README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/2300-2399/2364.Count%20Number%20of%20Bad%20Pairs/README_EN.md	1622	Biweekly Contest 84 Q2	Array Hash Table

2364. Count Number of Bad Pairs

中文文档

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

Solution 1: Equation Transformation + Hash Table

From the problem description, we know that for any $i < j$, if $j - i \neq nums[j] - nums[i]$, then $(i, j)$ is a bad pair.

We can transform the equation to $i - nums[i] \neq j - nums[j]$. This inspires us to use a hash table $cnt$ to count the occurrences of $i - nums[i]$.

We iterate through the array. For the current element $nums[i]$, we add $i - cnt[i - nums[i]]$ to the answer, then increment the count of $i - nums[i]$ by $1$.

Finally, we return the answer.

The time complexity is $O(n)$ and the space complexity is $O(n)$, where $n$ is the length of the array.

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}