Question 265361 in Javascript Question posted in
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Javascript – Sum of i j k with some especification

I am trying to resolve this situation in js:

Given three integers i, j, k, I want to calculate this sum:

 i+(i+1)+(i+2)+...+j+(j-1)+(j-2)+...+k

if i=0, j=5, k=-1, the result should be 24
if i=5, j=9, k=6, the result should be 56.
Because i increase until is equal to j and then decrease until is equal to k.

I tryed this:

 function getSequenceSum(i, j, k) {
      let sum = 0;
      if (i <= j) {
        while (i < j) {
          i = i + 1;
          sum = sum + i;
        }
        while (j > k) {
          j = j - 1;
          sum = sum + j;
        }
      }

      return sum;
    }

console.log(getSequenceSum(0, 5, -1));
console.log(getSequenceSum(5, 9, 6));

but the output was 51, any suggestions? Thanks

Tags:

3

Answers


  1. 0

    so, in reality your sum is:
    { i + (i+1)+(i+2)...(i+n) while((i+n) < j) }
    + { j + (j-1)+(j-2)...(j-m) while((j-m) >= k) }

    so your second while() should be while(j >= k), not while(j > k)

    console.log( getSequenceSum(0,5,-1)) // -> 24  
console.log( getSequenceSum(5,9,6))  // -> 56

function getSequenceSum(i, j, k)
  {
  if (i > j || j < k) return 0; 
  let sum = 0;
  while (i < j) sum += i++;
  while (j >= k) sum += j--;
  return sum;
  }
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  2. 0

    Here’s an O(1) solution.

    Input: i, j, k: i < j, j > k.

    The sum from i to j is: (i+0) + (i+1) + (i+2) + … + j

    Note that j = i + (j-i).

    So this is the sum from (i+0) to (i+j-i). This sum has (j-i+1) terms, so we can rewrite it as:

    (j-i+1)*i + (1+2+3+...+(j-i))
= (j-i+1)*i + (j-i)*(j-i+1)/2

    Then we need the sum from k to j-1. This is identical to the above, except we swap the i for k and the j for j-1.

    Helper method:

    f(a, b) = (b-a+1)*a + (b-a)*(b-a+1)/2

    Solution:

    g(i, j, k) = f(i, j) + f(k, j-1)

    Ruby implementation (sorry, don’t know javascript; ready as pseduocode)

    def f(a, b)
  return (b-a+1)*a + (b-a)*(b-a+1)/2
end

def g(i, j, k)
  return f(i, j) + f(k, j-1)
end

> g(0,5,-1)
=> 24
> g(5,9,6)
=> 56

    Javascript code courtesy of @Jabaa

    function f(a, b) {
    return (b-a+1)*a + (b-a)*(b-a+1)/2;
}

function g(i, j, k) {
    return f(i, j) + f(k, j-1);
}

console.log(g(0,5,-1));
console.log(g(5,9,6));
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  3. 0

    Your second loop is off by one and you should first add the number, then increment/decrement it, otherwise you miss the first element of the range:

    function getSequenceSum(i, j, k) {
  let sum = 0;
  if (i <= j) {
    while (i < j) {
      sum += i;
      ++i;
    }
    while (j >= k) {
      sum += j;
      --j;
    }
  }
  return sum;
}

console.log(getSequenceSum(0, 5, -1));
console.log(getSequenceSum(5, 9, 6));

    I would change the code to

    function getSequenceSum(i, j, k) {
  if (i > j || j < k) {
    return 0;
  }
  let sum = 0;
  while (i < j) {
    sum += i;
    ++i;
  }
  while (j >= k) {
    sum += j;
    --j;
  }
  return sum;
}

console.log(getSequenceSum(0, 5, -1));
console.log(getSequenceSum(5, 9, 6));

    but it has different behavior. It checks the ranges with early exit.

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