[ACCEPTED]-Sum of Parts of An Array - JavaScript-reduce

Accepted answer

Score: 11

There is no reason to compute the sum over 5 and over. On a long array this will be very 4 inefficient ( O(n²) ) and might explain 3 your timeout errors. Compute the sum at 2 the beginning and then subtract each element 1 from it in a loop.

ls = [0, 1, 3, 6, 10]

function partsSums(ls) {
    let sum = ls.reduce((sum, n) => sum + n, 0)
    res  = [sum]
    for (let i = 1; i <= ls.length; i++){
        sum -= ls[i-1]
        res.push(sum )
    }
    return res
}
console.log(partsSums(ls))

Score: 5

Another solution that passed all of the 1 tests:

function partsSums(ls) {
    let result = [0],
      l = ls.length - 1;
      
    for (let i = l; i >= 0; i--) {
        result.push(ls[i] + result[ l - i]);
    }
    return result.reverse();
}


console.log(partsSums([]));
console.log(partsSums([0, 1, 3, 6, 10])); 
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));

Score: 1

You could use for loop with slice and when i == 0 you 4 can slice len + 1 which is going to return you 3 empty array and sum will be 0.

function partsSums(arr) {
  const res = [], len = arr.length
  for (let i = len; i > -1; i--) {
    res.push(arr.slice(-i || len + 1).reduce((a, n) => a + n, 0))
  }
  return res;
}

console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));

You can also 2 use two double reduce and if there is no next 1 element push zero.

function partsSums(arr) {
  const sum = arr => arr.reduce((r, e) => r + e, 0);
  return arr.reduce((r, e, i, a) => {
    const res = sum(a.slice(i, a.length));
    return r.concat(!a[i + 1] ? [res, 0] : res)
  }, [])
}

console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));

Score: 1

You could iterate from the end and take 4 this value plus the last inserted value 3 of the result set.

This approach works with 2 a single loop and without calculating the 1 maximum sum in advance.

function partsSums(ls) {
  var result = [0],
      i = ls.length;
      
  while (i--) {
      result.unshift(ls[i] + result[0]);
  }
  return result;
}

console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([]));

.as-console-wrapper { max-height: 100% !important; top: 0; }

With push and reverse.

function partsSums(ls) {
  var result = [0],
      l = 0,
      i = ls.length;
      
  while (i--) result.push(l += ls[i]);
  return result.reverse();
}

console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([]));

.as-console-wrapper { max-height: 100% !important; top: 0; }

Score: 1

try this with recursion :

function partsSums(ls) {
  let sum = ls.reduce((a, b) => a + b, 0);
  return  ls.length > 0 ? [sum].concat(partsSums(ls.slice(1))) : [0];
}

console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));

Score: 1

Here's one thing you could do

function partsSums(ls) {
  if(!ls.length) return [0];
  let prevTotal = ls.reduce((a,b) => a + b);
  return [prevTotal, ...ls.map(val => prevTotal -= val)]
}

console.log(partsSums([0, 1, 3, 6, 10]));

Source: stackoverflow.com

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