Optimal technique for dividing an array based on a specified condition when iterating through the array in JavaScript

Question

Optimal technique for dividing an array based on a specified condition when iterating through the array in JavaScript

Looking for the best approach to solve this array problem.

[ 1, -1, -1, -1, 1, -1, 1, 1 ]

I'm debating between using the reduce method, a do..while loop, or possibly something else.

The task involves looping through the array and summing up as I go. If the sum equals 0, then add those elements to a new array.

For instance, if the first two elements in the array result in 0 when added together, like [1, -1], they should be placed into a new array:

[[1, -1], [-1, -1, 1, -1, 1, 1]]

My current attempt uses the reduce method:

let hikeArr = [ 1, -1, -1, -1, 1, -1, 1, 1 ];

let newArr = hikeArr.reduce((a, b) => {
        let sum = a + b;
        if( sum == 0) {
            a.push(b)
        }
        return a;
    }, []);
    console.log("newArr", newArr);

Any suggestions on a more efficient approach?

javascript arrays

Answer 1

Answer №1

Using the reduce method might not be the most efficient approach in this scenario, as it requires keeping track of multiple variables simultaneously. Instead, opting for a solution that utilizes external variables and a traditional for loop could prove to be more effective:

let hikeArr = [ 1, -1, -1, -1, 1, -1, 1, 1 ];

const output = [];
let sum = 0;
let subarr = [];
for (const num of hikeArr) {
  sum += num;
  subarr.push(num);
  if (sum === 0) {
    output.push(subarr);
    subarr = [];
  }
}
console.log(output);

Answer 2

Using the reduce method might not be the most efficient approach in this scenario, as it requires keeping track of multiple variables simultaneously. Instead, opting for a solution that utilizes external variables and a traditional for loop could prove to be more effective:

let hikeArr = [ 1, -1, -1, -1, 1, -1, 1, 1 ];

const output = [];
let sum = 0;
let subarr = [];
for (const num of hikeArr) {
  sum += num;
  subarr.push(num);
  if (sum === 0) {
    output.push(subarr);
    subarr = [];
  }
}
console.log(output);

Answer 3

Answer №2

In order to achieve the desired outcome, creating a temporary array with a closure and incorporating a variable for sum using the reduce function is essential.

const numbers = [3, 7, -2, 9, -5, 1],
    finalResult = numbers.reduce(((array, total) => (result, value) => {
        array.push(value);
        total += value;
        if (!total) {
            result.push(array);
            array = [];
        }
        return result;
    })([], 0), []);

console.log(finalResult);

Answer 4

In order to achieve the desired outcome, creating a temporary array with a closure and incorporating a variable for sum using the reduce function is essential.

const numbers = [3, 7, -2, 9, -5, 1],
    finalResult = numbers.reduce(((array, total) => (result, value) => {
        array.push(value);
        total += value;
        if (!total) {
            result.push(array);
            array = [];
        }
        return result;
    })([], 0), []);

console.log(finalResult);

Answer 5

Answer №3

This problem can be easily solved by utilizing the Array.reduce method.

Within the callback function of Array.reduce, each current value is added to a temporary array called subArr, and the sum is continuously calculated.

When the sum reaches 0, the contents of subArr are moved to the main result array and then reset to an initial state.

let hikeArr = [ 1, -1, -1, -1, 1, -1, 1, 1 ];


let subArr = [];
let sum = 0;
const result = hikeArr.reduce((acc, cur) => {
  subArr.push(cur);
  sum += cur;
  if (sum === 0) {
    acc.push(subArr);
    subArr = [];
    sum = 0;
  }
  return acc;
}, []);
console.log(result);

Answer 6

This problem can be easily solved by utilizing the Array.reduce method.

Within the callback function of Array.reduce, each current value is added to a temporary array called subArr, and the sum is continuously calculated.

When the sum reaches 0, the contents of subArr are moved to the main result array and then reset to an initial state.

let hikeArr = [ 1, -1, -1, -1, 1, -1, 1, 1 ];


let subArr = [];
let sum = 0;
const result = hikeArr.reduce((acc, cur) => {
  subArr.push(cur);
  sum += cur;
  if (sum === 0) {
    acc.push(subArr);
    subArr = [];
    sum = 0;
  }
  return acc;
}, []);
console.log(result);

Optimal technique for dividing an array based on a specified condition when iterating through the array in JavaScript

Answer №1

Answer №2

Answer №3

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