DEV Community

Cover image for A running sum
zuzexx
zuzexx

Posted on • Updated on

A running sum

Today I took a break from working with JS objects, and did a challenge involving an array instead.

/* Given an array nums. We define a running sum of an array as runningSum[i] = sum(nums[0]…nums[i]).

Return the running sum of nums.

Example:

Input: nums = [1,2,3,4]
Output: [1,3,6,10]
Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4].
*/
Enter fullscreen mode Exit fullscreen mode

Write the function that sums together array items

const runningSum = (array) => {
//loop through the array for the length of the array.
// sum together the previous item from the array to the current item of the array
//return the new array

};
Enter fullscreen mode Exit fullscreen mode



The start seems really easy, just apply a for loop to the array and the deed is done. Not so fast.


for(let i = 0; i<array.length; i++){

}
Enter fullscreen mode Exit fullscreen mode



When working with arrays, I usually start the index (i) at the value of 0. Let's think about how this would work in this example.

We need to put together the previous array item [i-1] and the current array item [i]. If we run the code as is, with let i = 0, then the beginning index is -1. That won't do. Javascript arrays are zero indexed and I needed to modify the value of i, so that [i-1] returns zero on the first loop. I just decided on changing the value to 1 and that solved the problem.

const runningSum = (array) => {
for(let i = 1; i<array.length; i++){

}

};
Enter fullscreen mode Exit fullscreen mode

After that it is just as easy as suming both values together, set the new value of the existing array to the said sum / new value and continue doing so until the end of the array. After that the new array is returned and that is it.

const runningSum = (array) => {
  for (let i = 1; i < array.length; i++) {
    let newValue = array[i - 1] + array[i];
    array[i] = newValue;
  }
  return array;
};
Enter fullscreen mode Exit fullscreen mode

I tried the code on a few different examples and got the wanted results. It looks like the code is working as intended.

console.log(runningSum([1, 2, 3, 4]));
console.log(runningSum([1, 1, 1, 1, 1, 1]));
console.log(runningSum([1, 2, 3, 4, 5, 6]));
console.log(runningSum([1, 2, 3, 3, 2, 1]));

/* Results:
[ 1, 3, 6, 10 ]
[ 1, 2, 3, 4, 5, 6 ]
[ 1, 3, 6, 10, 15, 21 ]
[ 1, 3, 6, 9, 11, 12 ]
*/
Enter fullscreen mode Exit fullscreen mode

Conclusion

That is it for today's challenge. It was nice short and simple, great for practicing things that I studied.

Top comments (0)