Enable ruff E741 rule (#11370)

* Enable ruff E741 rule

* [pre-commit.ci] auto fixes from pre-commit.com hooks

for more information, see https://pre-commit.ci

---------

Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com>

data_structures/binary_tree/non_recursive_segment_tree.py

@@ -87,12 +87,12 @@ class SegmentTree(Generic[T]):
             p = p // 2
             self.st[p] = self.fn(self.st[p * 2], self.st[p * 2 + 1])
 
-    def query(self, l: int, r: int) -> T | None:
+    def query(self, left: int, right: int) -> T | None:
         """
         Get range query value in log(N) time
-        :param l: left element index
-        :param r: right element index
-        :return: element combined in the range [l, r]
+        :param left: left element index
+        :param right: right element index
+        :return: element combined in the range [left, right]
 
         >>> st = SegmentTree([1, 2, 3, 4], lambda a, b: a + b)
         >>> st.query(0, 2)
@@ -104,15 +104,15 @@ class SegmentTree(Generic[T]):
         >>> st.query(2, 3)
         7
         """
-        l, r = l + self.N, r + self.N
+        left, right = left + self.N, right + self.N
 
         res: T | None = None
-        while l <= r:
-            if l % 2 == 1:
-                res = self.st[l] if res is None else self.fn(res, self.st[l])
-            if r % 2 == 0:
-                res = self.st[r] if res is None else self.fn(res, self.st[r])
-            l, r = (l + 1) // 2, (r - 1) // 2
+        while left <= right:
+            if left % 2 == 1:
+                res = self.st[left] if res is None else self.fn(res, self.st[left])
+            if right % 2 == 0:
+                res = self.st[right] if res is None else self.fn(res, self.st[right])
+            left, right = (left + 1) // 2, (right - 1) // 2
         return res
 
 

data_structures/binary_tree/segment_tree.py

@@ -35,13 +35,13 @@ class SegmentTree:
         """
         return idx * 2 + 1
 
-    def build(self, idx, l, r):
-        if l == r:
-            self.st[idx] = self.A[l]
+    def build(self, idx, left, right):
+        if left == right:
+            self.st[idx] = self.A[left]
         else:
-            mid = (l + r) // 2
-            self.build(self.left(idx), l, mid)
-            self.build(self.right(idx), mid + 1, r)
+            mid = (left + right) // 2
+            self.build(self.left(idx), left, mid)
+            self.build(self.right(idx), mid + 1, right)
             self.st[idx] = max(self.st[self.left(idx)], self.st[self.right(idx)])
 
     def update(self, a, b, val):
@@ -56,18 +56,18 @@ class SegmentTree:
         """
         return self.update_recursive(1, 0, self.N - 1, a - 1, b - 1, val)
 
-    def update_recursive(self, idx, l, r, a, b, val):
+    def update_recursive(self, idx, left, right, a, b, val):
         """
         update(1, 1, N, a, b, v) for update val v to [a,b]
         """
-        if r < a or l > b:
+        if right < a or left > b:
             return True
-        if l == r:
+        if left == right:
             self.st[idx] = val
             return True
-        mid = (l + r) // 2
-        self.update_recursive(self.left(idx), l, mid, a, b, val)
-        self.update_recursive(self.right(idx), mid + 1, r, a, b, val)
+        mid = (left + right) // 2
+        self.update_recursive(self.left(idx), left, mid, a, b, val)
+        self.update_recursive(self.right(idx), mid + 1, right, a, b, val)
         self.st[idx] = max(self.st[self.left(idx)], self.st[self.right(idx)])
         return True
 
@@ -83,17 +83,17 @@ class SegmentTree:
         """
         return self.query_recursive(1, 0, self.N - 1, a - 1, b - 1)
 
-    def query_recursive(self, idx, l, r, a, b):
+    def query_recursive(self, idx, left, right, a, b):
         """
         query(1, 1, N, a, b) for query max of [a,b]
         """
-        if r < a or l > b:
+        if right < a or left > b:
             return -math.inf
-        if l >= a and r <= b:
+        if left >= a and right <= b:
             return self.st[idx]
-        mid = (l + r) // 2
-        q1 = self.query_recursive(self.left(idx), l, mid, a, b)
-        q2 = self.query_recursive(self.right(idx), mid + 1, r, a, b)
+        mid = (left + right) // 2
+        q1 = self.query_recursive(self.left(idx), left, mid, a, b)
+        q2 = self.query_recursive(self.right(idx), mid + 1, right, a, b)
         return max(q1, q2)
 
     def show_data(self):

data_structures/heap/min_heap.py

@@ -66,14 +66,14 @@ class MinHeap:
     # this is min-heapify method
     def sift_down(self, idx, array):
         while True:
-            l = self.get_left_child_idx(idx)
-            r = self.get_right_child_idx(idx)
+            left = self.get_left_child_idx(idx)
+            right = self.get_right_child_idx(idx)
 
             smallest = idx
-            if l < len(array) and array[l] < array[idx]:
-                smallest = l
-            if r < len(array) and array[r] < array[smallest]:
-                smallest = r
+            if left < len(array) and array[left] < array[idx]:
+                smallest = left
+            if right < len(array) and array[right] < array[smallest]:
+                smallest = right
 
             if smallest != idx:
                 array[idx], array[smallest] = array[smallest], array[idx]

dynamic_programming/longest_common_subsequence.py

@@ -38,30 +38,30 @@ def longest_common_subsequence(x: str, y: str):
     n = len(y)
 
     # declaring the array for storing the dp values
-    l = [[0] * (n + 1) for _ in range(m + 1)]
+    dp = [[0] * (n + 1) for _ in range(m + 1)]
 
     for i in range(1, m + 1):
         for j in range(1, n + 1):
             match = 1 if x[i - 1] == y[j - 1] else 0
 
-            l[i][j] = max(l[i - 1][j], l[i][j - 1], l[i - 1][j - 1] + match)
+            dp[i][j] = max(dp[i - 1][j], dp[i][j - 1], dp[i - 1][j - 1] + match)
 
     seq = ""
     i, j = m, n
     while i > 0 and j > 0:
         match = 1 if x[i - 1] == y[j - 1] else 0
 
-        if l[i][j] == l[i - 1][j - 1] + match:
+        if dp[i][j] == dp[i - 1][j - 1] + match:
             if match == 1:
                 seq = x[i - 1] + seq
             i -= 1
             j -= 1
-        elif l[i][j] == l[i - 1][j]:
+        elif dp[i][j] == dp[i - 1][j]:
             i -= 1
         else:
             j -= 1
 
-    return l[m][n], seq
+    return dp[m][n], seq
 
 
 if __name__ == "__main__":

dynamic_programming/longest_increasing_subsequence_o_nlogn.py

@@ -7,14 +7,14 @@
 from __future__ import annotations
 
 
-def ceil_index(v, l, r, key):
-    while r - l > 1:
-        m = (l + r) // 2
-        if v[m] >= key:
-            r = m
+def ceil_index(v, left, right, key):
+    while right - left > 1:
+        middle = (left + right) // 2
+        if v[middle] >= key:
+            right = middle
         else:
-            l = m
-    return r
+            left = middle
+    return right
 
 
 def longest_increasing_subsequence_length(v: list[int]) -> int:

graphs/articulation_points.py

@@ -1,6 +1,6 @@
 # Finding Articulation Points in Undirected Graph
-def compute_ap(l):
-    n = len(l)
+def compute_ap(graph):
+    n = len(graph)
     out_edge_count = 0
     low = [0] * n
     visited = [False] * n
@@ -12,7 +12,7 @@ def compute_ap(l):
         visited[at] = True
         low[at] = at
 
-        for to in l[at]:
+        for to in graph[at]:
             if to == parent:
                 pass
             elif not visited[to]:
@@ -41,7 +41,7 @@ def compute_ap(l):
 
 
 # Adjacency list of graph
-data = {
+graph = {
     0: [1, 2],
     1: [0, 2],
     2: [0, 1, 3, 5],
@@ -52,4 +52,4 @@ data = {
     7: [6, 8],
     8: [5, 7],
 }
-compute_ap(data)
+compute_ap(graph)

graphs/dinic.py

@@ -37,7 +37,7 @@ class Dinic:
     # Here we calculate the flow that reaches the sink
     def max_flow(self, source, sink):
         flow, self.q[0] = 0, source
-        for l in range(31):  # l = 30 maybe faster for random data
+        for l in range(31):  # l = 30 maybe faster for random data  # noqa: E741
             while True:
                 self.lvl, self.ptr = [0] * len(self.q), [0] * len(self.q)
                 qi, qe, self.lvl[source] = 0, 1, 1

machine_learning/sequential_minimum_optimization.py

@@ -309,9 +309,9 @@ class SmoSVM:
         # calculate L and H  which bound the new alpha2
         s = y1 * y2
         if s == -1:
-            l, h = max(0.0, a2 - a1), min(self._c, self._c + a2 - a1)
+            l, h = max(0.0, a2 - a1), min(self._c, self._c + a2 - a1)  # noqa: E741
         else:
-            l, h = max(0.0, a2 + a1 - self._c), min(self._c, a2 + a1)
+            l, h = max(0.0, a2 + a1 - self._c), min(self._c, a2 + a1)  # noqa: E741
         if l == h:
             return None, None
 

maths/pi_generator.py

@@ -41,7 +41,7 @@ def calculate_pi(limit: int) -> str:
     t = 1
     k = 1
     n = 3
-    l = 3
+    m = 3
 
     decimal = limit
     counter = 0
@@ -65,11 +65,11 @@ def calculate_pi(limit: int) -> str:
             q *= 10
             r = nr
         else:
-            nr = (2 * q + r) * l
-            nn = (q * (7 * k) + 2 + (r * l)) // (t * l)
+            nr = (2 * q + r) * m
+            nn = (q * (7 * k) + 2 + (r * m)) // (t * m)
             q *= k
-            t *= l
-            l += 2
+            t *= m
+            m += 2
             k += 1
             n = nn
             r = nr

other/sdes.py

@@ -44,11 +44,11 @@ def function(expansion, s0, s1, key, message):
     right = message[4:]
     temp = apply_table(right, expansion)
     temp = xor(temp, key)
-    l = apply_sbox(s0, temp[:4])
-    r = apply_sbox(s1, temp[4:])
-    l = "0" * (2 - len(l)) + l
-    r = "0" * (2 - len(r)) + r
-    temp = apply_table(l + r, p4_table)
+    left_bin_str = apply_sbox(s0, temp[:4])
+    right_bin_str = apply_sbox(s1, temp[4:])
+    left_bin_str = "0" * (2 - len(left_bin_str)) + left_bin_str
+    right_bin_str = "0" * (2 - len(right_bin_str)) + right_bin_str
+    temp = apply_table(left_bin_str + right_bin_str, p4_table)
     temp = xor(left, temp)
     return temp + right
 

project_euler/problem_011/sol2.py

@@ -35,37 +35,47 @@ def solution():
     70600674
     """
     with open(os.path.dirname(__file__) + "/grid.txt") as f:
-        l = []
+        grid = []
         for _ in range(20):
-            l.append([int(x) for x in f.readline().split()])
+            grid.append([int(x) for x in f.readline().split()])
 
         maximum = 0
 
         # right
         for i in range(20):
             for j in range(17):
-                temp = l[i][j] * l[i][j + 1] * l[i][j + 2] * l[i][j + 3]
+                temp = grid[i][j] * grid[i][j + 1] * grid[i][j + 2] * grid[i][j + 3]
                 if temp > maximum:
                     maximum = temp
 
         # down
         for i in range(17):
             for j in range(20):
-                temp = l[i][j] * l[i + 1][j] * l[i + 2][j] * l[i + 3][j]
+                temp = grid[i][j] * grid[i + 1][j] * grid[i + 2][j] * grid[i + 3][j]
                 if temp > maximum:
                     maximum = temp
 
         # diagonal 1
         for i in range(17):
             for j in range(17):
-                temp = l[i][j] * l[i + 1][j + 1] * l[i + 2][j + 2] * l[i + 3][j + 3]
+                temp = (
+                    grid[i][j]
+                    * grid[i + 1][j + 1]
+                    * grid[i + 2][j + 2]
+                    * grid[i + 3][j + 3]
+                )
                 if temp > maximum:
                     maximum = temp
 
         # diagonal 2
         for i in range(17):
             for j in range(3, 20):
-                temp = l[i][j] * l[i + 1][j - 1] * l[i + 2][j - 2] * l[i + 3][j - 3]
+                temp = (
+                    grid[i][j]
+                    * grid[i + 1][j - 1]
+                    * grid[i + 2][j - 2]
+                    * grid[i + 3][j - 3]
+                )
                 if temp > maximum:
                     maximum = temp
         return maximum

pyproject.toml

@@ -2,7 +2,6 @@
 lint.ignore = [    # `ruff rule S101` for a description of that rule
   "B904",     # Within an `except` clause, raise exceptions with `raise ... from err` -- FIX ME
   "B905",     # `zip()` without an explicit `strict=` parameter -- FIX ME
-  "E741",     # Ambiguous variable name 'l' -- FIX ME
   "EM101",    # Exception must not use a string literal, assign to variable first
   "EXE001",   # Shebang is present but file is not executable -- DO NOT FIX
   "G004",     # Logging statement uses f-string

strings/jaro_winkler.py

@@ -28,12 +28,12 @@ def jaro_winkler(str1: str, str2: str) -> float:
     def get_matched_characters(_str1: str, _str2: str) -> str:
         matched = []
         limit = min(len(_str1), len(_str2)) // 2
-        for i, l in enumerate(_str1):
+        for i, char in enumerate(_str1):
             left = int(max(0, i - limit))
             right = int(min(i + limit + 1, len(_str2)))
-            if l in _str2[left:right]:
-                matched.append(l)
-                _str2 = f"{_str2[0:_str2.index(l)]} {_str2[_str2.index(l) + 1:]}"
+            if char in _str2[left:right]:
+                matched.append(char)
+                _str2 = f"{_str2[0:_str2.index(char)]} {_str2[_str2.index(char) + 1:]}"
 
         return "".join(matched)
 

strings/manacher.py

@@ -9,9 +9,9 @@ def palindromic_string(input_string: str) -> str:
 
     1. first this convert input_string("xyx") into new_string("x|y|x") where odd
         positions are actual input characters.
-    2. for each character in new_string it find corresponding length and store the
-        length and l,r to store previously calculated info.(please look the explanation
-        for details)
+    2. for each character in new_string it find corresponding length and
+        store the length and left,right to store previously calculated info.
+        (please look the explanation for details)
 
     3. return corresponding output_string by removing all "|"
     """
@@ -29,7 +29,7 @@ def palindromic_string(input_string: str) -> str:
 
     # we will store the starting and ending of previous furthest ending palindromic
     # substring
-    l, r = 0, 0
+    left, right = 0, 0
 
     # length[i] shows the length of palindromic substring with center i
     length = [1 for i in range(len(new_input_string))]
@@ -37,7 +37,7 @@ def palindromic_string(input_string: str) -> str:
     # for each character in new_string find corresponding palindromic string
     start = 0
     for j in range(len(new_input_string)):
-        k = 1 if j > r else min(length[l + r - j] // 2, r - j + 1)
+        k = 1 if j > right else min(length[left + right - j] // 2, right - j + 1)
         while (
             j - k >= 0
             and j + k < len(new_input_string)
@@ -47,11 +47,11 @@ def palindromic_string(input_string: str) -> str:
 
         length[j] = 2 * k - 1
 
-        # does this string is ending after the previously explored end (that is r) ?
-        # if yes the update the new r to the last index of this
-        if j + k - 1 > r:
-            l = j - k + 1
-            r = j + k - 1
+        # does this string is ending after the previously explored end (that is right) ?
+        # if yes the update the new right to the last index of this
+        if j + k - 1 > right:
+            left = j - k + 1
+            right = j + k - 1
 
         # update max_length and start position
         if max_length < length[j]:
@@ -78,8 +78,9 @@ if __name__ == "__main__":
 consider the string for which we are calculating the longest palindromic substring is
 shown above where ... are some characters in between and right now we are calculating
 the length of palindromic substring with center at a5 with following conditions :
-i) we have stored the length of palindromic substring which has center at a3 (starts at
-    l ends at r) and it is the furthest ending till now, and it has ending after a6
+i) we have stored the length of palindromic substring which has center at a3
+    (starts at left ends at right) and it is the furthest ending till now,
+    and it has ending after a6
 ii) a2 and a4 are equally distant from a3 so char(a2) == char(a4)
 iii) a0 and a6 are equally distant from a3 so char(a0) == char(a6)
 iv) a1 is corresponding equal character of a5 in palindrome with center a3 (remember
@@ -98,11 +99,11 @@ so we can say that palindrome at center a5 is at least as long as palindrome at
 a1 but this only holds if a0 and a6 are inside the limits of palindrome centered at a3
 so finally ..
 
-len_of_palindrome__at(a5) = min(len_of_palindrome_at(a1), r-a5)
-where a3 lies from l to r and we have to keep updating that
+len_of_palindrome__at(a5) = min(len_of_palindrome_at(a1), right-a5)
+where a3 lies from left to right and we have to keep updating that
 
-and if the a5 lies outside of l,r boundary we calculate length of palindrome with
-bruteforce and update l,r.
+and if the a5 lies outside of left,right boundary we calculate length of palindrome with
+bruteforce and update left,right.
 
 it gives the linear time complexity just like z-function
 """