README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/2000-2099/2006.Count%20Number%20of%20Pairs%20With%20Absolute%20Difference%20K/README_EN.md	1271	Biweekly Contest 61 Q1	Array Hash Table Counting

2006. Count Number of Pairs With Absolute Difference K

中文文档

Description

Given an integer array nums and an integer k, return the number of pairs (i, j) where i < j such that |nums[i] - nums[j]| == k.

The value of |x| is defined as:

• x if x >= 0.
• -x if x < 0.

 

Example 1:

Input: nums = [1,2,2,1], k = 1
Output: 4
Explanation: The pairs with an absolute difference of 1 are:
- [1,2,2,1]
- [1,2,2,1]
- [1,2,2,1]
- [1,2,2,1]

Example 2:

Input: nums = [1,3], k = 3
Output: 0
Explanation: There are no pairs with an absolute difference of 3.

Example 3:

Input: nums = [3,2,1,5,4], k = 2
Output: 3
Explanation: The pairs with an absolute difference of 2 are:
- [3,2,1,5,4]
- [3,2,1,5,4]
- [3,2,1,5,4]

 

Constraints:

• 1 <= nums.length <= 200
• 1 <= nums[i] <= 100
• 1 <= k <= 99

Solutions

Solution 1: Brute Force Enumeration

We notice that the length of the array $nums$ does not exceed $200$, so we can enumerate all pairs $(i, j)$, where $i < j$, and check if $|nums[i] - nums[j]|$ equals $k$. If it does, we increment the answer by one.

Finally, we return the answer.

The time complexity is $O(n^2)$, and the space complexity is $O(1)$. Here, $n$ is the length of the array $nums$.

Python3

class Solution:
    def countKDifference(self, nums: List[int], k: int) -> int:
        n = len(nums)
        return sum(
            abs(nums[i] - nums[j]) == k for i in range(n) for j in range(i + 1, n)
        )

Java

class Solution {
    public int countKDifference(int[] nums, int k) {
        int ans = 0;
        for (int i = 0, n = nums.length; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                if (Math.abs(nums[i] - nums[j]) == k) {
                    ++ans;
                }
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int countKDifference(vector<int>& nums, int k) {
        int n = nums.size();
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                ans += abs(nums[i] - nums[j]) == k;
            }
        }
        return ans;
    }
};

Go

func countKDifference(nums []int, k int) int {
	n := len(nums)
	ans := 0
	for i := 0; i < n; i++ {
		for j := i + 1; j < n; j++ {
			if abs(nums[i]-nums[j]) == k {
				ans++
			}
		}
	}
	return ans
}

func abs(x int) int {
	if x > 0 {
		return x
	}
	return -x
}

TypeScript

function countKDifference(nums: number[], k: number): number {
    let ans = 0;
    let cnt = new Map();
    for (let num of nums) {
        ans += (cnt.get(num - k) || 0) + (cnt.get(num + k) || 0);
        cnt.set(num, (cnt.get(num) || 0) + 1);
    }
    return ans;
}

Rust

impl Solution {
    pub fn count_k_difference(nums: Vec<i32>, k: i32) -> i32 {
        let mut res = 0;
        let n = nums.len();
        for i in 0..n - 1 {
            for j in i..n {
                if (nums[i] - nums[j]).abs() == k {
                    res += 1;
                }
            }
        }
        res
    }
}

Solution 2: Hash Table or Array

We can use a hash table or array to record the occurrence count of each number in the array $nums$. Then, we enumerate each number $x$ in the array $nums$, and check if $x + k$ and $x - k$ are in the array $nums$. If they are, we increment the answer by the sum of the occurrence counts of $x + k$ and $x - k$.

Finally, we return the answer.

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

Python3

class Solution:
    def countKDifference(self, nums: List[int], k: int) -> int:
        ans = 0
        cnt = Counter()
        for num in nums:
            ans += cnt[num - k] + cnt[num + k]
            cnt[num] += 1
        return ans

Java

class Solution {
    public int countKDifference(int[] nums, int k) {
        int ans = 0;
        int[] cnt = new int[110];
        for (int num : nums) {
            if (num >= k) {
                ans += cnt[num - k];
            }
            if (num + k <= 100) {
                ans += cnt[num + k];
            }
            ++cnt[num];
        }
        return ans;
    }
}

C++

class Solution {
public:
    int countKDifference(vector<int>& nums, int k) {
        int ans = 0;
        int cnt[110]{};
        for (int num : nums) {
            if (num >= k) {
                ans += cnt[num - k];
            }
            if (num + k <= 100) {
                ans += cnt[num + k];
            }
            ++cnt[num];
        }
        return ans;
    }
};

Go

func countKDifference(nums []int, k int) (ans int) {
	cnt := [110]int{}
	for _, num := range nums {
		if num >= k {
			ans += cnt[num-k]
		}
		if num+k <= 100 {
			ans += cnt[num+k]
		}
		cnt[num]++
	}
	return
}

Rust

impl Solution {
    pub fn count_k_difference(nums: Vec<i32>, k: i32) -> i32 {
        let mut arr = [0; 101];
        let mut res = 0;
        for num in nums {
            if num - k >= 1 {
                res += arr[(num - k) as usize];
            }
            if num + k <= 100 {
                res += arr[(num + k) as usize];
            }
            arr[num as usize] += 1;
        }
        res
    }
}