给定一个整数数组arr[]和一个整数K ,任务是找到非空子集 S 的数量,使得min(S) + max(S) < K 。
例子:
Input: arr[] = {2, 4, 5, 7} K = 8
Output: 4
Explanation:
The possible subsets are {2}, {2, 4}, {2, 4, 5} and {2, 5}
Input:: arr[] = {2, 4, 2, 5, 7} K = 10
Output: 26
方法
- 首先对输入数组进行排序。
- 现在使用双指针技术来计算子集的数量。
- 取左右两个指针并设置 left = 0 和 right = N-1。
if (arr[left] + arr[right] < K )
Increment the left pointer by 1 and add 2 j – i into answer, because the left and right values make up a potential end values of a subset. All the values from [i, j – 1] also make up end of subsets which will have the sum < K. So, we need to calculate all the possible subsets for left = i and right ∊ [i, j]. So, after suming up values 2 j – i + 1 + 2 j – i – 2 + … + 2 0 of the GP, we get 2 j – i .
if( arr[left] + arr[right] >= K )
Decrement the right pointer by 1.
- 重复下面的过程直到left <= right 。
下面是上述方法的实现:
C++
// C++ program to print count
// of subsets S such that
// min(S) + max(S) < K
#include
using namespace std;
// Function that return the
// count of subset such that
// min(S) + max(S) < K
int get_subset_count(int arr[], int K,
int N)
{
// Sorting the array
sort(arr, arr + N);
int left, right;
left = 0;
right = N - 1;
// ans stores total number of subsets
int ans = 0;
while (left <= right) {
if (arr[left] + arr[right] < K) {
// add all possible subsets
// between i and j
ans += 1 << (right - left);
left++;
}
else {
// Decrease the sum
right--;
}
}
return ans;
}
// Driver code
int main()
{
int arr[] = { 2, 4, 5, 7 };
int K = 8;
int N = sizeof(arr) / sizeof(arr[0]);
cout << get_subset_count(arr, K, N);
return 0;
}
Java
// Java program to print count
// of subsets S such that
// Math.min(S) + Math.max(S) < K
import java.util.*;
class GFG{
// Function that return the
// count of subset such that
// Math.min(S) + Math.max(S) < K
static int get_subset_count(int arr[], int K,
int N)
{
// Sorting the array
Arrays.sort(arr);
int left, right;
left = 0;
right = N - 1;
// ans stores total number
// of subsets
int ans = 0;
while (left <= right)
{
if (arr[left] + arr[right] < K)
{
// Add all possible subsets
// between i and j
ans += 1 << (right - left);
left++;
}
else
{
// Decrease the sum
right--;
}
}
return ans;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 2, 4, 5, 7 };
int K = 8;
int N = arr.length;
System.out.print(get_subset_count(arr, K, N));
}
}
// This code is contributed by Rajput-Ji
Python3
# Python3 program to print
# count of subsets S such
# that min(S) + max(S) < K
# Function that return the
# count of subset such that
# min(S) + max(S) < K
def get_subset_count(arr, K, N):
# Sorting the array
arr.sort()
left = 0;
right = N - 1;
# ans stores total number of subsets
ans = 0;
while (left <= right):
if (arr[left] + arr[right] < K):
# Add all possible subsets
# between i and j
ans += 1 << (right - left);
left += 1;
else:
# Decrease the sum
right -= 1;
return ans;
# Driver code
arr = [ 2, 4, 5, 7 ];
K = 8;
print(get_subset_count(arr, K, 4))
# This code is contributed by grand_master
C#
// C# program to print count
// of subsets S such that
// Math.Min(S) + Math.Max(S) < K
using System;
class GFG{
// Function that return the
// count of subset such that
// Math.Min(S) + Math.Max(S) < K
static int get_subset_count(int []arr, int K,
int N)
{
// Sorting the array
Array.Sort(arr);
int left, right;
left = 0;
right = N - 1;
// ans stores total number
// of subsets
int ans = 0;
while (left <= right)
{
if (arr[left] + arr[right] < K)
{
// Add all possible subsets
// between i and j
ans += 1 << (right - left);
left++;
}
else
{
// Decrease the sum
right--;
}
}
return ans;
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 2, 4, 5, 7 };
int K = 8;
int N = arr.Length;
Console.Write(get_subset_count(arr, K, N));
}
}
// This code is contributed by gauravrajput1
Javascript
4
时间复杂度: O(N* log N)
辅助空间: O(1)
如果您希望与专家一起参加现场课程,请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。