计算差异等于 K | 的所有不同对设置 2
给定一个整数数组arr[]和一个正整数 K，任务是计算所有差异等于K的不同对。
例子：
Input: arr[ ] = {1, 5, 3, 4, 2}, K = 3
Output: 2
Explanation: There are 2 distinct pairs with difference 3, the pairs are {1, 4} and {5, 2}
Input: arr[] = {8, 12, 16, 4, 0, 20}, K = 4
Output: 5
Explanation: There are 5 unique pairs with difference 4.
The pairs are {0, 4}, {4, 8}, {8, 12}, {12, 16} and {16, 20}
本文的Set 1中提到了朴素的方法和基于排序和二分查找的方法。
方法：通过在无序映射的帮助下使用散列，可以将这个问题的时间复杂度降低到平均情况下的线性复杂度，如下所示：
For forming such unique pairs, if traversed from the smallest element, an element (say x) will form such a pair with another element having value (x+K).
When the difference K = 0 then the elements having frequency more than 1 will be able to form pairs with itself.
请按照以下步骤解决问题：
- 初始化一个无序映射并将所有数组元素推送到映射中。
- 如果K的给定值为0 ：
- 如果当前元素x的频率大于 1，则将count加 1。
- 其他元素也可以尝试相同的方法。
- 如果K的给定值不为 0：
- 在地图中搜索x + K ，如果找到，则将计数加 1。
- 否则尝试其他元素。
- 返回计数。
下面是上述方法的实现。
C++
// C++ code to implement the above approach.
#include
using namespace std;
// Function to find total pairs
int TotalPairs(vector nums, int K)
{
// Initializing a map
unordered_map mp;
int cnt = 0;
for (int i = 0; i < nums.size(); i++) {
mp[nums[i]]++;
}
// Difference equal to zero
if (K == 0) {
for (auto i : mp) {
// Frequency of element is
// greater than one then
// distinct pair is possible
if (i.second > 1)
cnt++;
}
}
// Difference is not equal to zero
else {
for (auto i : mp) {
// Frequency of element + k
// is not zero then distinct
// pair is possible
if (mp.find(i.first + K)
!= mp.end()) {
cnt++;
}
}
}
return cnt;
}
// Driver Code
int main()
{
vector arr = { 8, 12, 16, 4, 0, 20 };
int K = 4;
// Function call
int ans = TotalPairs(arr, K);
cout << ans;
return 0;
} Java
// Java code to implement the above approach.
import java.io.*;
import java.util.*;
class GFG {
// Function to find total pairs
public static int TotalPairs(int nums[], int K)
{
// Initializing a map
Map mp
= new HashMap();
int cnt = 0;
for (int i = 0; i < nums.length; i++) {
if (mp.get(nums[i]) != null)
mp.put(nums[i], mp.get(nums[i]) + 1);
else
mp.put(nums[i], 1);
}
// Difference equal to zero
if (K == 0) {
for (Map.Entry it :
mp.entrySet()) {
// Frequency of element is
// greater than one then
// distinct pair is possible
if (it.getValue() > 1)
cnt++;
}
}
// Difference is not equal to zero
else {
for (Map.Entry it :
mp.entrySet()) {
// Frequency of element + k
// is not zero then distinct
// pair is possible
if (mp.get(it.getKey() + K) != null) {
cnt++;
}
}
}
return cnt;
}
public static void main(String[] args)
{
int arr[] = { 8, 12, 16, 4, 0, 20 };
int K = 4;
// Function call
int ans = TotalPairs(arr, K);
System.out.print(ans);
}
}
// This code is contributed by Rohit Pradhan Python3
# Python3 program for above approach
# function to find total pairs
def TotalPairs(nums, K):
# Initializing a map or dictionary
mp = dict()
cnt = 0
for i in range(len(nums)):
if nums[i] in mp:
mp[nums[i]] += 1
else:
mp[nums[i]] = 1
# Difference equal to zero
if K == 0:
for i in mp:
# Frequency of element is
# greater than one then
# distinct pair is possible
if mp[i] > 1:
cnt += 1
# Difference is not equal to zero
else:
for i in mp:
# Frequency of element + k
# is not zero then distinct
#pair is possible
if i + K in mp:
cnt += 1
return cnt
# Driver Code
arr = [8, 12, 16, 4, 0, 20]
K = 4
# Function call
ans = TotalPairs(arr, K)
print(ans)
# This code is contributed by phasing175时间复杂度： O(N) [在平均情况下，因为无序映射的平均情况时间复杂度是 O(1)]
辅助空间： O(N)