add(coding/2364): count-number-of-bad-pairs

source/coding/2364-count-number-of-bad-pairs.md

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+---
+title: 2364. Count Number of Bad Pairs
+notebook: coding
+tags:
+- medium
+date: 2025-02-09 14:15:31
+updated: 2025-02-09 14:15:31
+---
+## Problem
+
+You are given a **0-indexed** integer array `nums`. A pair of indices `(i, j)` is a **bad pair** if `i < j` and `j - i != nums[j] - nums[i]`.
+
+Return _the total number of **bad pairs** in_ `nums`.
+
+<https://leetcode.com/problems/count-number-of-bad-pairs/>
+
+**Example 1:**
+
+> Input: `nums = [4,1,3,3]`
+> Output: `5`
+> Explanation: The pair `(0, 1)` is a bad pair since `1 - 0 != 1 - 4`.
+> The pair `(0, 2)` is a bad pair since `2 - 0 != 3 - 4`, `2 != -1`.
+> The pair `(0, 3)` is a bad pair since `3 - 0 != 3 - 4`, `3 != -1`.
+> The pair `(1, 2)` is a bad pair since `2 - 1 != 3 - 1`, `1 != 2`.
+> The pair `(2, 3)` is a bad pair since `3 - 2 != 3 - 3`, `1 != 0`.
+> There are a total of 5 bad pairs, so we return 5.
+
+**Example 2:**
+
+> Input: `nums = [1,2,3,4,5]`
+> Output: `0`
+> Explanation: There are no bad pairs.
+
+**Constraints:**
+
+- `1 <= nums.length <= 10⁵`
+- `1 <= nums[i] <= 10⁹`
+
+## Test Cases
+
+``` python
+class Solution:
+    def countBadPairs(self, nums: List[int]) -> int:
+```
+
+{% asset_code coding/2364-count-number-of-bad-pairs/solution_test.py %}
+
+## Thoughts
+
+思路跟 [1014. Best Sightseeing Pair](1014-best-sightseeing-pair) 有点儿像。先把 `j - i != nums[j] - nums[i]` 改写成 `nums[i] - i != nums[j] - j`，把 i 和 j 的耦合打散。
+
+然后为了计算不相等 pair 的数量，可以用总数减去相等的数量。
+
+Pair 总数显然为 `n * (n - 1) / 2`。如果有 m 个位置的 `nums[i] - i` 彼此相等，那么他们贡献了 `m * (m - 1) / 2` 个相等的 pair。
+
+用字典记录 `nums[i] - i` 的每个不同结果对应的 i 的数量，从总数中减去所有相等的 pair 数量，所得记为题目结果。
+
+时间复杂度 `O(n)`，空间复杂度 `O(n)`。
+
+## Code
+
+{% asset_code coding/2364-count-number-of-bad-pairs/solution.py %}

source/coding/2364-count-number-of-bad-pairs/solution.py

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+from collections import defaultdict
+
+
+class Solution:
+    def countBadPairs(self, nums: list[int]) -> int:
+        counts: dict[int, int] = defaultdict(int) # nums[i] - i -> count
+        for i, num in enumerate(nums):
+            counts[num - i] += 1
+
+        n = len(nums)
+        total = n * (n - 1) // 2
+        for count in counts.values():
+            total -= count * (count - 1) // 2
+
+        return total

source/coding/2364-count-number-of-bad-pairs/solution_test.py

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+import pytest
+
+from solution import Solution
+
+
+@pytest.mark.parametrize('nums, expected', [
+    ([4,1,3,3], 5),
+    ([1,2,3,4,5], 0),
+])
+@pytest.mark.parametrize('sol', [Solution()])
+def test_solution(sol, nums, expected):
+    assert sol.countBadPairs(nums) == expected