Introduction to Prefix Sums

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1D Prefix Sums

Let's say we have a one-indexed integer array $\texttt{arr}$ of size $N$ and we want to compute the value of

for $Q$ different pairs $(a,b)$ satisfying $1\le a\le b\le N$. We'll use the following example with $N = 6$:

Naively, for every query, we can iterate through all entries from index $a$ to index $b$ to add them up. Since we have $Q$ queries and each query requires a maximum of $\mathcal{O}(N)$ operations to calculate the sum, our total time complexity is $\mathcal{O}(NQ)$. For most problems of this nature, the constraints will be $N, Q \leq 10^5$, so $NQ$ is on the order of $10^{10}$. This is not acceptable; it will almost certainly exceed the time limit.

Instead, we can use prefix sums to process these array sum queries. We designate a prefix sum array $\texttt{prefix}$. First, because we're 1-indexing the array, set $\texttt{prefix}[0]=0$, then for indices $k$ such that $1 \leq k \leq N$, define the prefix sum array as follows:

Basically, what this means is that the element at index $k$ of the prefix sum array stores the sum of all the elements in the original array from index $1$ up to $k$. This can be calculated easily in $\mathcal{O}(N)$ by the following formula for each $1\le k\le N$:

For the example case, our prefix sum array looks like this:

Now, when we want to query for the sum of the elements of $\texttt{arr}$ between (1-indexed) indices $a$ and $b$ inclusive, we can use the following formula:

Using our definition of the elements in the prefix sum array, we have

Since we are only querying two elements in the prefix sum array, we can calculate subarray sums in $\mathcal{O}(1)$ per query, which is much better than the $\mathcal{O}(N)$ per query that we had before. Now, after an $\mathcal{O}(N)$ preprocessing to calculate the prefix sum array, each of the $Q$ queries takes $\mathcal{O}(1)$ time. Thus, our total time complexity is $\mathcal{O}(N+Q)$, which should now pass the time limit.

Let's do an example query and find the subarray sum between indices $a = 2$ and $b = 5$, inclusive, in the 1-indexed $\texttt{arr}$. From looking at the original array, we see that this is

Using prefix sums:

These are also known as partial sums.

Solution - Static Range Sum

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#include <bits/stdc++.h>
using namespace std;

#define sz(x) (int)size(x)

using ll = long long;
using vl = vector<ll>;

vl psum(const vl &a) {
	vl psum(sz(a) + 1);

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import java.io.*;
import java.util.*;

public class Main {
	static int N, Q;
	public static void main(String[] args) throws IOException {
		BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter writer = new PrintWriter(System.out);
		StringTokenizer st = new StringTokenizer(reader.readLine());
		N = Integer.parseInt(st.nextToken());

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def psum(a):
	psum = [0]
	for i in a:
		psum.append(psum[-1] + i)
	return psum


N, Q = map(int, input().split())
a = list(map(int, input().split()))
p = psum(a)

for i in range(Q):
	l, r = map(int, input().split())
	print(p[r] - p[l])

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import itertools


def psum(a):
	return [0] + list(itertools.accumulate(a))


 Code Snippet: Same code as above (Click to expand)

Problems

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