# Easy interview question got harder: given numbers 1..100, find the missing number(s) given exactly k are missing

Here’s a summary of Dimitris Andreou’s link.

Remember sum of i-th powers, where i=1,2,..,k. This reduces the problem to solving the system of equations

a1 + a2 + … + ak = b1

a12 + a22 + … + ak2 = b2

a1k + a2k + … + akk = bk

Using Newton’s identities, knowing bi allows to compute

c1 = a1 + a2 + … ak

c2 = a1a2 + a1a3 + … + ak-1ak

ck = a1a2 … ak

If you expand the polynomial (x-a1)…(x-ak) the coefficients will be exactly c1, …, ck – see Viète’s formulas. Since every polynomial factors uniquely (ring of polynomials is an Euclidean domain), this means ai are uniquely determined, up to permutation.

This ends a proof that remembering powers is enough to recover the numbers. For constant k, this is a good approach.

However, when k is varying, the direct approach of computing c1,…,ck is prohibitely expensive, since e.g. ck is the product of all missing numbers, magnitude n!/(n-k)!. To overcome this, perform computations in Zq field, where q is a prime such that n <= q < 2n – it exists by Bertrand’s postulate. The proof doesn’t need to be changed, since the formulas still hold, and factorization of polynomials is still unique. You also need an algorithm for factorization over finite fields, for example the one by Berlekamp or Cantor-Zassenhaus.

High level pseudocode for constant k:

• Compute i-th powers of given numbers
• Subtract to get sums of i-th powers of unknown numbers. Call the sums bi.
• Use Newton’s identities to compute coefficients from bi; call them ci. Basically, c1 = b1; c2 = (c1b1 – b2)/2; see Wikipedia for exact formulas
• Factor the polynomial xk-c1xk-1 + … + ck.
• The roots of the polynomial are the needed numbers a1, …, ak.

For varying k, find a prime n <= q < 2n using e.g. Miller-Rabin, and perform the steps with all numbers reduced modulo q.

EDIT: The previous version of this answer stated that instead of Zq, where q is prime, it is possible to use a finite field of characteristic 2 (q=2^(log n)). This is not the case, since Newton’s formulas require division by numbers up to k.