给定一个由N 个正整数和一个正整数K组成的数组arr[] ,任务是找到其不同数字是K数字的子集的数组元素的计数。
例子:
Input: arr[] = { 1, 12, 1222, 13, 2 }, K = 12
Output: 4
Explanation:
Distinct Digits of K are { 1, 2 }
Distinct Digits of arr[0] are { 1 }, which is the subset of the digits of K.
Distinct Digits of arr[1] are { 1, 2 }, which is the subset of the digits of K.
Distinct Digits of arr[2] are { 1, 2 }, which is the subset of the digits of K.
Distinct Digits of arr[3] are { 1, 3 }, which is not the subset of the digits of K.
Distinct Digits of arr[4] are { 2 }, which is the subset of the digits of K.
Therefore, the required output is 4.
Input: arr = {1, 2, 3, 4, 1234}, K = 1234
Output: 5
朴素的方法:解决问题的最简单的方法是遍历数组arr[]并对于每个数组元素,检查其所有不同的数字是否都出现在K 中。如果发现为真,则增加计数。最后,打印获得的计数。
下面是上述方法的实现:
C++14
// C++ program for the above approach
#include
using namespace std;
// Function to check a digit occurs
// in the digit of K or not
static bool isValidDigit(int digit, int K)
{
// Iterate over all possible
// digits of K
while (K != 0)
{
// If current digit
// equal to digit
if (K % 10 == digit)
{
return true;
}
// Update K
K = K / 10;
}
return false;
}
// Function to find the count of array
// elements whose distinct digits are
// a subset of digits of K
int noOfValidNumbers(int K, int arr[], int n)
{
// Stores count of array elements
// whose distinct digits are subset
// of digits of K
int count = 0;
// Traverse the array, []arr
for(int i = 0; i < n; i++)
{
// Stores the current element
int no = arr[i];
// Check if all the digits arr[i]
// is a subset of the digits of
// K or not
bool flag = true;
// Iterate over all possible
// digits of arr[i]
while (no != 0)
{
// Stores current digit
int digit = no % 10;
// If current digit does not
// appear in K
if (!isValidDigit(digit, K))
{
// Update flag
flag = false;
break;
}
// Update no
no = no / 10;
}
// If all the digits arr[i]
// appear in K
if (flag == true)
{
// Update count
count++;
}
}
// Finally print count
return count;
}
// Driver Code
int main()
{
int K = 12;
int arr[] = { 1, 12, 1222, 13, 2 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << noOfValidNumbers(K, arr, n);
return 0;
}
// This code is contributed by susmitakundugoaldanga
Java
// Java program for the above approach
import java.io.*;
class GFG {
// Function to check a digit occurs
// in the digit of K or not
static boolean isValidDigit(int digit, int K)
{
// Iterate over all possible
// digits of K
while (K != 0) {
// If current digit
// equal to digit
if (K % 10 == digit) {
return true;
}
// Update K
K = K / 10;
}
return false;
}
// Function to find the count of array elements
// whose distinct digits are a subset of digits of K
static int noOfValidNumbers(int K, int arr[])
{
// Stores count of array elements
// whose distinct digits are subset
// of digits of K
int count = 0;
// Traverse the array, arr[]
for (int i = 0; i < arr.length; i++) {
// Stores the current element
int no = arr[i];
// Check if all the digits arr[i]
// is a subset of the digits of
// K or not
boolean flag = true;
// Iterate over all possible
// digits of arr[i]
while (no != 0) {
// Stores current digit
int digit = no % 10;
// If current digit does not
// appear in K
if (!isValidDigit(digit, K)) {
// Update flag
flag = false;
break;
}
// Update no
no = no / 10;
}
// If all the digits arr[i] appear in K
if (flag == true) {
// Update count
count++;
}
}
// Finally print count
return count;
}
// Driver Code
public static void main(String[] args)
{
int K = 12;
int arr[] = { 1, 12, 1222, 13, 2 };
System.out.println(noOfValidNumbers(K, arr));
}
}
Python3
# Python3 program for the above approach
# Function to check a digit occurs
# in the digit of K or not
def isValidDigit(digit, K):
# Iterate over all possible
# digits of K
while (K != 0):
# If current digit
# equal to digit
if (K % 10 == digit):
return True
# Update K
K = K // 10
return False
# Function to find the count of array
# elements whose distinct digits are
# a subset of digits of K
def noOfValidNumbers(K, arr, n):
# Stores count of array elements
# whose distinct digits are subset
# of digits of K
count = 0
# Traverse the array, []arr
for i in range(n):
# Stores the current element
no = arr[i]
# Check if all the digits arr[i]
# is a subset of the digits of
# K or not
flag = True
# Iterate over all possible
# digits of arr[i]
while (no != 0):
# Stores current digit
digit = no % 10
# If current digit does not
# appear in K
if (not isValidDigit(digit, K)):
# Update flag
flag = False
break
# Update no
no = no // 10
# If all the digits arr[i]
# appear in K
if (flag == True):
# Update count
count += 1
# Finally print count
return count
# Driver Code
if __name__ == "__main__":
K = 12
arr = [1, 12, 1222, 13, 2]
n = len(arr)
print(noOfValidNumbers(K, arr, n))
# This code is contributed by chitranayal.
C#
// C# program for the above approach
using System;
class GFG{
// Function to check a digit occurs
// in the digit of K or not
static bool isValidDigit(int digit, int K)
{
// Iterate over all possible
// digits of K
while (K != 0)
{
// If current digit
// equal to digit
if (K % 10 == digit)
{
return true;
}
// Update K
K = K / 10;
}
return false;
}
// Function to find the count of array elements
// whose distinct digits are a subset of digits of K
static int noOfValidNumbers(int K, int []arr)
{
// Stores count of array elements
// whose distinct digits are subset
// of digits of K
int count = 0;
// Traverse the array, []arr
for(int i = 0; i < arr.Length; i++)
{
// Stores the current element
int no = arr[i];
// Check if all the digits arr[i]
// is a subset of the digits of
// K or not
bool flag = true;
// Iterate over all possible
// digits of arr[i]
while (no != 0)
{
// Stores current digit
int digit = no % 10;
// If current digit does not
// appear in K
if (!isValidDigit(digit, K))
{
// Update flag
flag = false;
break;
}
// Update no
no = no / 10;
}
// If all the digits arr[i] appear in K
if (flag == true)
{
// Update count
count++;
}
}
// Finally print count
return count;
}
// Driver Code
public static void Main(String[] args)
{
int K = 12;
int []arr = { 1, 12, 1222, 13, 2 };
Console.WriteLine(noOfValidNumbers(K, arr));
}
}
// This code is contributed by shikhasingrajput
Javascript
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to the count of array elements whose
// distinct digits are a subset of the digits of K
static int noOfValidKbers(int K, vector arr)
{
// Stores distinct digits of K
map set;
// Iterate over all the digits of K
while (K != 0)
{
// Insert current digit into set
set[K % 10] = 1;
// Update K
K = K / 10;
}
// Stores the count of array elements
// whose distinct digits are a subset
// of the digits of K
int count = 0;
// Traverse the array, arr[]
for (int i = 0; i < arr.size(); i++)
{
// Stores current element
int no = arr[i];
// Check if all the digits of arr[i]
// are present in K or not
bool falg = true;
// Iterate over all the digits of arr[i]
while (no != 0)
{
// Stores current digit
int digit = no % 10;
// If digit not present in the set
if (set.find(digit)==set.end())
{
// Update flag
falg = false;
break;
}
// Update no
no = no / 10;
}
// If all the digits of
// arr[i] present in set
if (falg == true)
{
// Update count
count++;
}
}
return count;
}
// Driver Code
int main()
{
int K = 12;
vector arr = { 1, 12, 1222, 13, 2 };
cout<<(noOfValidKbers(K, arr));
}
// This code is contributed by mohit kumar 29
Java
// Java program to implement
// the above approach
import java.io.*;
import java.util.*;
class GFG {
// Function to the count of array elements whose
// distinct digits are a subset of the digits of K
static int noOfValidKbers(int K, int arr[])
{
// Stores distinct digits of K
HashSet set = new HashSet<>();
// Iterate over all the digits of K
while (K != 0) {
// Insert current digit into set
set.add(K % 10);
// Update K
K = K / 10;
}
// Stores the count of array elements
// whose distinct digits are a subset
// of the digits of K
int count = 0;
// Traverse the array, arr[]
for (int i = 0; i < arr.length; i++) {
// Stores current element
int no = arr[i];
// Check if all the digits of arr[i]
// are present in K or not
boolean falg = true;
// Iterate over all the digits of arr[i]
while (no != 0) {
// Stores current digit
int digit = no % 10;
// If digit not present in the set
if (!set.contains(digit)) {
// Update flag
falg = false;
break;
}
// Update no
no = no / 10;
}
// If all the digits of
// arr[i] present in set
if (falg == true) {
// Update count
count++;
}
}
return count;
}
// Driver Code
public static void main(String[] args)
{
int K = 12;
int arr[] = { 1, 12, 1222, 13, 2 };
System.out.println(noOfValidKbers(K, arr));
}
}
Python3
# Python 3 program to implement
# the above approach
# Function to the count of array elements whose
# distinct digits are a subst of the digits of K
def noOfValidKbers(K, arr):
# Stores distinct digits of K
st = {}
# Iterate over all the digits of K
while(K != 0):
# Insert current digit into st
if(K % 10 in st):
st[K % 10] = 1
else:
st[K % 10] = st.get(K % 10, 0) + 1
# Update K
K = K // 10
# Stores the count of array elements
# whose distinct digits are a subst
# of the digits of K
count = 0
# Traverse the array, arr[]
for i in range(len(arr)):
# Stores current element
no = arr[i]
# Check if all the digits of arr[i]
# are present in K or not
falg = True
# Iterate over all the digits of arr[i]
while(no != 0):
# Stores current digit
digit = no % 10
# If digit not present in the st
if (digit not in st):
# Update flag
falg = False
break
# Update no
no = no//10
# If all the digits of
# arr[i] present in st
if (falg == True):
# Update count
count += 1
return count
# Driver Code
if __name__ == '__main__':
K = 12
arr = [1, 12, 1222, 13, 2]
print(noOfValidKbers(K, arr))
# This code is contributed by SURENDRA_GANGWAR.
C#
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to the count of array elements whose
// distinct digits are a subset of the digits of K
static int noOfValidKbers(int K, int []arr)
{
// Stores distinct digits of K
HashSet set = new HashSet();
// Iterate over all the digits of K
while (K != 0)
{
// Insert current digit into set
set.Add(K % 10);
// Update K
K = K / 10;
}
// Stores the count of array elements
// whose distinct digits are a subset
// of the digits of K
int count = 0;
// Traverse the array, []arr
for(int i = 0; i < arr.Length; i++)
{
// Stores current element
int no = arr[i];
// Check if all the digits of arr[i]
// are present in K or not
bool falg = true;
// Iterate over all the digits of arr[i]
while (no != 0)
{
// Stores current digit
int digit = no % 10;
// If digit not present in the set
if (!set.Contains(digit))
{
// Update flag
falg = false;
break;
}
// Update no
no = no / 10;
}
// If all the digits of
// arr[i] present in set
if (falg == true)
{
// Update count
count++;
}
}
return count;
}
// Driver Code
public static void Main(String[] args)
{
int K = 12;
int []arr = { 1, 12, 1222, 13, 2 };
Console.WriteLine(noOfValidKbers(K, arr));
}
}
// This code is contributed by 29AjayKumar
Javascript
4
时间复杂度: O(N * log 10 (K) * log 10 (Max)),Max为最大数组元素
辅助空间: O(1)
高效的方法:上述方法可以使用 HashSet 进行优化。请按照以下步骤解决问题:
- 迭代K的数字并将所有数字插入一个 HashSet say set
- 遍历数组arr[]并对每个数组元素,迭代当前元素的所有数字并检查所有数字是否都存在于集合中。如果发现为真,则增加计数。
- 最后,打印获得的计数。
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to the count of array elements whose
// distinct digits are a subset of the digits of K
static int noOfValidKbers(int K, vector arr)
{
// Stores distinct digits of K
map set;
// Iterate over all the digits of K
while (K != 0)
{
// Insert current digit into set
set[K % 10] = 1;
// Update K
K = K / 10;
}
// Stores the count of array elements
// whose distinct digits are a subset
// of the digits of K
int count = 0;
// Traverse the array, arr[]
for (int i = 0; i < arr.size(); i++)
{
// Stores current element
int no = arr[i];
// Check if all the digits of arr[i]
// are present in K or not
bool falg = true;
// Iterate over all the digits of arr[i]
while (no != 0)
{
// Stores current digit
int digit = no % 10;
// If digit not present in the set
if (set.find(digit)==set.end())
{
// Update flag
falg = false;
break;
}
// Update no
no = no / 10;
}
// If all the digits of
// arr[i] present in set
if (falg == true)
{
// Update count
count++;
}
}
return count;
}
// Driver Code
int main()
{
int K = 12;
vector arr = { 1, 12, 1222, 13, 2 };
cout<<(noOfValidKbers(K, arr));
}
// This code is contributed by mohit kumar 29
Java
// Java program to implement
// the above approach
import java.io.*;
import java.util.*;
class GFG {
// Function to the count of array elements whose
// distinct digits are a subset of the digits of K
static int noOfValidKbers(int K, int arr[])
{
// Stores distinct digits of K
HashSet set = new HashSet<>();
// Iterate over all the digits of K
while (K != 0) {
// Insert current digit into set
set.add(K % 10);
// Update K
K = K / 10;
}
// Stores the count of array elements
// whose distinct digits are a subset
// of the digits of K
int count = 0;
// Traverse the array, arr[]
for (int i = 0; i < arr.length; i++) {
// Stores current element
int no = arr[i];
// Check if all the digits of arr[i]
// are present in K or not
boolean falg = true;
// Iterate over all the digits of arr[i]
while (no != 0) {
// Stores current digit
int digit = no % 10;
// If digit not present in the set
if (!set.contains(digit)) {
// Update flag
falg = false;
break;
}
// Update no
no = no / 10;
}
// If all the digits of
// arr[i] present in set
if (falg == true) {
// Update count
count++;
}
}
return count;
}
// Driver Code
public static void main(String[] args)
{
int K = 12;
int arr[] = { 1, 12, 1222, 13, 2 };
System.out.println(noOfValidKbers(K, arr));
}
}
蟒蛇3
# Python 3 program to implement
# the above approach
# Function to the count of array elements whose
# distinct digits are a subst of the digits of K
def noOfValidKbers(K, arr):
# Stores distinct digits of K
st = {}
# Iterate over all the digits of K
while(K != 0):
# Insert current digit into st
if(K % 10 in st):
st[K % 10] = 1
else:
st[K % 10] = st.get(K % 10, 0) + 1
# Update K
K = K // 10
# Stores the count of array elements
# whose distinct digits are a subst
# of the digits of K
count = 0
# Traverse the array, arr[]
for i in range(len(arr)):
# Stores current element
no = arr[i]
# Check if all the digits of arr[i]
# are present in K or not
falg = True
# Iterate over all the digits of arr[i]
while(no != 0):
# Stores current digit
digit = no % 10
# If digit not present in the st
if (digit not in st):
# Update flag
falg = False
break
# Update no
no = no//10
# If all the digits of
# arr[i] present in st
if (falg == True):
# Update count
count += 1
return count
# Driver Code
if __name__ == '__main__':
K = 12
arr = [1, 12, 1222, 13, 2]
print(noOfValidKbers(K, arr))
# This code is contributed by SURENDRA_GANGWAR.
C#
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to the count of array elements whose
// distinct digits are a subset of the digits of K
static int noOfValidKbers(int K, int []arr)
{
// Stores distinct digits of K
HashSet set = new HashSet();
// Iterate over all the digits of K
while (K != 0)
{
// Insert current digit into set
set.Add(K % 10);
// Update K
K = K / 10;
}
// Stores the count of array elements
// whose distinct digits are a subset
// of the digits of K
int count = 0;
// Traverse the array, []arr
for(int i = 0; i < arr.Length; i++)
{
// Stores current element
int no = arr[i];
// Check if all the digits of arr[i]
// are present in K or not
bool falg = true;
// Iterate over all the digits of arr[i]
while (no != 0)
{
// Stores current digit
int digit = no % 10;
// If digit not present in the set
if (!set.Contains(digit))
{
// Update flag
falg = false;
break;
}
// Update no
no = no / 10;
}
// If all the digits of
// arr[i] present in set
if (falg == true)
{
// Update count
count++;
}
}
return count;
}
// Driver Code
public static void Main(String[] args)
{
int K = 12;
int []arr = { 1, 12, 1222, 13, 2 };
Console.WriteLine(noOfValidKbers(K, arr));
}
}
// This code is contributed by 29AjayKumar
Javascript
4
时间复杂度: O(N * log 10 (Max)),其中 Max 是最大的数组元素
辅助空间: O(10)
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。