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Sum of Parts of An Array - JavaScript

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6

1

Trying to solve this challenge on codewars. According to the challenge, the parts of array:

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

Are

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

ls = [1, 3, 6, 10]

ls = [3, 6, 10]

ls = [6, 10]

ls = [10]

ls = []

And we need to return an array with the sums of those parts.

So my code is as follows:

function partsSums(ls) {

  let arrayOfSums = []; 

  while(ls.length > 0) {

    let sum = ls.reduce((a, b) => a + b);

    arrayOfSums.push(sum);

    ls.shift();

  }

return arrayOfSums;

}



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

The issue is that it wants us to add the last sum 0 when the array is empty. So we should be getting:

[ 20, 20, 19, 16, 10, 0 ]

Instead of

[ 20, 20, 19, 16, 10]

So I tried this:

function partsSums(ls) {

  let arrayOfSums = []; 

  while(ls.length > 0) {

    let sum = ls.reduce((a, b) => a + b);

    arrayOfSums.push(sum);

    ls.shift();

  }

arrayOfSums.push(0);

return arrayOfSums;

}

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

And this:

function partsSums(ls) {

  ls.push(0);

  let arrayOfSums = []; 

  while(ls.length > 0) {

    let sum = ls.reduce((a, b) => a + b);

    arrayOfSums.push(sum);

    ls.shift();

  }

return arrayOfSums;

}

But these caused execution time-out errors on Codewars:

Execution Timed Out (12000 ms)

So I also tried:

function partsSums(ls) {

  let arrayOfSums = []; 

  while(ls.length > -1) {

    let sum = ls.reduce((a, b) => a + b);

    arrayOfSums.push(sum);

    ls.shift();

  }

return arrayOfSums;

}

But now this causes a TypeError:

TypeError: Reduce of empty array with no initial value

I am not understanding the concept of how to get 0 into the array when all of the values have been shifted out. The challenge seems to want 0 as the final "sum" of the array, even when the array is empty. But you cannot reduce an empty array - what else can I do here?

EDIT: Tried adding initial value to the reduce method:

function partsSums(ls) {

  let arrayOfSums = []; 

  while(ls.length > 0) {

    let sum = ls.reduce((a, b) => a + b, 0);

    arrayOfSums.push(sum);

    ls.shift();

  }

return arrayOfSums;

}

Unfortunately this still fails the basic test :

expected [] to deeply equal [ 0 ]

javascript arrays sum reduce

share|improve this question

edited 11 hours ago

HappyHands31

asked 11 hours ago

HappyHands31HappyHands31

1,39033 gold badges2222 silver badges4141 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  don't you think you should be solving it yourself?
  
  – Arpit Pandey
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I have tried to solve it myself. I don't know what else to try. I am not understanding the concept of how to add the sum of an empty array to the final array. Because you can't use reduce on an empty array. Any other ideas?
  
  – HappyHands31
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Cant you just push 0 after the loop?
  
  – Kobe
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  This: But you cannot reduce an empty array - that is not what that error message (TypeError: Reduce of empty array with no initial value) says. In fact, it is very specific: Please re-read the documentation for reduce: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… - and then reread the error message.
  
  – Randy Casburn
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  BINGO! But consider the iterator point too.
  
  – Randy Casburn
  11 hours ago
  

  
  
  
  

 | 
show 10 more comments

6

1

Trying to solve this challenge on codewars. According to the challenge, the parts of array:

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

Are

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

ls = [1, 3, 6, 10]

ls = [3, 6, 10]

ls = [6, 10]

ls = [10]

ls = []

And we need to return an array with the sums of those parts.

So my code is as follows:

function partsSums(ls) {

  let arrayOfSums = []; 

  while(ls.length > 0) {

    let sum = ls.reduce((a, b) => a + b);

    arrayOfSums.push(sum);

    ls.shift();

  }

return arrayOfSums;

}



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

The issue is that it wants us to add the last sum 0 when the array is empty. So we should be getting:

[ 20, 20, 19, 16, 10, 0 ]

Instead of

[ 20, 20, 19, 16, 10]

So I tried this:

function partsSums(ls) {

  let arrayOfSums = []; 

  while(ls.length > 0) {

    let sum = ls.reduce((a, b) => a + b);

    arrayOfSums.push(sum);

    ls.shift();

  }

arrayOfSums.push(0);

return arrayOfSums;

}

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

And this:

function partsSums(ls) {

  ls.push(0);

  let arrayOfSums = []; 

  while(ls.length > 0) {

    let sum = ls.reduce((a, b) => a + b);

    arrayOfSums.push(sum);

    ls.shift();

  }

return arrayOfSums;

}

But these caused execution time-out errors on Codewars:

Execution Timed Out (12000 ms)

So I also tried:

function partsSums(ls) {

  let arrayOfSums = []; 

  while(ls.length > -1) {

    let sum = ls.reduce((a, b) => a + b);

    arrayOfSums.push(sum);

    ls.shift();

  }

return arrayOfSums;

}

But now this causes a TypeError:

TypeError: Reduce of empty array with no initial value

I am not understanding the concept of how to get 0 into the array when all of the values have been shifted out. The challenge seems to want 0 as the final "sum" of the array, even when the array is empty. But you cannot reduce an empty array - what else can I do here?

EDIT: Tried adding initial value to the reduce method:

function partsSums(ls) {

  let arrayOfSums = []; 

  while(ls.length > 0) {

    let sum = ls.reduce((a, b) => a + b, 0);

    arrayOfSums.push(sum);

    ls.shift();

  }

return arrayOfSums;

}

Unfortunately this still fails the basic test :

expected [] to deeply equal [ 0 ]

javascript arrays sum reduce

share|improve this question

edited 11 hours ago

HappyHands31

asked 11 hours ago

HappyHands31HappyHands31

1,39033 gold badges2222 silver badges4141 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  don't you think you should be solving it yourself?
  
  – Arpit Pandey
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I have tried to solve it myself. I don't know what else to try. I am not understanding the concept of how to add the sum of an empty array to the final array. Because you can't use reduce on an empty array. Any other ideas?
  
  – HappyHands31
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Cant you just push 0 after the loop?
  
  – Kobe
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  This: But you cannot reduce an empty array - that is not what that error message (TypeError: Reduce of empty array with no initial value) says. In fact, it is very specific: Please re-read the documentation for reduce: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… - and then reread the error message.
  
  – Randy Casburn
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  BINGO! But consider the iterator point too.
  
  – Randy Casburn
  11 hours ago
  

  
  
  
  

 | 
show 10 more comments

Trying to solve this challenge on codewars. According to the challenge, the parts of array:

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

Are

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

ls = [1, 3, 6, 10]

ls = [3, 6, 10]

ls = [6, 10]

ls = [10]

ls = []

And we need to return an array with the sums of those parts.

So my code is as follows:

function partsSums(ls) {

  let arrayOfSums = []; 

  while(ls.length > 0) {

    let sum = ls.reduce((a, b) => a + b);

    arrayOfSums.push(sum);

    ls.shift();

  }

return arrayOfSums;

}



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

The issue is that it wants us to add the last sum 0 when the array is empty. So we should be getting:

[ 20, 20, 19, 16, 10, 0 ]

Instead of

[ 20, 20, 19, 16, 10]

So I tried this:

function partsSums(ls) {

  let arrayOfSums = []; 

  while(ls.length > 0) {

    let sum = ls.reduce((a, b) => a + b);

    arrayOfSums.push(sum);

    ls.shift();

  }

arrayOfSums.push(0);

return arrayOfSums;

}

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

And this:

function partsSums(ls) {

  ls.push(0);

  let arrayOfSums = []; 

  while(ls.length > 0) {

    let sum = ls.reduce((a, b) => a + b);

    arrayOfSums.push(sum);

    ls.shift();

  }

return arrayOfSums;

}

But these caused execution time-out errors on Codewars:

Execution Timed Out (12000 ms)

So I also tried:

function partsSums(ls) {

  let arrayOfSums = []; 

  while(ls.length > -1) {

    let sum = ls.reduce((a, b) => a + b);

    arrayOfSums.push(sum);

    ls.shift();

  }

return arrayOfSums;

}

But now this causes a TypeError:

TypeError: Reduce of empty array with no initial value

I am not understanding the concept of how to get 0 into the array when all of the values have been shifted out. The challenge seems to want 0 as the final "sum" of the array, even when the array is empty. But you cannot reduce an empty array - what else can I do here?

EDIT: Tried adding initial value to the reduce method:

function partsSums(ls) {

  let arrayOfSums = []; 

  while(ls.length > 0) {

    let sum = ls.reduce((a, b) => a + b, 0);

    arrayOfSums.push(sum);

    ls.shift();

  }

return arrayOfSums;

}

Unfortunately this still fails the basic test :

expected [] to deeply equal [ 0 ]

javascript arrays sum reduce

share|improve this question

edited 11 hours ago

HappyHands31

asked 11 hours ago

HappyHands31HappyHands31

1,39033 gold badges2222 silver badges4141 bronze badges

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

ls = [1, 3, 6, 10]

ls = [3, 6, 10]

ls = [6, 10]

ls = [10]

ls = []

function partsSums(ls) {

  let arrayOfSums = []; 

  while(ls.length > 0) {

    let sum = ls.reduce((a, b) => a + b);

    arrayOfSums.push(sum);

    ls.shift();

  }

return arrayOfSums;

}



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

function partsSums(ls) {

  let arrayOfSums = []; 

  while(ls.length > 0) {

    let sum = ls.reduce((a, b) => a + b);

    arrayOfSums.push(sum);

    ls.shift();

  }

arrayOfSums.push(0);

return arrayOfSums;

}

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

function partsSums(ls) {

  ls.push(0);

  let arrayOfSums = []; 

  while(ls.length > 0) {

    let sum = ls.reduce((a, b) => a + b);

    arrayOfSums.push(sum);

    ls.shift();

  }

return arrayOfSums;

}

function partsSums(ls) {

  let arrayOfSums = []; 

  while(ls.length > -1) {

    let sum = ls.reduce((a, b) => a + b);

    arrayOfSums.push(sum);

    ls.shift();

  }

return arrayOfSums;

}

function partsSums(ls) {

  let arrayOfSums = []; 

  while(ls.length > 0) {

    let sum = ls.reduce((a, b) => a + b, 0);

    arrayOfSums.push(sum);

    ls.shift();

  }

return arrayOfSums;

}

function partsSums(ls) {

  let arrayOfSums = []; 

  while(ls.length > 0) {

    let sum = ls.reduce((a, b) => a + b);

    arrayOfSums.push(sum);

    ls.shift();

  }

return arrayOfSums;

}



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

function partsSums(ls) {

  let arrayOfSums = []; 

  while(ls.length > 0) {

    let sum = ls.reduce((a, b) => a + b);

    arrayOfSums.push(sum);

    ls.shift();

  }

return arrayOfSums;

}



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

function partsSums(ls) {

  let arrayOfSums = []; 

  while(ls.length > 0) {

    let sum = ls.reduce((a, b) => a + b);

    arrayOfSums.push(sum);

    ls.shift();

  }

arrayOfSums.push(0);

return arrayOfSums;

}

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

function partsSums(ls) {

  let arrayOfSums = []; 

  while(ls.length > 0) {

    let sum = ls.reduce((a, b) => a + b);

    arrayOfSums.push(sum);

    ls.shift();

  }

arrayOfSums.push(0);

return arrayOfSums;

}

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

function partsSums(ls) {

  ls.push(0);

  let arrayOfSums = []; 

  while(ls.length > 0) {

    let sum = ls.reduce((a, b) => a + b);

    arrayOfSums.push(sum);

    ls.shift();

  }

return arrayOfSums;

}

function partsSums(ls) {

  ls.push(0);

  let arrayOfSums = []; 

  while(ls.length > 0) {

    let sum = ls.reduce((a, b) => a + b);

    arrayOfSums.push(sum);

    ls.shift();

  }

return arrayOfSums;

}

function partsSums(ls) {

  let arrayOfSums = []; 

  while(ls.length > -1) {

    let sum = ls.reduce((a, b) => a + b);

    arrayOfSums.push(sum);

    ls.shift();

  }

return arrayOfSums;

}

function partsSums(ls) {

  let arrayOfSums = []; 

  while(ls.length > -1) {

    let sum = ls.reduce((a, b) => a + b);

    arrayOfSums.push(sum);

    ls.shift();

  }

return arrayOfSums;

}

share|improve this question

edited 11 hours ago

HappyHands31

asked 11 hours ago

HappyHands31HappyHands31

1,39033 gold badges2222 silver badges4141 bronze badges

share|improve this question

edited 11 hours ago

HappyHands31

asked 11 hours ago

HappyHands31HappyHands31

1,39033 gold badges2222 silver badges4141 bronze badges

asked 11 hours ago

HappyHands31HappyHands31

1,39033 gold badges2222 silver badges4141 bronze badges

asked 11 hours ago

HappyHands31HappyHands31

1,39033 gold badges2222 silver badges4141 bronze badges

6 Answers
6

active

oldest

votes

7

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

share|improve this answer

edited 11 hours ago

answered 11 hours ago

Mark MeyerMark Meyer

47.6k33 gold badges4141 silver badges7474 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It works. There are some things about it that I don't understand. What is ( O(n^2) )? XOR, right?
  
  – HappyHands31
  11 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  O(n²) means that the number of calculations required grows exponentially as the length of your array grows. The reduce needs to look at every element in the array and you do that for every element in the array.
  
  – Mark Meyer
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @NinaScholz Are you talking about the initial reduce()? At least on chrome the overhead of the initial sum is much smaller than the penalty for unshift() in my tests.
  
  – Mark Meyer
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  So, to break this down a bit more, we have the first sum as 20, then within the for-loop, we create the "result" array, initializing it to 20. Then we subtract ls[i - 1] from sum. So the first sum to get pushed into the result array from the for-loop will also be 20, because ls[i - 1] in this case is 0. Then 20 - 1, then 19 - 3, etc. Makes sense - thank you.
  
  – HappyHands31
  10 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @HappyHands31 per Mark Meyer's comment above, if you need to ask "What is ( O(n^2))" then you probably can benefit by being told explicitly this is called "Big-O" notation.
  
  – JimLohse
  3 hours ago
  
  
  

  
  
  
  

 | 
show 1 more comment

3

Another solution that passed all of the 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]));

share|improve this answer

edited 10 hours ago

answered 10 hours ago

FractionFraction

1,83322 silver badges1616 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  That is very clever yet simple. Thx.
  
  – HappyHands31
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You could also do result.unshift(ls[i] + result[0]) and avoid reverse.
  
  – georg
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @georg unshift doesn't pass the test because it is slower than push+reverse
  
  – Fraction
  9 hours ago
  

  
  
  
  

add a comment | 

1

You could use for loop with slice and when i == 0 you can slice len + 1 which is going to return you 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 use two double reduce and if there is no next 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]));

share|improve this answer

edited 11 hours ago

answered 11 hours ago

Nenad VracarNenad Vracar

76.6k1212 gold badges6565 silver badges8888 bronze badges

add a comment | 

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]));

share|improve this answer

answered 10 hours ago

Elma CherbElma Cherb

19611 silver badge1010 bronze badges

add a comment | 

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]));

share|improve this answer

answered 10 hours ago

Alexandre FradetteAlexandre Fradette

7999 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I like this also - can you please help me understand how the if(!ls.length) return [0] gets added to the end of the returned array? There aren't any loops that are reducing the length of ls?
  
  – HappyHands31
  10 hours ago
  
  
  

  
  
  
  

add a comment | 

1

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

This approach works with a single loop and without calculating the 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; }

share|improve this answer

edited 10 hours ago

answered 11 hours ago

Nina ScholzNina Scholz

212k1616 gold badges127127 silver badges192192 bronze badges

add a comment | 

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7

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

share|improve this answer

edited 11 hours ago

answered 11 hours ago

Mark MeyerMark Meyer

47.6k33 gold badges4141 silver badges7474 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It works. There are some things about it that I don't understand. What is ( O(n^2) )? XOR, right?
  
  – HappyHands31
  11 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  O(n²) means that the number of calculations required grows exponentially as the length of your array grows. The reduce needs to look at every element in the array and you do that for every element in the array.
  
  – Mark Meyer
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @NinaScholz Are you talking about the initial reduce()? At least on chrome the overhead of the initial sum is much smaller than the penalty for unshift() in my tests.
  
  – Mark Meyer
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  So, to break this down a bit more, we have the first sum as 20, then within the for-loop, we create the "result" array, initializing it to 20. Then we subtract ls[i - 1] from sum. So the first sum to get pushed into the result array from the for-loop will also be 20, because ls[i - 1] in this case is 0. Then 20 - 1, then 19 - 3, etc. Makes sense - thank you.
  
  – HappyHands31
  10 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @HappyHands31 per Mark Meyer's comment above, if you need to ask "What is ( O(n^2))" then you probably can benefit by being told explicitly this is called "Big-O" notation.
  
  – JimLohse
  3 hours ago
  
  
  

  
  
  
  

 | 
show 1 more comment

7

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

share|improve this answer

edited 11 hours ago

answered 11 hours ago

Mark MeyerMark Meyer

47.6k33 gold badges4141 silver badges7474 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It works. There are some things about it that I don't understand. What is ( O(n^2) )? XOR, right?
  
  – HappyHands31
  11 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  O(n²) means that the number of calculations required grows exponentially as the length of your array grows. The reduce needs to look at every element in the array and you do that for every element in the array.
  
  – Mark Meyer
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @NinaScholz Are you talking about the initial reduce()? At least on chrome the overhead of the initial sum is much smaller than the penalty for unshift() in my tests.
  
  – Mark Meyer
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  So, to break this down a bit more, we have the first sum as 20, then within the for-loop, we create the "result" array, initializing it to 20. Then we subtract ls[i - 1] from sum. So the first sum to get pushed into the result array from the for-loop will also be 20, because ls[i - 1] in this case is 0. Then 20 - 1, then 19 - 3, etc. Makes sense - thank you.
  
  – HappyHands31
  10 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @HappyHands31 per Mark Meyer's comment above, if you need to ask "What is ( O(n^2))" then you probably can benefit by being told explicitly this is called "Big-O" notation.
  
  – JimLohse
  3 hours ago
  
  
  

  
  
  
  

 | 
show 1 more comment

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

share|improve this answer

edited 11 hours ago

answered 11 hours ago

Mark MeyerMark Meyer

47.6k33 gold badges4141 silver badges7474 bronze badges

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))

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))

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))

share|improve this answer

edited 11 hours ago

answered 11 hours ago

Mark MeyerMark Meyer

47.6k33 gold badges4141 silver badges7474 bronze badges

answered 11 hours ago

Mark MeyerMark Meyer

47.6k33 gold badges4141 silver badges7474 bronze badges

answered 11 hours ago

Mark MeyerMark Meyer

47.6k33 gold badges4141 silver badges7474 bronze badges

3

Another solution that passed all of the 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]));

share|improve this answer

edited 10 hours ago

answered 10 hours ago

FractionFraction

1,83322 silver badges1616 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  That is very clever yet simple. Thx.
  
  – HappyHands31
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You could also do result.unshift(ls[i] + result[0]) and avoid reverse.
  
  – georg
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @georg unshift doesn't pass the test because it is slower than push+reverse
  
  – Fraction
  9 hours ago
  

  
  
  
  

add a comment | 

3

Another solution that passed all of the 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]));

share|improve this answer

edited 10 hours ago

answered 10 hours ago

FractionFraction

1,83322 silver badges1616 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  That is very clever yet simple. Thx.
  
  – HappyHands31
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You could also do result.unshift(ls[i] + result[0]) and avoid reverse.
  
  – georg
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @georg unshift doesn't pass the test because it is slower than push+reverse
  
  – Fraction
  9 hours ago
  

  
  
  
  

add a comment | 

Another solution that passed all of the 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]));

share|improve this answer

edited 10 hours ago

answered 10 hours ago

FractionFraction

1,83322 silver badges1616 bronze badges

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]));

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]));

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]));

share|improve this answer

edited 10 hours ago

answered 10 hours ago

FractionFraction

1,83322 silver badges1616 bronze badges

answered 10 hours ago

FractionFraction

1,83322 silver badges1616 bronze badges

answered 10 hours ago

FractionFraction

1,83322 silver badges1616 bronze badges

1

You could use for loop with slice and when i == 0 you can slice len + 1 which is going to return you 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 use two double reduce and if there is no next 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]));

share|improve this answer

edited 11 hours ago

answered 11 hours ago

Nenad VracarNenad Vracar

76.6k1212 gold badges6565 silver badges8888 bronze badges

add a comment | 

1

You could use for loop with slice and when i == 0 you can slice len + 1 which is going to return you 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 use two double reduce and if there is no next 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]));

share|improve this answer

edited 11 hours ago

answered 11 hours ago

Nenad VracarNenad Vracar

76.6k1212 gold badges6565 silver badges8888 bronze badges

add a comment | 

You could use for loop with slice and when i == 0 you can slice len + 1 which is going to return you 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 use two double reduce and if there is no next 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]));

share|improve this answer

edited 11 hours ago

answered 11 hours ago

Nenad VracarNenad Vracar

76.6k1212 gold badges6565 silver badges8888 bronze badges

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]));

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]));

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]));

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]));

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]));

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]));

share|improve this answer

edited 11 hours ago

answered 11 hours ago

Nenad VracarNenad Vracar

76.6k1212 gold badges6565 silver badges8888 bronze badges