给定一个数组arr[]和一个在数组中至少出现一次的数字K ,任务是找到数组中子集的数量,使得每个子集只包含给定的值K 。
例子:
Input: arr[] = {1, 0, 0, 1, 0, 1, 2, 5, 2, 1}, K = 0
Output: 4
Explanation:
From the two 0’s present in the array at the index 2 and 3, 3 subsequences can be formed: {0}, {0}, {0, 0}
From the 0 present in the array at the index 5, 1 subsequence can be formed: {0}
Therefore, a total of 4 subsequences are formed.
Input: arr[] = {1, 0, 0, 1, 1, 0, 0, 2, 3, 5}, K = 1
Output: 4
方法:为了找到子集的数量,需要对给定集合中不同数量的元素形成的子集数量进行一次观察。
因此,让 N 是我们需要为其找到子集的元素数。
那么,如果:
N = 1: Only one subset can be formed.
N = 2: Three subsets can be formed.
N = 3: Six subsets can be formed.
N = 4: Ten subsets can be formed.
.
.
.
N = K: (K * (K + 1))/2 subsets can be formed.
由于我们正在计算值 K 的连续出现所形成的子集的数量,因此想法是找到给定数组中存在的连续 K 的计数,并使用给定的公式找到计数。
下面是上述方法的实现:
C++
// C++ implementation to find the
// number of subsets formed by
// the given value K
#include
using namespace std;
// Function to find the number
// of subsets formed by the
// given value K
int count(int arr[], int N, int K)
{
// Count is used to maintain the
// number of continuous K's
int count = 0, ans = 0;
// Iterating through the array
for (int i = 0; i < N; i++) {
// If the element in the array
// is equal to K
if (arr[i] == K) {
count = count + 1;
}
else {
// count*(count+1)/2 is the
// total number of subsets
// with only K as their element
ans += (count * (count + 1)) / 2;
// Change count to 0 because
// other element apart from
// K has been found
count = 0;
}
}
// To handle the last set of K's
ans = ans + (count * (count + 1)) / 2;
return ans;
}
// Driver code
int main()
{
int arr[] = { 1, 0, 0, 1, 1, 0, 0 };
int N = sizeof(arr) / sizeof(int);
int K = 0;
cout << count(arr, N, K);
}
Java
// Java implementation to find the
// number of subsets formed by
// the given value K
class GFG{
// Function to find the number
// of subsets formed by the
// given value K
static int count(int arr[], int N, int K)
{
// Count is used to maintain the
// number of continuous K's
int count = 0, ans = 0;
// Iterating through the array
for (int i = 0; i < N; i++) {
// If the element in the array
// is equal to K
if (arr[i] == K) {
count = count + 1;
}
else {
// count*(count+1)/2 is the
// total number of subsets
// with only K as their element
ans += (count * (count + 1)) / 2;
// Change count to 0 because
// other element apart from
// K has been found
count = 0;
}
}
// To handle the last set of K's
ans = ans + (count * (count + 1)) / 2;
return ans;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 0, 0, 1, 1, 0, 0 };
int N = arr.length;
int K = 0;
System.out.print(count(arr, N, K));
}
}
// This code is contributed by 29AjayKumar
Python3
# Python 3 implementation to find the
# number of subsets formed by
# the given value K
# Function to find the number
# of subsets formed by the
# given value K
def count(arr, N, K):
# Count is used to maintain the
# number of continuous K's
count = 0
ans = 0
# Iterating through the array
for i in range(N):
# If the element in the array
# is equal to K
if (arr[i] == K):
count = count + 1
else:
# count*(count+1)/2 is the
# total number of subsets
# with only K as their element
ans += (count * (count + 1)) // 2
# Change count to 0 because
# other element apart from
# K has been found
count = 0
# To handle the last set of K's
ans = ans + (count * (count + 1)) // 2
return ans
# Driver code
if __name__ == '__main__':
arr = [1, 0, 0, 1, 1, 0, 0]
N = len(arr)
K = 0
print(count(arr, N, K))
# This code is contributed by Surendra_Gangwar
C#
// C# implementation to find the
// number of subsets formed by
// the given value K
using System;
class GFG{
// Function to find the number
// of subsets formed by the
// given value K
static int count(int []arr, int N, int K)
{
// Count is used to maintain the
// number of continuous K's
int count = 0, ans = 0;
// Iterating through the array
for(int i = 0; i < N; i++)
{
// If the element in the array
// is equal to K
if (arr[i] == K)
{
count = count + 1;
}
else
{
// count*(count+1)/2 is the
// total number of subsets
// with only K as their element
ans += (count * (count + 1)) / 2;
// Change count to 0 because
// other element apart from
// K has been found
count = 0;
}
}
// To handle the last set of K's
ans = ans + (count * (count + 1)) / 2;
return ans;
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 1, 0, 0, 1, 1, 0, 0 };
int N = arr.Length;
int K = 0;
Console.Write(count(arr, N, K));
}
}
//This is contributed by shivanisinghss2110
Javascript
6
时间复杂度: O(N) ,其中 N 是数组的大小。
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