给定一个数组arr[]和整数K ，任务是将数组拆分为大小为K 的子集，使得每个子集由K个连续元素组成。
例子：
Input: arr[] = {1, 2, 3, 6, 2, 3, 4, 7, 8}, K = 3
Output: true
Explanation:
The given array of length 9 can be split into 3 subsets {1, 2, 3}, {2, 3, 4} and {6, 7, 8} such that each subset consists of 3 consecutive elements.
Input: arr[] = [1, 2, 3, 4, 5], K = 4
Output: false
Explanation:
The given array of length 5 cannot be split into subsets of 4.
方法
请按照以下步骤解决问题：
- 将所有数组元素的频率存储在 HashMap 中
- 遍历HashMap 。
- 对于HashMap 中存在的每个元素，检查是否所有出现的元素都可以与其下一个 (K – 1) 个连续元素一起分组到一个子集中。如果是，则在HashMap中相应地降低包含在子集中的元素的频率并继续前进。
- 如果发现任何不能被分组为 K 个连续元素的子集的元素，则打印 False。否则打印 True。
下面是上述方法的实现：
C++
// C++ Program to implement the
// above approach
#include
using namespace std;
// Function to check if a given array can
// be split into subsets of K consecutive
// elements
bool groupInKConsecutive(vector& arr,
int K)
{
// Stores the frequencies of
// array elements
map count;
for (int h : arr) {
++count[h];
}
// Traverse the map
for (auto c : count) {
int cur = c.first;
int n = c.second;
// Check if all its occurrences can
// be grouped into K subsets
if (n > 0) {
// Traverse next K elements
for (int i = 1; i < K; ++i) {
// If the element is not
// present in the array
if (!count.count(cur + i)) {
return false;
}
count[cur + i] -= n;
// If it cannot be split into
// required number of subsets
if (count[cur + i] < 0)
return false;
}
}
}
return true;
}
// Driver Code
int main()
{
vector arr = { 1, 2, 3, 6, 2,
3, 4, 7, 8 };
int k = 3;
if (groupInKConsecutive(arr, k)) {
cout << "True";
}
else {
cout << "False";
}
} Java
// Java Program to implement the
// above approach
import java.util.*;
class GFG{
// Function to check if a given array can
// be split into subsets of K consecutive
// elements
static boolean groupInKConsecutive(int[] arr,
int K)
{
// Stores the frequencies of
// array elements
HashMap count = new HashMap();
for (int h : arr)
{
if(count.containsKey(h))
count.put(h, count.get(h) + 1);
else
count.put(h, 1);
}
// Traverse the map
for (Map.Entry c : count.entrySet())
{
int cur = c.getKey();
int n = c.getValue();
// Check if all its occurrences can
// be grouped into K subsets
if (n > 0)
{
// Traverse next K elements
for (int i = 1; i < K; ++i)
{
// If the element is not
// present in the array
if (!count.containsKey(cur + i))
{
return false;
}
count.put(cur + i, count.get(cur + i) - n);
// If it cannot be split into
// required number of subsets
if (count.get(cur + i) < 0)
return false;
}
}
}
return true;
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 1, 2, 3, 6, 2,
3, 4, 7, 8 };
int k = 3;
if (groupInKConsecutive(arr, k))
{
System.out.print("True");
}
else
{
System.out.print("False");
}
}
}
// This code contributed by sapnasingh4991 Python3
# Python3 program to implement the
# above approach
from collections import defaultdict
# Function to check if a given array can
# be split into subsets of K consecutive
# elements
def groupInKConsecutive(arr, K):
# Stores the frequencies of
# array elements
count = defaultdict(int)
for h in arr:
count[h] += 1
# Traverse the map
for key, value in count.items():
cur = key
n = value
# Check if all its occurrences can
# be grouped into K subsets
if (n > 0):
# Traverse next K elements
for i in range(1, K):
# If the element is not
# present in the array
if ((cur + i) not in count):
return False
count[cur + i] -= n
# If it cannot be split into
# required number of subsets
if (count[cur + i] < 0):
return False
return True
# Driver Code
if __name__ == "__main__":
arr = [ 1, 2, 3, 6, 2,
3, 4, 7, 8 ]
k = 3
if (groupInKConsecutive(arr, k)):
print("True")
else:
print("False")
# This code is contributed by chitranayalC#
// C# program to implement the
// above approach
using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
class GFG{
// Function to check if a given array can
// be split into subsets of K consecutive
// elements
static bool groupInKConsecutive(int[] arr,
int K)
{
// Stores the frequencies of
// array elements
Dictionary count = new Dictionary();
foreach(int h in arr)
{
if (count.ContainsKey(h))
count[h]++;
else
count[h] = 1;
}
// Traverse the map
foreach(int c in count.Keys.ToList())
{
int cur = c;
int n = count;
// Check if all its occurrences can
// be grouped into K subsets
if (n > 0)
{
// Traverse next K elements
for(int i = 1; i < K; ++i)
{
// If the element is not
// present in the array
if (!count.ContainsKey(cur + i))
{
return false;
}
count[cur + i] -= n;
// If it cannot be split into
// required number of subsets
if (count[cur + i] < 0)
return false;
}
}
}
return true;
}
// Driver Code
public static void Main(string[] args)
{
int[] arr = { 1, 2, 3, 6, 2,
3, 4, 7, 8 };
int k = 3;
if (groupInKConsecutive(arr, k))
{
Console.Write("True");
}
else
{
Console.Write("False");
}
}
}
// This code is contributed by rutvik_56 Javascript
输出：
True时间复杂度： O(N*log(N))
辅助空间： O(N)
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