From f4fe48bf621c9eeb3c69d0d69e84bdb291503dd8 Mon Sep 17 00:00:00 2001
From: Sanjay Muthu <sanjaymuthu12@gmail.com>
Date: Mon, 24 Feb 2025 20:00:08 +0530
Subject: [PATCH 1/2] Create prefix_sum.py

---
 dynamic_programming/prefix_sum.py | 90 +++++++++++++++++++++++++++++++
 1 file changed, 90 insertions(+)
 create mode 100644 dynamic_programming/prefix_sum.py

diff --git a/dynamic_programming/prefix_sum.py b/dynamic_programming/prefix_sum.py
new file mode 100644
index 000000000000..ccf7ec6987d2
--- /dev/null
+++ b/dynamic_programming/prefix_sum.py
@@ -0,0 +1,90 @@
+"""
+Author  : Sanjay Muthu <https://github.com/XenoBytesX>
+
+This is an implementation of the Dynamic Programming solution to the Range Sum Query problem.
+
+The problem statement is:
+    Given an array and q queries,
+    each query stating you to find the sum of elements l to r (inclusive) (l and r are given in the query)
+
+Example:
+    arr = [1, 4, 6, 2, 61, 12]
+    queries = 3
+    l_1 = 2, r_1 = 5
+    l_2 = 1, r_2 = 5
+    l_3 = 3, r_3 = 4
+
+    as input will return
+
+    [81, 85, 63]
+
+    as output
+
+    [0-indexing]
+
+NOTE:- 0-indexing means the indexing of the array starts from 0
+Example:- a = [1, 2, 3, 4, 5, 6]
+          Here, the 0th index of a is 1,
+                the 1st index of a is 2,
+                and so forth
+
+Time Complexity:- O(N + Q)
+-- O(N) pre-calculation time to calculate the prefix sum array
+-- and O(1) time per each query = O(1 * Q) = O(Q) time
+
+Space Complexity:- O(N)
+-- O(N) to store the prefix sum
+
+Algorithm:-
+So first we calculate the prefix sum (dp) of the array.
+The prefix sum of the index i is the sum of all elements indexed from 0 to i (inclusive)
+The prefix sum of the index i is the prefix sum of index i-1 + the current element.
+So, The state of the dp is dp[i] = dp[i-1] + a[i];
+
+After we calculate the prefix sum,
+For each query [l, r]
+The answer is dp[r]-dp[l-1] ( we need to be careful because l might be 0 )
+For example take this array:-
+    [4, 2, 1, 6, 3]
+The prefix sum calculated for this array would be:-
+    [4, 4+2, 4+2+1, 4+2+1+6, 4+2+1+6+3]
+==> [4, 6, 7, 13, 16]
+If the query was l=3, r=4,
+The answer would be 6+3 = 9 but this would require O(r-l) time ≈ O(N) time
+If we use prefix sums we can find it in O(1) time by using the formula prefix[r]-prefix[l-1]
+This formula works because prefix[r] is the sum of elements from [0, r] and prefix[l-1] is the sum of elements from [0, l-1],
+so if we do prefix[r]-prefix[l-1] it will be [0, r] - [0, l-1] = [0, l-1] + [l, r] - [0, l-1] = [l, r]
+"""
+
+from __future__ import annotations
+
+
+def prefix_sum(array: list[int], queries: list[tuple[int, int]]) -> list[int]:
+    """
+    Some examples
+    >>> prefix_sum([1, 4, 6, 2, 61, 12], [(2, 5), (1, 5), (3, 4)])
+    [81, 85, 63]
+    >>> prefix_sum([4, 2, 1, 6, 3], [(3, 4), (1, 3), (0, 2)])
+    [9, 9, 7]
+    """
+    # The prefix sum array
+    dp = [0] * len(array)
+    dp[0] = array[0]
+    for i in range(1, len(array)):
+        dp[i] = dp[i-1] + array[i]
+
+    # Read Algorithm section (Line 38)
+    result = []
+    for query in queries:
+        res = dp[query[1]]
+        if query[0] != 0:
+            res -= dp[query[0]-1]
+        result.append(res)
+
+    return result
+
+
+if __name__ == "__main__":
+    import doctest
+
+    doctest.testmod()

From 463e6aaece9962d7b9f934082e03776d2c3c75fe Mon Sep 17 00:00:00 2001
From: "pre-commit-ci[bot]"
 <66853113+pre-commit-ci[bot]@users.noreply.github.com>
Date: Mon, 24 Feb 2025 14:31:08 +0000
Subject: [PATCH 2/2] [pre-commit.ci] auto fixes from pre-commit.com hooks

for more information, see https://pre-commit.ci
---
 dynamic_programming/prefix_sum.py | 4 ++--
 1 file changed, 2 insertions(+), 2 deletions(-)

diff --git a/dynamic_programming/prefix_sum.py b/dynamic_programming/prefix_sum.py
index ccf7ec6987d2..d2dc0f0394b4 100644
--- a/dynamic_programming/prefix_sum.py
+++ b/dynamic_programming/prefix_sum.py
@@ -71,14 +71,14 @@ def prefix_sum(array: list[int], queries: list[tuple[int, int]]) -> list[int]:
     dp = [0] * len(array)
     dp[0] = array[0]
     for i in range(1, len(array)):
-        dp[i] = dp[i-1] + array[i]
+        dp[i] = dp[i - 1] + array[i]
 
     # Read Algorithm section (Line 38)
     result = []
     for query in queries:
         res = dp[query[1]]
         if query[0] != 0:
-            res -= dp[query[0]-1]
+            res -= dp[query[0] - 1]
         result.append(res)
 
     return result