I asked a question before where I was looking to understand the complexity class of the following problem:

Given a range of integers $\{a,a+1,...,b-1,b\}$, find a subset of size $k$ such that the sum is equal to $s$.

I had found a $O(k)$ algorithm as hinted at by the accepted answer. After playing around with the results of this algorithm, I found that it was not 100% what I was looking for.


Given a range of integers $\{a,a+1,...,b-1,b\}$, find the range of the $i^{th}$ element of all subsets of size $k$, sorted in increasing order, that sum to $s$:

For example, given $k = 3$, $a = 1$, $b = 8$, and $s = 11$ we get the following list of all possible solutions sorted in increasing order:

{ 1 | 2 | 8 }
{ 1 | 3 | 7 }
{ 1 | 4 | 6 }
{ 2 | 3 | 6 }
{ 2 | 4 | 5 }

By grouping each column, we can see the ranges of values for each of the $i^{th}$ elements in the set:

  1. $\{1,2\}$
  2. $\{2,3,4\}$
  3. $\{5,6,7,8\}$


I would like an algorithm that in constant time could generate the range of values for a $i^{th}$ element for all subsets summing to $s$ for a given $a$ and $b$.

 get_ith_range(int k, int a, int b, int s, int i)
    int lower_bound, upper_bound
    // Some cool O(1) algorithm...
    return (lower_bound, upper_bound)

I do not mind if it is not $O(1)$, but I feel as though there should be a way to get there at least in $O(k)$.


Currently, I have a way to generate the smallest for the $1^{st}$ range and the largest for the $k^{th}$ range (not in constant time though). But, I have not been able to even get to where I could generate the full $i^{th}$ range in any case.

Using my solution from my previous question, I can in $O(k)$ time produce the smallest possible value and the largest:

while k:
    k -= 1
    temp = int((k)*(2*a+k-1)/2)
    if s > b+temp:
        yield b
        b -= 1
        s -= b
        yield s-temp
        while k:
            k -= 1
            yield a+k

The last value generated by the above algorithm will be the smallest possible value. Also, the first value generated is the largest possible value.

Although, the largest can be done trivially (which is how I came up with the above greedy algorithm):

# (number of values)*(smallest + largest)/2 => sum {smallest,...,largest}
temp = int((k-1)*(2*a + k - 2)/2)
if s < b+temp:
    max = b-temp
    max = b
  • $\begingroup$ I don't see why the problem is useful, but if you want to approach it, you might try to solve the following problem: given $a\le x \le b$ and $u,v$ and $z$, determine whether there is a size-$u$ subset of $a,a+1,\dots,x-1$ and a size-$v$ subset of $x+1,x+2,\dots,b$ that together sum to $z$. If you can solve that, you have determined whether $x$ can appear as the $i$th element of a size-$k$ subset of $a,\dots,b$ that sums to $s$ (by letting $u=i-1$, $v=k-i-1$, and $z=s-x$), then you can iterate over $x$. (continued) $\endgroup$
    – D.W.
    Apr 4, 2019 at 21:05
  • $\begingroup$ It might be possible to extend the reasoning of cs.stackexchange.com/a/103314/755 to solve that problem. $\endgroup$
    – D.W.
    Apr 4, 2019 at 21:07
  • $\begingroup$ @D.W., The reason for the algorithm is pretty useless. I like to play around with obscure/esoteric problems. Like code-golfing with weird programming languages:) $\endgroup$
    – tkellehe
    Apr 4, 2019 at 22:54

1 Answer 1


Suppose that the value at location $i$ is $j$. According to the continuity of $k$-sums (mentioned in an answer to the linked question), the range of possible sums of values at locations $1,\ldots,i-1$ is $$(i-1)a + \binom{i-1}{2},\ldots,(i-1)(j-1) - \binom{i-1}{2}.$$ Similarly, the range of possible sums of values at locations $i+1,\ldots,k$ is $$ (k-i)(j+1) + \binom{k-i}{2}, \ldots, (k-i)b - \binom{k-i}{2}. $$ Therefore the range of possible sums of all $k$ values is precisely $$ (i-1)a + \binom{i-1}{2} + j + (k-i)(j+1) + \binom{k-i}{2}, \ldots, \\ (i-1)(j-1) - \binom{i-1}{2} + j + (k-i)b - \binom{k-i}{2}. $$ We want to find for which values of $j$, the value $s$ is in this range. We get the following inequalities: $$ (i-1)a + \binom{i-1}{2} + \binom{k-i+1}{2} + (k-i+1)j \leq s, \\ (k-i)b - \binom{i}{2} - \binom{k-i}{2} + ij \geq s. $$ Therefore the range of $j$ is $$ \max\left(\left\lceil \frac{s-(k-i)b + \binom{i}{2} + \binom{k-i}{2}}{i} \right \rceil,a+i-1\right) \leq j \leq \\ \min\left(\left\lfloor \frac{s - (i-1)a - \binom{i-1}{2} - \binom{k-i+1}{2}}{k-i+1} \right \rfloor,b-k+i\right). $$


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