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This function needs to find the first occurrence of consecutive numbers whose sum equals 10, and return them as an array. My idea was to keep summing the numbers until the accumulator, in this case x becomes greater than 10, at which point the value of x is reset to zero, and i is reset back to index 1, by being assigned to the ct variable that is supposed to grow by one on each reset, so the loop would go back and start at the next element, until it gets the correct value. But for some reason ct is stuck at 1 and doesn't grow and the loop doesn't move past the second array element at index 1. Is there an error in the logic or the implementation?
Warning: This code may cause the page to crash or become unresponsive
const arr = [2, 3, 1, 2, 7, 2, 5, 4, 5, 1, 2, 1];
function test(arr) {
let x = 0;
for (let i = 0; i < arr.length; i++) {
let ct = 0;
x = x + arr[i]
if (x > 10) {
x = 0;
i = ct;
ct++
}
console.log(x)
}
}
console.log(test(arr))
The problem is that you've defined ct in the wrong place. It's inside the loop, meaning that every time you get to a point where i = ct, i is reset to 0 and ct++ has no effect (because every iteration, ct is reset to 0). You need to define ct in the same scope as x:
const arr = [2, 3, 1, 2, 7, 2, 5, 4, 5, 1, 2, 1];
function test(arr) {
let x = 0;
let ct = 0;
for (let i = 0; i < arr.length; i++) {
x = x + arr[i];
if (x > 10) {
x = 0;
i = ct;
ct++;
}
console.log(x);
}
}
test(arr);
I've also filled in the rest of the logic of the function for you in case you'd like it:
const arr = [2, 3, 1, 2, 7, 2, 5, 4, 5, 1, 2, 1];
function test(arr) {
let x = 0;
let ct = 0;
let result = [];
for (let i = 0; i < arr.length; i++) {
x = x + arr[i];
if (x > 10) {
x = 0;
i = ct;
ct++;
}
if (x == 10) {
result = arr.slice(ct, i + 1);
return result;
}
console.log(x);
}
return result;
}
console.log(test(arr));
I did this module on functions and execution context - all questions have gone well but there is one challenge I have spent a lot of time on and still can't figure it out. Any help will be greatly appreciated. Thank you
Challenge question says:
Write a function addingAllTheWeirdStuff which adds the sum of all the odd numbers in array2 to each element under 10 in array1.
Similarly, addingAllTheWeirdStuff should also add the sum of all the even numbers in array2 to those elements over 10 in array1.
BONUS: If any element in array2 is greater than 20, add 1 to every element in array1.
// Uncomment these to check your work!
// console.log(addingAllTheWeirdStuff([1, 3, 5, 17, 15], [1, 2, 3, 4, 5])); // expected log [10, 12, 14, 23, 21]
// console.log(addingAllTheWeirdStuff([1, 3, 5, 17, 15, 1], [1, 2, 3, 4, 5, 22])); // expected log [11, 13, 15, 46, 44, 11]
// my attempt so far:
function addingAllTheWeirdStuff(array1, array2) {
// ADD CODE HERE
let result = []
for (let i = 0; i < array2.length; i++) {
if (array2[i] > 20) {
result = array1[i] += 1
}
}
for (let i = 0; i < array2.length; i++) {
if (array2[i] % 2 === 0 && array1[i] > 10) {
result = array1[i] + array2[i]
}
}
for (let i = 0; i < array2.length; i++) {
if (array2[i] % 2 !== 0 && array1[i] < 10) {
result = array1[i] + array2[i]
}
}
return result
}
You can easily achieve this using reduce and map array method, with the ternary operator:
const array1 = [1, 3, 5, 17, 15];
const array2 = [1, 2, 3, 4, 5];
function addingAllTheWeirdStuff(array1, array2) {
const oddSum = array2.reduce((sum, current) => current % 2 ? current + sum : 0 + sum, 0)
const oddEven = array2.reduce((sum, current) => current % 2 == 0 ? current + sum : 0 + sum, 0)
return array1.map(num => num < 10 ? num + oddSum : num + oddEven)
}
console.log(addingAllTheWeirdStuff(array1, array2))
If you break the challenge into smaller pieces, you can deconstruct it better and come up with your solutions.
This is what I did... I will be adding more comments shortly with more explanations
I chose to keep using loops as I assumed this was the point of the challenges (to practice for loops, multiple conditions, etc) - In other words, I chose to not use map / reduce on purpose but if that's allowed, use the answer by #charmful0x as it results in less code :)
// function to get sum of all odd numbers in array
function getSumOfAllOddNumbersInArray( elementArray ){
var sumOfOddNumbers = 0;
for (let i = 0; i < elementArray.length; i++) {
// use remainder operator to find out if element is odd or not
if (elementArray[i] % 2 !== 0 ) {
sumOfOddNumbers += elementArray[i];
}
}
return sumOfOddNumbers;
}
// function to get sum of all EVEN numbers in array
function getSumOfAllEvenNumbersInArray( elementArray ){
var sumOfEvenNumbers = 0;
for (let i = 0; i < elementArray.length; i++) {
// use remainder operator to find out if element is odd or not
if (elementArray[i] % 2 === 0 ) {
sumOfEvenNumbers += elementArray[i];
}
}
return sumOfEvenNumbers;
}
// Return true if there is at least one element in array that is greater than 20
function hasElementOverTwenty( elementArray ){
for (let i = 0; i < elementArray.length; i++) {
if (elementArray[i] > 20 ) {
// no need to keep looping, we found one - exit function
return true;
}
}
return false;
}
function addingAllTheWeirdStuff( firstArray, secondArray ){
var sumOfOddNumbersInArray = getSumOfAllOddNumbersInArray( secondArray );
var sumOfEvenNumbersInArray = getSumOfAllEvenNumbersInArray( secondArray );
var needToAddOne = hasElementOverTwenty( secondArray );
for (let i = 0; i < firstArray.length; i++) {
// Challenge One
if (firstArray[i] < 10) {
firstArray[i] = firstArray[i] + sumOfOddNumbersInArray;
} else if (firstArray[i] > 10) {
// Challenge Two
firstArray[i] = firstArray[i] + sumOfEvenNumbersInArray;
}
// bonus
if( needToAddOne ){
firstArray[i]++;
}
}
return firstArray;
}
// Uncomment these to check your work!
console.log(addingAllTheWeirdStuff([1, 3, 5, 17, 15], [1, 2, 3, 4, 5]));
console.log('expected:' + [10, 12, 14, 23, 21] );
console.log(addingAllTheWeirdStuff([1, 3, 5, 17, 15, 1], [1, 2, 3, 4, 5, 22]));
console.log('expected:' + [11, 13, 15, 46, 44, 11] );
Challenge question says: Write a function addingAllTheWeirdStuff which adds the sum of all the odd numbers in array2 to each element under 10 in array1.
Similarly, addingAllTheWeirdStuff should also add the sum of all the even numbers in array2 to those elements over 10 in array1.
BONUS: If any element in array2 is greater than 20, add 1 to every element in array1.
Added maximum number according to the input length should be returned.
For example, if the length is 2 then the max among arr[0] + arr[1], arr[1] + arr[2], arr[2] + arr[3] should be returned.
Input is array and length.
I solved this in a real job interview but I think there will be a way not to use nested loop.
const add = (arr, len) => {
let rtnVal = 0
for (let i = len - 1; i < arr.length; i++) {
let temp_idx = i;
let temp_sum = 0;
for (let j = 0; j < len; j++) {
temp_sum = (temp_sum || 0) + arr[temp_idx]
temp_idx -= 1
}
if (temp_sum > rtnVal) {
rtnVal = temp_sum
}
}
return rtnVal
}
console.log(add([1, 2, 3, 4, 5, 6, 7, 8, 9], 4))
I expect the output 30
// enhanced nested loop
const add = (arr, len) => {
let rtnVal = 0;
for(let i=len-1;i<arr.length;i++){
let sum = 0
for(let j=i;j>=i-len+1;j--){
sum += arr[j]
}
if (sum > rtnVal) rtnVal = sum
}
return rtnVal
}
console.log(add([9, 9, 9, 4, 5, 6, 7, 8, 9], 3))
Use a moving window. Add up len numbers starting at the beginning. Then continue through the array adding the next number and subtracting the trailing number.
const add = (arr, len) => {
return arr.reduce((a,v,i,arr) => {
a.running += v;
if(i >= len) {
a.running -= arr[i-len];
}
if(i+1 >= len && a.largest < a.running) {
a.largest = a.running;
}
return a;
}, {
largest: Number.NEGATIVE_INFINITY,
running: 0
}).largest;
}
console.log(add([1,2,3,4,5,6,7,8,9],4)); // 30
console.log(add([-1,-1,-1,-1],2)); // -2
console.log(add([1],1)); // 1
Assuming its sorted like your example. You can use negative slice to select from end and then reduce the array.
const add = (arr, len) => arr.slice(len > arr.len ? arr.len : -len).reduce((total, num) => total + num)
console.log(add([1, 2, 3, 4, 5, 6, 7, 8, 9], 4))
i'm starting studying javascript. I'm stuck with this exercise: Create a function that takes an array as argument and return the sum of all elements of the array.
I have written this code:
function sum(table) {
let x = table[0];
let y = [table.length - 1];
let totale = 0;
for (count = x; count <= y; count++) {
total += x;
x += 1;
};
return total;
};
console.log(sum[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]);
I don't know why the result is undefined instead of 55. Thanks for helping.
total not totale
the function call needs parentheses as sum([…])
y = table.length - 1 you want the length, not an array that holds the length
x = 0 you don't need to set it to the first element of the array
total += table[count]; you don't need to set it to x or to deal with x at all
function sum(table) {
let x = 0;
let y = table.length - 1;
let total = 0;
for (count = x; count <= y; count++) {
total += table[count];
};
return total;
};
console.log(sum([1, 2, 3, 4, 5, 6, 7, 8, 9, 10]));
Quicker Solution
function sum(table) {
return table.reduce((p,c)=>p+c,0);
};
console.log(sum([1, 2, 3, 4, 5, 6, 7, 8, 9, 10]));
Here are couple of things.
You need to execute the sum function by calling like sum([array elements]) .Note the ( & ) braces
Secondly count <= y will give undefined as it will exceed the length of the array. The array index starts from 0;
There is a typo here totale.
You can avoid these set of lines
let x = table[0];
let y = [table.length - 1];
if you just initialize the loop conditional statement like this
for (let count = 0; count < table.length; count++)
function sum(table) {
let x = 0
for (let count = 0; count < table.length; count++) {
x += table[count];
};
return x;
};
console.log(sum([1, 2, 3, 4, 5, 6, 7, 8, 9, 10]));
Another option is to use the reduce method
function sum(table) {
return table.reduce(function(acc, curr) {
return acc += curr;
}, 0) // 0 is the initial value
};
console.log(sum([1, 2, 3, 4, 5, 6, 7, 8, 9, 10]));
my question is actually similar to: Extracting the most duplicate value from an array in JavaScript (with jQuery)
Solution (the one I found as the best, and slightly changed by me):
var arr = [3, 7, 7, 7, 7, 10, 10, 8, 5, 5, 5, 5, 20, 20, 1],
result = {},
max = 0,
res;
for( var i = 0, total = arr.length; i < total; ++i ) {
var val = arr[i],
inc = ( result[val] || 0 ) + 1;
result[val] = inc;
if( inc > max ) {
max = inc;
res = val;
}
}
alert(res);
I would like to add an extra which is : if we have, say two numbers with the same number of occurrences, how could we find the minimum of them (the above should alert 5 and not 7, which is the case)? The current solution works for getting only the first most duplicate found, but it does not deal with repetition.
Thanks!
Sort the array before you count the incidences:
var arr = [3, 7, 7, 7, 7, 10, 10, 8, 5, 5, 5, 5, 20, 20, 1];
function min_most_duplicate (arr) {
var result = {},
max = 0,
res;
arr = arr.slice(0); // make a copy of the array
arr.sort(function(a, b) { return (a - b); }); // sort it numerically
for( var i = 0, total = arr.length; i < total; ++i ) {
var val = arr[i],
inc = ( result[val] || 0 ) + 1;
result[val] = inc;
if( inc > max ) {
max = inc;
res = val;
}
}
return res;
}
min_most_duplicate(arr); // returns 5
This works because the way the for loop is written it's going to return the first one that it finds that has the most duplicates, so if the array is sorted, the smallest number will come first, so it'll be the one that the for loop finds.
This would work, example:
var arr = [3, 7, 7, 7, 7, 10, 10, 8, 5, 5, 5, 5, 20, 20, 1],
result = {},
max = 0,
res, key, min;
for (var i = 0, total = arr.length; i < total; ++i) {
var val = arr[i],
inc = (result[val] || 0) + 1;
result[val] = inc;
if (inc > max) {
max = inc;
res = val;
}
}
for (var i in result) {
if (result[i] === max) {
key = parseInt(i, 10);
min = min || key;
console.log(min)
if (min > key) {
min = key;
}
}
}
res = min;
alert(res);
let getMostDuplicated = array => {
let duplicated = '';
let max = 0;
for (let i = 0; i < array.length; i++) {
let counter = 0;
for (let j = 1; j < array.length - 1; j++) {
if (array[i] === array[j])
counter++;
}
if (counter > max) {
duplicated = array[i];
max = counter;
}
}
return duplicated;
};
let array = [3, 7, 7, 7, 7, 10, 10, 8, 5, 5, 5, 5, 20, 20, 1];
getMostDuplicated(array);