给定一个大小为N的数组arr[]和一个整数K ,任务是将数组划分为最少数量的子集,使得每个子集中的最大对和小于或等于K 。
例子:
Input: arr[] = {1, 2, 3, 4, 5}, K = 5
Output: 3
Explanation:
Subset having maximum pair sum less than or equal to K(= 5) are: {{2, 3}, {1, 4}, {5}}.
Therefore, the required output is 3.
Input: arr[] = {2, 6, 8, 10, 20, 25}, K = 26
Output: 3
Explanation:
Subset having maximum pair sum less than or equal to K(=26) are: {{2, 6, 8, 10}, {20}, {25}}.
Therefore, the required output is 3.
方法:这个问题可以使用两个指针技术来解决。这个想法是对数组进行分区,以使每个子集的最大对和最小化。请按照以下步骤解决问题:
- 对给定的数组进行排序。
- 初始化一个变量res来存储满足给定条件的最小子集数。
- 初始化两个变量,比如start和end分别存储排序数组的开始和结束索引。
- 遍历已排序的数组并检查arr[start] + arr[end] ≤ K是否。如果发现为真,则将start的值增加1 。
- 否则,将end的值减少1并将res增加1 。
- 最后,打印res的值。
下面是上述方法的实现:
C++
// C++ program to implement
// the above appraoch
#include
using namespace std;
// Function to get the minimum
// count of subsets that satisfy
// the given condition
int cntMinSub(int arr[],
int N, int K)
{
// Store the minimum count
// of subsets that satisfy
// the given condition
int res = 0;
// Stores start index
// of the sorted array.
int start = 0;
// Stores end index
// of the sorted array
int end = N - 1;
// Sort the given array
sort(arr, arr + N);
// Traverse the array
while (end - start > 1) {
if (arr[start] + arr[end]
<= K) {
start++;
}
else {
res++;
end--;
}
}
// If only two elements
// of sorted array left
if (end - start == 1) {
if (arr[start] + arr[end]
<= K) {
res++;
start++;
end--;
}
else {
res++;
end--;
}
}
// If only one elements
// left in the array
if (start == end) {
res++;
}
return res;
}
// Driver Code
int main()
{
int arr[] = { 2, 6, 8, 10, 20, 25 };
int N = sizeof(arr) / sizeof(arr[0]);
int K = 26;
cout << cntMinSub(arr, N, K);
}
Java
// Java program to implement
// the above appraoch
import java.util.*;
class GFG{
// Function to get the minimum
// count of subsets that satisfy
// the given condition
static int cntMinSub(int arr[],
int N, int K)
{
// Store the minimum count
// of subsets that satisfy
// the given condition
int res = 0;
// Stores start index
// of the sorted array.
int start = 0;
// Stores end index
// of the sorted array
int end = N - 1;
// Sort the given array
Arrays.sort(arr);
// Traverse the array
while (end - start > 1)
{
if (arr[start] +
arr[end] <= K)
{
start++;
}
else
{
res++;
end--;
}
}
// If only two elements
// of sorted array left
if (end - start == 1)
{
if (arr[start] +
arr[end] <= K)
{
res++;
start++;
end--;
}
else
{
res++;
end--;
}
}
// If only one elements
// left in the array
if (start == end)
{
res++;
}
return res;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = {2, 6, 8, 10, 20, 25};
int N = arr.length;
int K = 26;
System.out.print(cntMinSub(arr, N, K));
}
}
// This code is contributed by shikhasingrajput
Python3
# Python3 program to implement
# the above appraoch
# Function to get the minimum
# count of subsets that satisfy
# the given condition
def cntMinSub(arr, N, K):
# Store the minimum count
# of subsets that satisfy
# the given condition
res = 0
# Stores start index
# of the sorted array.
start = 0
# Stores end index
# of the sorted array
end = N - 1
# Sort the given array
arr = sorted(arr)
# Traverse the array
while (end - start > 1):
if (arr[start] + arr[end] <= K):
start += 1
else:
res += 1
end -= 1
# If only two elements
# of sorted array left
if (end - start == 1):
if (arr[start] + arr[end] <= K):
res += 1
start += 1
end -= 1
else:
res += 1
end -= 1
# If only one elements
# left in the array
if (start == end):
res += 1
return res
# Driver Code
if __name__ == '__main__':
arr = [ 2, 6, 8, 10, 20, 25 ]
N = len(arr)
K = 26
print(cntMinSub(arr, N, K))
# This code is contributed by mohit kumar 29
C#
// C# program to implement
// the above appraoch
using System;
class GFG{
// Function to get the minimum
// count of subsets that satisfy
// the given condition
static int cntMinSub(int []arr,
int N, int K)
{
// Store the minimum count
// of subsets that satisfy
// the given condition
int res = 0;
// Stores start index
// of the sorted array.
int start = 0;
// Stores end index
// of the sorted array
int end = N - 1;
// Sort the given array
Array.Sort(arr);
// Traverse the array
while (end - start > 1)
{
if (arr[start] +
arr[end] <= K)
{
start++;
}
else
{
res++;
end--;
}
}
// If only two elements
// of sorted array left
if (end - start == 1)
{
if (arr[start] +
arr[end] <= K)
{
res++;
start++;
end--;
}
else
{
res++;
end--;
}
}
// If only one elements
// left in the array
if (start == end)
{
res++;
}
return res;
}
// Driver Code
public static void Main(String[] args)
{
int []arr = {2, 6, 8, 10, 20, 25};
int N = arr.Length;
int K = 26;
Console.Write(cntMinSub(arr, N, K));
}
}
// This code is contributed by 29AjayKumar
Javascript
输出
3
时间复杂度: O(N)
辅助空间: O(1)
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