math + observation

Author: A-stick-bug

3495. Minimum Operations to Make Array Elements Zero.py

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+# https://leetcode.com/problems/minimum-operations-to-make-array-elements-zero
+# Math + observation
+# The number of divisions by 4 it takes to get a number to zero is floor(log4(n)) + 1
+# Observation: you will always be able to do 2 simultaneous operations unless the sum is odd
+# This is the case since the log4 representation goes up by at most 1 each time
+
+from itertools import accumulate
+from collections import Counter
+
+MN = 16  # log4(10^9)
+p4 = list(accumulate([0] + [3 * 4 ** i for i in range(MN)]))
+covers = [[p4[i] + 1, p4[i + 1]] for i in range(MN)]  # range of floor values
+
+
+# print(covers)
+
+def minOperations(queries: list[list[int]]) -> int:
+    def get_prefix_count(mx):
+        freq = Counter()
+        for i in range(MN):
+            l, r = covers[i]
+            if mx >= r:
+                freq[i + 1] += r - l + 1  # fully covered
+            else:
+                freq[i + 1] += mx - l + 1  # partially covered
+                break
+        return freq
+
+    def get_range_count(l, r):
+        return get_prefix_count(r) - get_prefix_count(l - 1)
+
+    ans = []
+    for l, r in queries:
+        freq = get_range_count(l, r)
+        # (number of operations) * (frequency), use ceil division
+        ans.append((sum(k * v for k, v in freq.items()) + 1) // 2)
+
+    # print(ans)
+    return sum(ans)
+
+
+print(minOperations(queries=[[1, 2], [2, 4], [2, 6]]))  # 1,2,4
+
+# # code used to make observations
+# def count_moves(n):
+#     c = 0
+#     while n > 0:
+#         n = n // 4
+#         c += 1
+#     return c
+#
+#
+# res = []
+# for i in range(1, 100):
+#     from math import log, floor
+#
+#     print(i, count_moves(i), floor(log(i, 4)) + 1)
+#     res.append(count_moves(i))
+#
+# from collections import Counter
+#
+# print(Counter(res))  # frequencies: [3, 12, 48, ..., 3*4^x]