Question

2006. Count Number of Pairs With Absolute Difference K

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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 $nums$.

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 $nums$.

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
  }
}