Problem:
Given an array of integers, return a new array such that each element at index i of the new array is the product of all the numbers in the original array except the one at i.
For example:
if our input was [1, 2, 3, 4, 5], the expected output would be [120, 60, 40, 30, 24].
If our input was [3, 2, 1], the expected output would be [2, 3, 6].
Solution 1 (With Nested loops): I'm able to solve this by nested loops like below:
const input = [1, 2, 3, 4, 5];
function output(items) {
const finalArray = [];
for (let i = 0; i < items.length; i++) {
let multipliedNum = 1;
items.forEach((item, indx) => {
if (i !== indx) {
multipliedNum = multipliedNum * item;
}
});
finalArray.push(multipliedNum)
}
return finalArray;
}
console.log(output(input))
I'm trying to find out another solution without nested loops inside output function? Any help or suggestion really appreciated.
If there are no zero values, you can loop through all the values once to get the product. Then just return the array where each the product is divided by each entry.
However, if there are zeros then there is a little more to be done to check how many there are. One zero is fine but more than 1 means that the value is zero for each entry.
const input = [1, 2, 3, 4, 5];
const input2 = [1, 2, 3, 4, 0];
const input3 = [1, 2, 3, 0, 0];
function output(items) {
let zeroCount = 0;
let totalProduct = 1;
for (let i = 0; i < items.length; i++) {
if (items[i] === 0) {
if (++zeroCount > 1) break;
continue;
}
totalProduct *= items[i];
}
if (zeroCount > 1) {
// more than 1 zero -> all values are 0
return new Array(items.length).fill(0);
} else if (zeroCount === 1) {
// only 1 zero -> only the value that is zero will be the totalProduct
return items.map(item => item === 0 ? totalProduct : 0);
}
// no zero in array -> divide the totalProduct by each item
return items.map(item => totalProduct / item);
}
console.log(output(input))
console.log(output(input2))
console.log(output(input3))Based on what @Mike said in the comment here's the answer.
const input = [1, 2, 3, 4, 5];
const mulValues = input.reduce((acc, next) => acc * next);
const output = input.map(i => mulValues/i)
console.log(output)
you can do something like that (assuming array doesn't contain zero):
const input = [1, 2, 3, 4, 5];
function output(items) {
const finalArray = [];
const multipliedNum=1;
for (let i = 0; i < items.length; i++) {
multipliedNum *= item[i];
}
for (let i = 0; i < items.length; i++) {
finalArray.push(multipliedNum/item[i]);
}
return finalArray;
}
console.log(output(input))