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Commit f528ce3

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XenoBytesXpre-commit-ci[bot]MaximSmolskiy
authored
Added dynamic_programming/range_sum_query.py (TheAlgorithms#12592)
* Create prefix_sum.py * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci * Fix pre-commit and ruff errors * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci * Rename prefix_sum.py to range_sum_query.py * Refactor description * Fix * Refactor code * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci * Fix --------- Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com> Co-authored-by: Maxim Smolskiy <mithridatus@mail.ru>
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"""
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Author: Sanjay Muthu <https://github.com/XenoBytesX>
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This is an implementation of the Dynamic Programming solution to the Range Sum Query.
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The problem statement is:
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Given an array and q queries,
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each query stating you to find the sum of elements from l to r (inclusive)
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Example:
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arr = [1, 4, 6, 2, 61, 12]
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queries = 3
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l_1 = 2, r_1 = 5
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l_2 = 1, r_2 = 5
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l_3 = 3, r_3 = 4
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as input will return
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[81, 85, 63]
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as output
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0-indexing:
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NOTE: 0-indexing means the indexing of the array starts from 0
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Example: a = [1, 2, 3, 4, 5, 6]
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Here, the 0th index of a is 1,
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the 1st index of a is 2,
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and so forth
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Time Complexity: O(N + Q)
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* O(N) pre-calculation time to calculate the prefix sum array
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* and O(1) time per each query = O(1 * Q) = O(Q) time
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Space Complexity: O(N)
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* O(N) to store the prefix sum
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Algorithm:
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So, first we calculate the prefix sum (dp) of the array.
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The prefix sum of the index i is the sum of all elements indexed
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from 0 to i (inclusive).
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The prefix sum of the index i is the prefix sum of index (i - 1) + the current element.
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So, the state of the dp is dp[i] = dp[i - 1] + a[i].
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After we calculate the prefix sum,
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for each query [l, r]
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the answer is dp[r] - dp[l - 1] (we need to be careful because l might be 0).
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For example take this array:
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[4, 2, 1, 6, 3]
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The prefix sum calculated for this array would be:
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[4, 4 + 2, 4 + 2 + 1, 4 + 2 + 1 + 6, 4 + 2 + 1 + 6 + 3]
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==> [4, 6, 7, 13, 16]
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If the query was l = 3, r = 4,
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the answer would be 6 + 3 = 9 but this would require O(r - l + 1) time ≈ O(N) time
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If we use prefix sums we can find it in O(1) by using the formula
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prefix[r] - prefix[l - 1].
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This formula works because prefix[r] is the sum of elements from [0, r]
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and prefix[l - 1] is the sum of elements from [0, l - 1],
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so if we do prefix[r] - prefix[l - 1] it will be
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[0, r] - [0, l - 1] = [0, l - 1] + [l, r] - [0, l - 1] = [l, r]
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"""
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def prefix_sum(array: list[int], queries: list[tuple[int, int]]) -> list[int]:
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"""
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>>> prefix_sum([1, 4, 6, 2, 61, 12], [(2, 5), (1, 5), (3, 4)])
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[81, 85, 63]
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>>> prefix_sum([4, 2, 1, 6, 3], [(3, 4), (1, 3), (0, 2)])
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[9, 9, 7]
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"""
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# The prefix sum array
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dp = [0] * len(array)
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dp[0] = array[0]
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for i in range(1, len(array)):
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dp[i] = dp[i - 1] + array[i]
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# See Algorithm section (Line 44)
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result = []
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for query in queries:
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left, right = query
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res = dp[right]
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if left > 0:
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res -= dp[left - 1]
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result.append(res)
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return result
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if __name__ == "__main__":
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import doctest
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doctest.testmod()

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