LeetCode 2364. Count Number of Bad Pairs Solution in Java, C++, Python & More | Explanation + Code

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 ≠ nums[j] - nums[i], then (i, j) is a bad pair.

We can transform the equation into i - nums[i] ≠ j - nums[j]. This suggests using a hash table cnt to count the occurrences of i - nums[i].

While iterating through the array, for the current element nums[i], we add i - cnt[i - nums[i]] to the answer, and 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 nums.

PythonJavaC++GoTypeScriptRust

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(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)

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; }(code-box)

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 } }(code-box)