From e28d832f2605b63b3d0caacb9c62fd235031ba7d Mon Sep 17 00:00:00 2001
From: Michael Zegar <85312805+3centroids@users.noreply.github.com>
Date: Wed, 24 Jan 2024 09:54:20 +0100
Subject: [PATCH] Update sqrt_decomposition.md
the sentence sounded unnatural
---
src/data_structures/sqrt_decomposition.md | 2 +-
1 file changed, 1 insertion(+), 1 deletion(-)
diff --git a/src/data_structures/sqrt_decomposition.md b/src/data_structures/sqrt_decomposition.md
index 7362733f7..ff9c3e351 100644
--- a/src/data_structures/sqrt_decomposition.md
+++ b/src/data_structures/sqrt_decomposition.md
@@ -106,7 +106,7 @@ Sqrt decomposition can be applied in a similar way to a whole class of other pro
Another class of problems appears when we need to **update array elements on intervals**: increment existing elements or replace them with a given value.
-For example, let's say we can do two types of operations on an array: add a given value $\delta$ to all array elements on interval $[l, r]$ or query the value of element $a[i]$. Let's store the value which has to be added to all elements of block $k$ in $b[k]$ (initially all $b[k] = 0$). During each "add" operation we need to add $\delta$ to $b[k]$ for all blocks which belong to interval $[l, r]$ and to add $\delta$ to $a[i]$ for all elements which belong to the "tails" of the interval. The answer a query $i$ is simply $a[i] + b[i/s]$. This way "add" operation has $O(\sqrt{n})$ complexity, and answering a query has $O(1)$ complexity.
+For example, let's say we can do two types of operations on an array: add a given value $\delta$ to all array elements on interval $[l, r]$ or query the value of element $a[i]$. Let's store the value which has to be added to all elements of block $k$ in $b[k]$ (initially all $b[k] = 0$). During each "add" operation we need to add $\delta$ to $b[k]$ for all blocks which belong to interval $[l, r]$ and to add $\delta$ to $a[i]$ for all elements which belong to the "tails" of the interval. The answer to query $i$ is simply $a[i] + b[i/s]$. This way "add" operation has $O(\sqrt{n})$ complexity, and answering a query has $O(1)$ complexity.
Finally, those two classes of problems can be combined if the task requires doing **both** element updates on an interval and queries on an interval. Both operations can be done with $O(\sqrt{n})$ complexity. This will require two block arrays $b$ and $c$: one to keep track of element updates and another to keep track of answers to the query.
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