this function return the sum of all elements in the array
const array = [4, 7, 24, 7, 0, 10];
const number = 7;
function addTheSameNumbers1(number, array) {
let count = array.length;
for (let i = 0; i < count; i++) {
if (array[i] === number) {
return array.reduce((a,b) => a + b, 0);
}
}
return null;
}
console.log(addTheSameNumbers1(number, array));```
Your reduce() is summing all the values. This is how to sum a single number:
const array = [4, 7, 24, 7, 0, 10];
const number = 7;
function addTheSameNumbers1(number, array) {
return array.reduce((accum, val) =>
val === number ? accum + val : accum
, 0);
}
const result = addTheSameNumbers1(number, array);
console.log(result);If I understand you correctly, you want to get the sum of the elements that have values which are the same to the length of their parent array. This code should do it.
const filterAndSum = (numbers, query) => {
return numbers
.filter((n) => n === query)
.reduce((n, total) => total + n, 0);
};
console.log(filterAndSum([1,2,3,4,5,4,3,2,1,2,3,2,1], 3))
First, it filters the elements which are not equal to the length of the parent array then gets the sum of the remaining elements. Note that validation is not implemented here, but I believe you could easily do that.
You can get the count the occurrence of variable and the multiple with the number
<script>
const countOccurrences = (arr, val) => arr.reduce((a, v) => (v === val ? a + 1 : a), 0);
const array = [4, 7, 24, 7, 0, 10];
const number = 7;
var accur = countOccurrences(array, number);
console.log(accur);
console.log(accur * number);
</script>