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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

Solution 1: Equation Transformation + Hash Table

According to the problem description, for any \(i \lt j\), if \(j - i \neq \textit{nums}[j] - \textit{nums}[i]\), then \((i, j)\) is a bad pair.

We can transform the equation into \(i - \textit{nums}[i] \neq j - \textit{nums}[j]\). This suggests using a hash table \(cnt\) to count the occurrences of \(i - \textit{nums}[i]\).

While iterating through the array, for the current element \(\textit{nums}[i]\), we add \(i - cnt[i - \textit{nums}[i]]\) to the answer, and then increment the count of \(i - \textit{nums}[i]\) 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 \(\textit{nums}\).

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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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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.merge(x, 1, Integer::sum) + 1;
        }
        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]++;
        }
        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;
}

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use std::collections::HashMap;

impl Solution {
    pub fn count_bad_pairs(nums: Vec<i32>) -> i64 {
        let mut cnt: HashMap<i32, i64> = HashMap::new();
        let mut ans: i64 = 0;
        for (i, &num) in nums.iter().enumerate() {
            let x = i as i32 - num;
            let count = *cnt.get(&x).unwrap_or(&0);
            ans += i as i64 - count;
            *cnt.entry(x).or_insert(0) += 1;
        }
        ans
    }
}

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