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Search An Array Consisting of Sub-Arrays For the Largest Number and Return in a New Array
(9 answers)
Closed 6 years ago.
What would be the most efficient way to take n smallest numbers from Array in Javascript
[[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]
Result is
[1,13,32,1]
Try Math.min(). It gets numbers and returns the minimum from them. I used the map() function to iterate over each nested array, then have used Math.min() with the ..., which will destroy the nested array and pass it to the function like Math.min(4,5,1,3) for the first nested one
ES6
var arr = [[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]];
var minArr = arr.map(item => Math.min(...item));
console.log(minArr);
ES5
var arr = [[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]];
var minArr = arr.map(function(array){
return Math.min.apply(null,array)
});
console.log(minArr);
Related
I am working through freecodecamp's basic algorithm challenges. The challenge is to return the largest number in an array. Here's the code.
function largestOfFour(arr) {
// You can do this!
let largestWord = [0,0,0,0];
for(let i = 0; i < arr.length; i++) {
for(let j = 0; j < arr[i].length; j++) {
if(arr[i][j] > largestWord[i]) {
largestWord[i] = arr[i][j];
}
}
}
return largestWord;
}
largestOfFour([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]);
I have passed three of the criteria:
largestOfFour([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]) should return an array.
Passed
largestOfFour([[13, 27, 18, 26], [4, 5, 1, 3], [32, 35, 37, 39], [1000, 1001, 857, 1]]) should return [27, 5, 39, 1001].
Passed
largestOfFour([[4, 9, 1, 3], [13, 35, 18, 26], [32, 35, 97, 39], [1000000, 1001, 857, 1]]) should return [9, 35, 97, 1000000].
All except largestOfFour([[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]) should return [25, 48, 21, -3].
Where did I go wrong?
So in your code you went wrong in this block.
if(arr[i][j] > largestWord[i]) {
largestWord[i] = arr[i][j];
}
The thing you forgot was you have a starting value for the largest number which is 0 and you compare array elements with that value. Only a value exceeds 0 going to replace that. So I would suggest just to initialize the array without values and do a check for that
//initialize the array without a value
let largestWord = new Array(4);
for(let i = 0; i < arr.length; i++) {
for(let j = 0; j < arr[i].length; j++) {
//also add a check for initial value
if(arr[i][j] > largestWord[i] || largestWord[i] == undefined) {
largestWord[i] = arr[i][j];
}
}
}
Seems you can do this a lot easier with .map() and Math.max()
function largestOfFour(arr) {
return arr.map(x => Math.max(...x));
}
console.log(largestOfFour([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]));
console.log(largestOfFour([[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]));
I want to get the biggest number on each array looping [0,1,2,3] with for.
I am not managing to figure what to do in order to have the desired output.
function largestOfFour(arr) {
for (let i = 0; i < arr.length; i++){
console.log(Math.max(...arr[i])) \\returns => 5, 27, 39, 1001
}
}
console.log(largestOfFour([[4, 5, 1], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]));
console.log(Math.max(...arr[i])) returns 5, 27, 39, 1001. When I remove console.log and use a return at this line, in many different approaches, the iteration is not done. The loop stops and returns just the biggest number at arr[0] instead of looping completely.
The desired output is to get the biggest number of each array = 5, 27, 39, 1001
you can't return more than once from a function
Your code, if it returned in the body of the for loop, would return from the function in the first iteration
use Array map function
function largestOfFour(arr) {
return arr.map(a => Math.max(...a))
}
console.log(largestOfFour([[4, 5, 1], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]));
function largestOfFour(arr) {
let result = [];
for (let i = 0; i < arr.length; i++){
result.push(Math.max(...arr[i]));
}
return result;
}
console.log(largestOfFour([[4, 5, 1], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]));
So I've been fighting with this a few hours now- the goal is to create a new array of the highest numbers in each array of 4. However, I can't seem to get it to loop more than once. How am I screwing up this for loop?
function largestOfFour(arr) {
for (var i = 0; i < arr.length; i++) {
var allTop = "";
var top = arr[i].sort(function(a, b) {
return b - a;
});
i++;
allTop.push(top[0]);
}
}
largestOfFour([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]);
The variable allTop should be defined before the loop as an array, and returned after the loop ends:
function largestOfFour(arr) {
var allTop = [];
for (var i = 0; i < arr.length; i++) {
var top = arr[i].sort(function(a, b) {
return b - a;
});
allTop.push(top[0]);
}
return allTop;
}
console.log(largestOfFour([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]));
A better approach is using the function map along with the function Math.max
function largestOfFour(arr) {
return arr.map(function(a) {
return Math.max.apply(null, a);
});
}
var result = largestOfFour([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]);
console.log(result);
Full ES6:
var largestOfFour = (arr) => arr.map(a => Math.max(...a));
var result = largestOfFour([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]);
console.log(result);
Try this:
function largestOfFour(arr) {
let allTop = [];
arr.forEach(a => {
allTop.push(Math.max.apply(Math, a));
});
return allTop;
}
console.log(
largestOfFour([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]])
);
Other solution would be to use function reduce along with the function Math.max
function largestOfFour(arr) {
return arr.reduce((a, x) => {
a.push(Math.max.apply(null,x));
return a;
}, []);
}
console.log(largestOfFour([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]));
function largestOfFour(arr) {
var largestNumArray = [];
for(i = 0; i < arr.length; i++) {
arr = arr[i].sort(function(a, b) {
return b - a;
});
}
}
largestOfFour([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]);
Where am I wrong and could you explain about it.
while running it shows the type error arr[i].sort(); is not a
function.
You are mutating the arr by assigning the result of arr[i].sort() to arr
Make it
arr[i]=arr[i].sort(function(a,b){
return b-a;
});
Edit
sort function mutates the array by itself, so no reason to store it back to arr[i] anyways.
arr[i].sort(function(a,b){
return b-a;
});
Here's a non-mutating version:
const largestOfFour = arr => arr.map(four => four.slice(0).sort((a, b) => a - b))
const foo = [[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]
largestOfFour(foo);
//=> [[4, 5, 1, 3], [13, 18, 26, 27], [32, 35, 37, 39], [1, 857, 1000, 1001]]
// but foo is unchanged
Using map together with slice(0), we can avoid mutation.
somehow arr[i].sort() not sorting last nested array and i getting bad result.
tryed FOR and WHILE, different operators, nothing helped. what im doing wrong?
it have return biggest numbers in array.
function largestOfFour(arr) {
var i = 0;
while (i != arr.length) {
arr[i] = arr[i].sort().pop();
i++;
}
return arr;
}
largestOfFour([
[4, 5, 8, 1, 3],
[13, 27, 18, 26],
[32, 35, 37, 39],
[1000, 1001, 817, 1]
]);
By default sort() orders the elements as strings, not numbers, and "1001" < "817". If you want a numeric sort you need to provide a comparison function that performs numeric ordering.
function largestOfFour(arr) {
var i = 0;
while (i != arr.length) {
arr[i] = arr[i].sort(function(a, b) {
return a - b;
}).pop();
i++;
}
return arr;
}
console.log(largestOfFour([
[4, 5, 8, 1, 3],
[13, 27, 18, 26],
[32, 35, 37, 39],
[1000, 1001, 817, 1]
]));
Not the cause of the problem you are seeing (which #Barmar answered), but you can use Array.map with Math.max.apply to get the max number of each list:
function largestOfFour(arr) {
return arr.map(function(list) {
return Math.max.apply(null, list);
});
}
console.log(largestOfFour([
[4, 5, 8, 1, 3],
[13, 27, 18, 26],
[32, 35, 37, 39],
[1000, 1001, 817, 1]
]));
Or a bit more concise with ES6:
const largestOfFour = arr => arr.map(list => Math.max(...list))
The problem is in sort defaults - it sorts the array as strings - so "8..." > "1..."
Try following:
function cmp(a,b){
return a-b;
}
function largestOfFour(arr) {
var i = 0;
while (i != arr.length) {
arr[i] = arr[i].sort(cmp).pop();
i++;
}
return arr;
}
largestOfFour([
[4, 5, 8, 1, 3],
[13, 27, 18, 26],
[32, 35, 37, 39],
[1000, 1001, 817, 1]
]);
Output: [ 8, 27, 39, 1001 ]