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 <= 2001 <= nums[i] <= 1001 <= 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(n2), and the space complexity is O(1). Here, n is the length of the array nums.
PythonJavaC++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)
)(code-box)
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;
}
}(code-box)
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;
}
};(code-box)
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
}(code-box)
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;
}(code-box)
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
}
}(code-box)
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.
PythonJavaC++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(code-box)
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;
}
}(code-box)
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;
}
};(code-box)
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
}(code-box)
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
}
}(code-box)
