I would like to find the sum of values from a given number array, where the repetition of numbers are allowed, closest to a target but the sum cannot exceed the target. If there are more solution, I'd prefer the one with the minimum element count.



Given values: [500, 1000, 2000, 5000]

Target: 7000

Result: [2000,5000]


Given values: [500, 1000, 2000, 5000]

Target: 7990

Result: [500, 2000, 5000]


Given values: [222,333]

Target: 444

Result: [222,222]


Given values: [222,333]

Target: 777

Result: [222,222,333]

Later on I would like to implement this algorithm in JavaScript, and make it run in a browser. I have tried:

-Knapsack algorithm

-Generating all combination of the numbers and find one with the minimum difference

but both are very slow when implemented, and used with big numbers.

  • $\begingroup$ This problem is NP-hard since it is a generalization of partition. The best you can hope for is a pseudopolynomial-time algorithm. $\endgroup$
    – Steven
    Oct 29, 2019 at 15:56
  • $\begingroup$ Your problem is known as the change-making problem if you require the numbers to add up exactly. As noted above, it is weakly NP-hard. If the set of numbers form a matroid, a greedy approach is possible (see greedy solution). $\endgroup$
    – Daniel
    Oct 29, 2019 at 16:21
  • $\begingroup$ @Steven How is this a generalization of partition? Can you show it's NP-hard? $\endgroup$ Oct 30, 2019 at 10:09
  • $\begingroup$ @YuvalFilmus See my answer :) $\endgroup$
    – Steven
    Oct 30, 2019 at 13:01

1 Answer 1


This problem is NP-hard by a reduction from partition. Let $X = \{x_1, \dots, x_n\}$, be an instance of partition and let $2M = \sum_{x \in X} x$ (assume that $M$ is an integer as otherwise the instance is trivial). Assume that each $x_i$ is positive and let $\ell = \lceil \log 2M \rceil$.

I will now construct some values based on $X$. I will denote a generic value by a tuple $v$ of $n+2$ entries $(v^{*}, v^{(0)}, v^{(1)}, \dots, v^{(n)})$, where all entries $v^{(i)}$, $i=0,\dots,n$ are between $0$ and $2^\ell - 1$. The tuple $v$ will actually correspond to integer whose binary representation is obtained by encoding $v^{*}$ in binary (using a variable number of bits), each of $v^{(0)}, \dots, v^{(n)}$ using $\ell$-bits, and finally concatenating the resulting bitstrings, in order. An example of the tuple $v=(5, 2, 7, 9)$ with $\ell = 4$ is as follows: $$ (\underbrace{101}_{v^*} \underbrace{0010}_{v^{(0)}} \underbrace{0111}_{v^{(1)}} \underbrace{1001}_{v^{(2)}})_2 = (21113)_{10} $$

Notice that the value obtained in this way has an encoding of polynomial length w.r.t. the length of the encoding of the original partition instance, as long as $v^{(*)}$ is small enough (as will be the case). The sum of two tuples is the tuple whose integer value equals the sum of the integer values of the summands.

Let then $y_i = (1, x_i, 0, \dots, 0, 1, 0, \dots, 0)$, where the second entry set to $1$ is $y_i^{(i)}$. Similarly, let $z_i = (1, 0, 0, \dots, 0,1,0, \dots, 0)$, where $z_i^{(i)} = 1$.

The instance of the problem in the question consists of the set of numbers $S = \{y_i : x_i \in X\} \cup \{z_i : x_i \in X\}$ and of the target value $T=(n,M,1, \dots, 1)$.

Given a solution $X' \subseteq X$ to the partition problem, it is easy to construct a solution $S' \subseteq S$ for our problem such that $\sum_{x \in S'} x = T$. Simply select $S' = \{ y_i : x_i \in X'\} \cup \{ z_i : x_i \in X \setminus X' \}$ (notice that it suffices for $S'$ to be a set rather than a multi-set).

I will now argue that if a (multi-)set $S' \subseteq S$ such that $\sum_{x \in S'} x = T$ exists, then the partition instance $X$ has answer "yes". From now on let $S'$ be such a set and notice that, by our choice of $T$, the overall multiplicity of all the elements of $S'$ cannot be larger than $n$ (as otherwise $T^{(*)}$ would be exceeded).

Claim: Let $m(x)$ be multiplicity in $S'$ of $x \in S$. $\forall i=1,\dots,n$ we have $m(y_i) + m(z_i) = 1$.

Proof: If $m(y_i) + m(z_i) = 0$ then, since $T^{(i)}=1$, $S'$ must contain at least $2^\ell \ge 2M > n$ elements $y_j$ or $z_j$ such that $j > i$, contradicting $\sum_{x \in S'} m(x) \le n$.

Assume then that $m(y_i) + m(z_i) > 1$, and consider the tuple $h$ corresponding to the sum of exactly 2 copies of $y_i$ and/or $z_i$ in $S'$. We have $h^{(i)}=2$ and, since $T^{(i)}=1$, we conclude that there must be at least $2^\ell - 2$ other copies of some $y_j$ or $z_j$ for $j \ge i$ in $S'$. The total multiplicity of the elements in $S'$ is therefore at least $2 + 2^\ell - 2 > n$, a contradiction. $\;\; \square$

Let $X' = \{x_i : y_i \in S' \}$ and call $h = \sum_{x \in S'} x$. Since there can be at most one copy of each $z_i$ in $S'$ (see the above Claim), the only way to affect $h^{(0)}$ is through the values $y_i \in S'$. Since there can be at most one copy of each $y_i$ in $S'$ (see the above Claim), we must have $\sum_{x_i \in X'} x_i = \sum_{y_i \in S'} y_i^{(0)} = h^{(0)} = T^{(0)} = M$.

  • $\begingroup$ Thank you for the detailed answer, I am speechless. $\endgroup$ Oct 31, 2019 at 14:52

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