3.5 KiB
| slug | title | author | order |
|---|---|---|---|
| /silver/prefix-sums | Prefix Sums | Eric Wei | 6 |
Given an array
A_1,A_2,\ldots,A_N, answerQqueries of the following form: computeA_L+A_{L+1}+\cdots+A_R.
Standard
Tutorials
This technique is also known as cumulative sum or partial sums.
- Intro to USACO 11
- CPH 9.1
Extensions
Max Subarray Sum
Maximum Subarray Sum (Note: This problem has a solution known as Kadane's Algorithm. Please don't use that solution; try to solve it with prefix sums.)
2D Prefix Sums
Given a 2-dimensional array of size NxM, answer Q queries of the following form: Find the sum of all elements within the rectangle of indices (x1,y1) to (x2,y2).
Difference Array
Task: Given an array of size N, do the following operation Q times: add X to the values between i and j. Afterwards, print the final array.
Solution: Consider the array formed by a_i-a_{i-1}. When processing a range addition, only two values in this difference array change! At the end, we can recover the original array using prefix sums. (The math is left as an exercise to the reader.)
Prefix Minimum, XOR, etc.
Similar to prefix sums, you can also take prefix minimum or maximum; but you cannot answer min queries over an arbitrary range with prefix minimum. (This is because minimum doesn't have an inverse operation, the way subtraction is to addition.) On the other hand, XOR is its own inverse operation...
More Complex Applications
Instead of storing just the values themselves, you can also take a prefix sum over i\cdot a_i, or 10^i \cdot a_i, for instance. Some math is usually helpful for these problems; don't be afraid to get dirty with algebra!
For instance, let's see how to quickly answer the following type of query: Find 1\cdot a_l+2\cdot a_{l+1}+3\cdot a_{l+2}+\cdots+(r-l+1)\cdot a_{r}.
First, define the following:
ps[i] = a_1+a_2+a_3+a_4+\cdots+a_i
ips[i] = 1\cdot a_1+2\cdot a_2+\cdots+i\cdot a_i
Then, we have:
l\cdot a_l + (l+1) \cdot a_{l+1} + \cdots + r \cdot a_r = ips[r]-ips[l-1]
(l-1) \cdot a_l + (l-1) \cdot a_{l+1} + \cdot + (l-1) \cdot a_r = (l-1)(ps[r]-ps[l-1])
And so,
1\cdot a_l + 2 \cdot a_{l+1} + \cdots + (r-l+1) \cdot a_r = ips[r]-ips[l-1]-(l-1)(ps[r]-ps[l-1])
Which is what we were looking for!
- AtCoder Multiple of 2019 (You may want to solve the below problem "Subsequences Summing to Seven" before doing this one.)
- Google Kick Start Candies (only Test Set 1.)