dynamic_programming.range_sum_query

Author: Sanjay Muthu <https://github.com/XenoBytesX>

This is an implementation of the Dynamic Programming solution to the Range Sum Query.

The problem statement is:

Given an array and q queries, each query stating you to find the sum of elements from l to r (inclusive)

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 + 1) time ≈ O(N) time

If we use prefix sums we can find it in O(1) 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]

Functions

prefix_sum(→ list[int])

Module Contents

dynamic_programming.range_sum_query.prefix_sum(array: list[int], queries: list[tuple[int, int]]) → list[int]
>>> 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]