Lottery Coupons: Most Frequent Digit-Sum Values
Problem
You are running a lottery with consecutively numbered coupons from 1 to n (inclusive). A coupon wins if the sum of its decimal digits equals a chosen target value s. For example, coupon 23 has digit sum 2+3=5.
For a given n, consider every possible target digit sum s in the range 1≤s≤9⋅d, where d is the number of digits of n. Each value of s produces some number of winning coupons among 1…n (possibly zero). Let M be the maximum number of winners that any single s achieves.
Implement a function lotteryCoupons(n) that returns how many distinct values of s achieve exactly M winners — i.e. the number of digit sums that tie for "most popular." Then explain the algorithm's time and space complexity.
Constraints & Assumptions
-
n≥1
, base-10 representation; assume
n
can be as large as
1018
(so an
O(n)
scan is too slow).
-
d
= number of decimal digits of
n
; the relevant digit-sum range is
1≤s≤9d
.
-
Values of
s
that no coupon in
1…n
can reach simply have
0
winners and never affect the maximum (assuming at least one coupon exists, the max is
≥1
).
-
Counts can be large; assume they fit in a 64-bit integer for
n≤1018
(use arbitrary-precision integers if
n
may exceed that).
Clarifying Questions to Ask
-
Is the numbering inclusive of both endpoints (
1
and
n
), and is
0
ever a coupon? (Here: inclusive of
1…n
;
0
is not a coupon.)
-
What is the upper bound on
n
— does it fit in 64 bits, or could it be an arbitrarily long number? (Drives whether
O(n)
is acceptable and which integer type to use.)
-
Are coupons always base-10, and are leading zeros ever printed on a coupon (which would change its digit sum)? (Here: base-10, no printed leading zeros.)
-
If two or more values of
s
tie for the maximum, do we return the
count
of such values, or the values themselves? (Here: the count.)
-
Should the function return the count of winners or the count of
winning targets s
? (The latter — confirm the deliverable.)
What a Strong Answer Covers
-
Reframing to a histogram problem:
recognizing that only per-digit-sum frequencies matter, so the output is a property of the
count[s]
distribution (its max and the multiplicity of that max), not of individual coupons.
-
An efficient counting method:
a digit DP (or equivalent combinatorial argument) that counts integers in
[1,n]
by digit sum in time polynomial in the number of digits, with a correct
tight/loose
transition that never lets a prefix exceed
n
.
-
Edge-case correctness:
excluding the spurious
0
introduced by leading zeros; using
9d
(the exact digit-count bound) rather than
9⌈log10n⌉
; handling single-digit
n
,
n
being a power of
10
, and
n=1
; choosing a wide enough integer type.
-
Complexity stated and justified:
time and space in terms of
d
(and why the naive
O(n)
enumeration is unacceptable at the stated scale), plus a worked trace on a small
n
(e.g.
n=23
) to demonstrate the answer.
Follow-up Questions
-
How would you adapt the DP to count coupons in an arbitrary range
[L,R]
rather than
[1,n]
? (Hint:
f(R)−f(L−1)
.)
-
Generalize to base
b
instead of base 10 — what changes in the transition and in the max-sum bound?
-
For a "full block"
n=10d−1
, the digit-sum distribution is the coefficients of
(1+x+⋯+x9)d
. What does that tell you about how many sums can tie for the maximum, and is that bound preserved for arbitrary
n
?
-
If you needed the answer for many queries with different
n
sharing the same digit length, how could you precompute or memoize to amortize the work?