
{odd: 0, even: 0},
);
}
I would argue that this is not very clear. Since performance is probably not critical, a clearer alternative would be:
function getOddEven(nums) {
return {
odd: nums.filter(num => num % 2 == 1).reduce((acc, num) => acc + num),
even: nums.filter(num => num % 2 == 0).reduce((acc, num) => acc + num),
};
}
based on your code, you need to return acc
let nums = [1, 2, 3, 4, 5, 6, 7, 8, 9]
function getOddEven(numbers) {
let {
even,
odd
} = numbers.reduce((acc, cuu) => {
if(cuu%2 == 0)
{
acc.even += cuu
}
else{
acc.odd += cuu
}
return acc
},
{
even: 0,
odd: 0
})
return {
even,
odd
}
}
console.log(getOddEven(nums))
I am trying to get better at understanding recursion so that I can get better at implementing dynamic programming principles. I am aware this problem can be solved using Kadane's algorithm; however, I would like to solve it using recursion.
Problem statement:
Given an array of integers, find the subset of non-adjacent elements with the maximum sum. Calculate the sum of that subset.
I have written the following partial solution:
const maxSubsetSum = (arr) => {
let max = -Infinity
const helper = (arr, len) => {
if (len < 0) return max
let pointer = len
let sum = 0
while (pointer >= 0) {
sum += arr[pointer]
pointer -= 2
}
return max = Math.max(sum, helper(arr, len - 1))
}
return helper(arr, arr.length - 1)
}
If I had this data:
console.log(maxSubsetSum([3, 5, -7, 8, 10])) //15
//Our subsets are [3,-7,10], [3,8], [3,10], [5,8], [5,10] and [-7,10].
My algorithm calculates 13. I know it's because when I start my algorithm my (n - 2) values are calculated, but I am not accounting for other subsets that are (n-3) or more that still validate the problem statement's condition. I can't figure out the logic to account for the other values, please guide me on how I can accomplish that.
The code is combining recursion (the call to helper inside helper) with iteration (the while loop inside helper). You should only be using recursion.
For each element of the array, there are two choices:
Skip the current element. In this case, the sum is not changed, and the next element can be used. So the recursive call is
sum1 = helper(arr, len - 1, sum)
Use the current element. In this case, the current element is added to the sum, and the next element must be skipped. So the recursive call is
sum2 = helper(arr, len - 2, sum + arr[len])
So the code looks like something this:
const maxSubsetSum = (arr) => {
const helper = (arr, len, sum) => {
if (len < 0) return sum
let sum1 = helper(arr, len - 1, sum)
let sum2 = helper(arr, len - 2, sum + arr[len])
return Math.max(sum1, sum2)
}
return helper(arr, arr.length - 1, 0)
}
Your thinking is right in that you need to recurse from (n-2) once you start with a current index. But you don't seem to understand that you don't need to run through your array to get sum and then recurse.
So the right way is to
either include the current item and recurse on the remaining n-2 items or
not include the current item and recurse on the remaining n-1 items
Lets look at those two choices:
Choice 1:
You chose to include the item at the current index. Then you recurse on the remaining n-2 items. So your maximum could be item itself without adding to any of remaining n-2 items or add to some items from n-2 items.
So Math.max( arr[idx], arr[idx] + recurse(idx-2)) is the maximum for this choice. If recurse(idx-2) gives you -Infinity, you just consider the item at the current index.
Choice 2:
You didn't choose to include the item at the current index. So just recurse on the remaining n-1 items - recurse(n-1)
The final maximum is maximum from those two choices.
Code is :
const maxSubsetSum = (arr) => {
let min = -Infinity
const helper = (arr, idx) => {
if ( idx < 0 ) return min
let inc = helper(arr, idx-2)
let notInc = helper(arr, idx-1)
inc = inc == min ? arr[idx] : Math.max(arr[idx], arr[idx] + inc)
return Math.max( inc, notInc )
}
return helper(arr, arr.length - 1)
}
console.log(maxSubsetSum([-3, -5, -7, -8, 10]))
console.log(maxSubsetSum([-3, -5, -7, -8, -10]))
console.log(maxSubsetSum([-3, 5, 7, -8, 10]))
console.log(maxSubsetSum([3, 5, 7, 8, 10]))
Output :
10
-3
17
20
For the case where all items are negative:
In this case you can say that there are no items to combine together to get a maximum sum. If that is the requirement the result should be zero. In that case just return 0 by having 0 as the default result. Code in that case is :
const maxSubsetSum = (arr) => {
const helper = (arr, idx) => {
if ( idx < 0 ) return 0
let inc = arr[idx] + helper(arr, idx-2)
let notInc = helper(arr, idx-1)
return Math.max( inc, notInc )
}
return helper(arr, arr.length - 1)
}
With memoization:
You could memoize this solution for the indices you visited during recursion. There is only one state i.e. the index so your memo is one dimensional. Code with memo is :
const maxSubsetSum = (arr) => {
let min = -Infinity
let memo = new Array(arr.length).fill(min)
const helper = (arr, idx) => {
if ( idx < 0 ) return min
if ( memo[idx] !== min) return memo[idx]
let inc = helper(arr, idx-2)
let notInc = helper(arr, idx-1)
inc = inc == min ? arr[idx] : Math.max(arr[idx], arr[idx] + inc)
memo[idx] = Math.max( inc, notInc )
return memo[idx]
}
return helper(arr, arr.length - 1)
}
A basic version is simple enough with the obvious recursion. We either include the current value in our sum or we don't. If we do, we need to skip the next value, and then recur on the remaining values. If we don't then we need to recur on all the values after the current one. We choose the larger of these two results. That translates almost directly to code:
const maxSubsetSum = ([n, ...ns]) =>
n == undefined // empty array
? 0
: Math .max (n + maxSubsetSum (ns .slice (1)), maxSubsetSum (ns))
Update
That was missing a case, where our highest sum is just the number itself. That's fixed here (and in the snippets below)
const maxSubsetSum = ([n, ...ns]) =>
n == undefined // empty array
? 0
: Math .max (n, n + maxSubsetSum (ns .slice (1)), maxSubsetSum (ns))
console.log (maxSubsetSum ([3, 5, -7, 8, 10])) //15
But, as you note in your comments, we really might want to memoize this for performance reasons. There are several ways we could choose to do this. One option would be to turn the array we're testing in one invocation of our function into something we can use as a key in an Object or a Map. It might look like this:
const maxSubsetSum = (ns) => {
const memo = {}
const mss = ([n, ...ns]) => {
const key = `${n},${ns.join(',')}`
return n == undefined
? 0
: key in memo
? memo [key]
: memo [key] = Math .max (n, n + maxSubsetSum (ns .slice (1)), maxSubsetSum (ns))
}
return mss(ns)
}
console.log (maxSubsetSum ([3, 5, -7, 8, 10])) //15
We could also do this with a helper function that acted on the index and memoized using the index for a key. It would be about the same level of complexity.
This is a bit ugly, however, and perhaps we can do better.
There's one issue with this sort of memoization: it only lasts for the current run. It I'm going to memoize a function, I would rather it holds that cache for any calls for the same data. That means memoization in the definition of the function. I usually do this with a reusable external memoize helper, something like this:
const memoize = (keyGen) => (fn) => {
const cache = {}
return (...args) => {
const key = keyGen (...args)
return cache[key] || (cache[key] = fn (...args))
}
}
const maxSubsetSum = memoize (ns => ns .join (',')) (([n, ...ns]) =>
n == undefined
? 0
: Math .max (n, n + maxSubsetSum (ns .slice (1)), maxSubsetSum (ns)))
console.log (maxSubsetSum ([3, 5, -7, 8, 10])) //15
memoize takes a function that uses your arguments to generate a String key, and returns a function that accepts your function and returns a memoized version of it. It runs by calling the key generation on your input, checks whether that key is in the cache. If it is, we simply return it. If not, we call your function, store the result under that key and return it.
For this version, the key generated is simply the string created by joining the array values with ','. There are probably other equally-good options.
Note that we cannot do
const recursiveFunction = (...args) => /* some recursive body */
const memomizedFunction = memoize (someKeyGen) (recursiveFunction)
because the recursive calls in memoizedFunction would then be to the non-memoized recursiveFunction. Instead, we always have to use it like this:
const memomizedFunction = memoize (someKeyGen) ((...args) => /* some recursive body */)
But that's a small price to pay for the convenience of being able to simply wrap up the function definition with a key-generator to memoize a function.
This code was accepted:
function maxSubsetSum(A) {
return A.reduce((_, x, i) =>
A[i] = Math.max(A[i], A[i-1] | 0, A[i] + (A[i-2] | 0)));
}
But trying to recurse that far, (I tried submitting Scott Sauyet's last memoised example), I believe results in run-time errors since we potentially pass the recursion limit.
For fun, here's bottom-up that gets filled top-down :)
function f(A, i=0){
if (i > A.length - 3)
return A[i] = Math.max(A[i] | 0, A[i+1] | 0);
// Fill the table
f(A, i + 1);
return A[i] = Math.max(A[i], A[i] + A[i+2], A[i+1]);
}
var As = [
[3, 7, 4, 6, 5], // 13
[2, 1, 5, 8, 4], // 11
[3, 5, -7, 8, 10] // 15
];
for (let A of As){
console.log('' + A);
console.log(f(A));
}
I'm trying to solve this task of finding the unique element inside an array.
So far I managed to solve 95%, but I'm failing on 0. I get an error saying that expected 0 and got 1.
I should get //10, which it does, but after I'm failing the test online. For all other values it has passed.
Any ideas about how to solve this and what I'm missing here?
function findOne(arr) {
let x = arr[0];
for (let i of arr) {
if (i === x) {
continue;
} else {
x = i;
}
return x;
}
}
console.log(findOne([3, 10, 3, 3, 3]));
I don't really understand your code. You start with the first value in the array, then you loop through the array, skipping anything that's the same, and then return the first one that's not the same. That won't find unique values, it'll just find the first value that doesn't equal the first value. So for instance, try it on the array [1,2,2,2,2] and you'll get a result of 2 instead of 1, even though that's clearly wrong.
Instead, you can create a map of each value and its incidence, then filter by the ones that equal 1 at the end.
function findOne(arr) {
const incidences = arr.reduce((map, val) => {
map[val] = (map[val] || 0) + 1;
return map;
}, {});
const values = Object.keys(incidences);
for (let i = 0; i < values.length; ++i) {
if (incidences[values[i]] === 1) { return values[i]; }
}
return null;
}
EDIT The above won't preserve the type of the value (i.e. it'll convert it to a string always, even if it was originally a number). To preserve the type, you can use an actual Map instead of an object:
function findOne(arr) {
const incidences = arr.reduce((map, val) => {
map.set(val, (map.get(val) || 0) + 1);
return map;
}, new Map());
const singletons = Array.from(incidences).filter(entry => entry[1] === 1);
return singletons.map(singleton => singleton[0]);
}
Consider the following:
Recall that a span = max - min + 1;
Let Partition P1 be span from 0..span-1;
Let Partition P2 be span from span..(2*span)-1:
Place a number in P1 if it is not in P2.
Place a number in P2 if it is already in P1.
Once the number is in P2, do not consider it again.
If a number is in P1 then it is unique.
You can get all values that appear once, by using a map to count how many times each element has appeared. You can then reduce that map into an array of unique values:
const findUnique = arr => {
const mapEntries = [...arr.reduce((a, v) => a.set(v, (a.get(v) || 0) + 1), new Map()).entries()]
return mapEntries.reduce((a, v) => (v[1] === 1 && a.push(v[0]), a), [])
}
console.log(findUnique([3, 10, 3, 3, 3]))
console.log(findUnique([1, 2, 3, 2, 4]))
console.log(findUnique([4, 10, 4, 5, 3]))
If you don't care about multiple unique values, you can just sort the array and use logic, rather than checking every value, provided the array only contains 2 different values, and has a length greater than 2:
const findUnique = arr => {
a = arr.sort((a, b) => a - b)
if (arr.length < 3 || new Set(a).size === 1) return null
return a[0] === a[1] ? a[a.length-1] : a[0]
}
console.log(findUnique([3, 10, 3, 3, 3]))
console.log(findUnique([3, 3, 1]))
console.log(findUnique([3, 1]))
console.log(findUnique([3, 3, 3, 3, 3]))
Your code is complex, Try this
function findOne(arr) {
const uniqueItems = [];
arr.forEach(item => {
const sameItems = arr.filter(x => x === item);
if (sameItems.length === 1) {
uniqueItems.push(item);
}
});
return uniqueItems;
}
console.log(findOne([0, 1, 1, 3, 3, 3, 4]));
I'm getting all unique items from passed array, It may have multiple unique item
this is a way simpler and fast:
function findOne(arr) {
const a = arr.reduce((acc, e) => {
e in acc || (acc[e] = 0)
acc[e]++
return acc
}, {})
return Object.keys(a).filter(k => a[k] === 1)[0] || null
}
So I need to solve this problem STRICTLY using recursion
// 2. Compute the sum of an array of integers.
// sum([1,2,3,4,5,6]); // 21
And then I'm testing this solution in PythonLive
var sum = function(array) {
if(array.length===0){
return array
}
return array.slice(0,array.length)+sum(array.pop())
};
sum([1,2,3,4,5,6]);
Then at step 6 it says "TypeError: array.slice is not a function"
I don't understand why if it already worked taking 6 off the array and returning the remaining array...
Can someone explain me what am I doing wrong please?
thanks! :)
If you look at the return values you will see that you are always returning an array. This can't be right when you want a number as a final result. When array.length === 0 you can safely return 0 because that's the same of an empty array. That's your edge condition. After that you just want the sum of one element plus the rest.
You can also just return the array length when it's zero making for a very succinct solution. && shortcircuits returning the left element if it's false (like 0) otherwise the second:
var sum = (array) => array.length && array.pop() + sum(array)
console.log(sum([1,2,3,4,5,6]));
If you prefer slice you could also this, which is basically the same:
var sum = (array) => array.length && array[0] + sum(array.slice(1))
console.log(sum([1, 2, 3, 4, 5, 6]));
Recursive sum function:
const sum = list => {
if (!list.length) return 0;
return list.pop() + sum(list);
};
Because .pop mutates the array, you don't need to use slice. For a non-destructive version (doesn't alter the original array):
const sum = ([first, ...rest]) => {
if (first === undefined) return 0;
return first + sum(rest);
};
The issue with your code is that you are processing the values the wrong way around, it should be
return sum(array.slice(0,array.length-1)) + array.pop();
In fact since array.pop() removes the element, you can just do it this way around:
return array.pop() + sum(array);
You also need to return 0 when array.length===0, otherwise the sum will fail.
if (array.length===0) return 0;
However it's much simpler just do this with reduce:
let arr = [1,2,3,4,5,6];
console.log(arr.reduce((t, v) => { return t + v; }, 0));
Another encoding using an explicit empty and pure expressions
const empty = x =>
x === empty
const sum = ([ x = empty, ...rest ]) =>
empty (x)
? 0
: x + sum (rest)
console.log
( sum ([ 1, 2, 3, 4, 5 ]) // 15
, sum ([]) // 0
)
Your other question that asks how to sum a nested array was put on-hold for reasons I don't understand. You can adapt the above implementation to support input of nested arrays
const empty = x =>
x === empty
const sum = ([ x = empty, ...rest ]) =>
empty (x)
? 0
: Array.isArray (x)
? sum (x) + sum (rest)
: x + sum (rest)
console.log
( sum ([ 1, [ 2, [ 3, 4 ], 5 ]]) // 15
, sum ([ 1, 2, 3, 4, 5 ]) // 15
, sum ([[[]]]) // 0
, sum ([]) // 0
)
Looking for a way to solve this problem by recursing sum(). Right now, the code works, but I am supposed to call sum() more than once, and it should not mutate the input array.
var sum = function(array) {
if(array.length === 0){
return 0;
}
function add(array, i){
console.log(array[i]);
if(i === array.length-1){
return array[i];
}
return array[i] + add(array, i+1);
}
return add(array, 0);
};
sum([1, 2, 3, 4, 5, 6]) //21
A one-liner that meets all your requirements:
var sum = function(array) {
return (array.length === 0) ? 0 : array[0] + sum(array.slice(1));
}
// or in ES6
var sum = (array) => (array.length === 0) ? 0 : array[0] + sum(array.slice(1));
// Test cases
sum([1,2,3]); // 6
var s = [1,2,3];
sum(s); // 6
sum(s); // 6
Reasoning
In a recursive call, you need to model your task as reduction to a base case. The simplest base case in this case is the empty array - at that point, your function should return zero.
What should the reduction step be? Well you can model a sum of an array as the result of adding the first element to the sum of the remainder of the array - at some point, these successive calls will eventually result in a call to sum([]), the answer to which you already know. That is exactly what the code above does.
array.slice(1) creates a shallow copy of the array starting from the first element onwards, and no mutation ever occurs on the original array. For conciseness, I have used a ternary expression.
Breakdown:
sum([1,2,3])
-> 1 + sum([2,3])
-> 1 + 2 + sum([3])
-> 1 + 2 + 3 + sum([])
-> 1 + 2 + 3 + 0
-> 6
You're on the right track, but consider that sum could take an optional second argument (that defaults to zero) that indicates the position to start summing from...
function sum(array, n) {
n ||= 0;
if (n === array.length) {
return 0;
} else {
return array[n] + sum(array, n + 1);
}
}
function sumNumbersRecursively(input){
if (input.length == 0){
return 0;
} else{
return input.shift() + sumNumbersRecursively(input);
}
}
console.log(sumNumbersRecursively([2,3,4]))
You don't really need the add function inside your sum function just inline the function and initiate with, 0 as a starting point, or optionally check the i variable for undefined and initialize it to 0!
var sum = function(array, i) {
if(array.length === 0){
return 0;
}
console.log(array[i]);
if(i === array.length-1){
return array[i];
}
return array[i] + sum(array, i+1);
};
console.log(sum([1, 2, 3, 4, 5, 6],0)) //21
You have two solutions:
you can use .reduce() method
or perform a simple tail recursion
With reduction:
function sum(a, b) {
return a + b;
}
const array = [1, 2, 3, 4, 5, 6];
//In our reduce, we will apply our sum function,
//and pass the result as the next value
const res = array.reduce(sum);
With recursion:
function sumRec(array, acc = 0, index) {
//We will update our accumulator, and increment
// the value of our current index
return index === array.length
? acc
: sumRec(array, acc += array[index], ++index);
}
console.log(sumRec(array, 0, 0));
Personally, I find the first solution more elegant.
arr = [1,2,3,4]
sum = arr.reduce((acc, curr)=> acc+curr)
function sumArr(arr){
if(arr.length>1){
return arr.pop()+sumArr(arr);
}else{
return arr[0];
}
}
If you have to call sum more than once, then use the binary approach: split the array in half and recur on each piece. When you get to a length of 1, return the single value.
Does this work for you? I'm afraid I don't recall the JS syntax for array slices, so my recursion statement may be wrong in the details.
var sum = function(array) {
if(array.length === 1){
return array[0];
}
mid = array.length / 2
return sum(array[0:mid-1]) + sum(array[mid:array.length-1])
};
sum([1, 2, 3, 4, 5, 6]) //21