2006. Count Number of Pairs With Absolute Difference K

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 \(nums\).

Python3JavaC++GoTypeScriptRust

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)
        )

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;
    }
}

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;
    }
};

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
}

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;
}

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 \(nums\).

Python3JavaC++GoRust

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

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;
    }
}

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;
    }
};

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
}

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
    }
}

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