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2364. Count Number of Bad Pairs
Description
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
.
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
Solutions
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class Solution { public long countBadPairs(int[] nums) { Map<Integer, Integer> cnt = new HashMap<>(); long ans = 0; for (int i = 0; i < nums.length; ++i) { int x = i - nums[i]; ans += i - cnt.getOrDefault(x, 0); cnt.merge(x, 1, Integer::sum); } return ans; } }
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class Solution { public: long long countBadPairs(vector<int>& nums) { unordered_map<int, int> cnt; long long ans = 0; for (int i = 0; i < nums.size(); ++i) { int x = i - nums[i]; ans += i - cnt[x]; ++cnt[x]; } return ans; } };
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class Solution: def countBadPairs(self, nums: List[int]) -> int: cnt = Counter() ans = 0 for i, x in enumerate(nums): ans += i - cnt[i - x] cnt[i - x] += 1 return ans
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func countBadPairs(nums []int) (ans int64) { cnt := map[int]int{} for i, x := range nums { x = i - x ans += int64(i - cnt[x]) cnt[x]++ } return }
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function countBadPairs(nums: number[]): number { const cnt = new Map<number, number>(); let ans = 0; for (let i = 0; i < nums.length; ++i) { const x = i - nums[i]; ans += i - (cnt.get(x) ?? 0); cnt.set(x, (cnt.get(x) ?? 0) + 1); } return ans; }