给定一个整数N ,任务是检查给定的数字是否为全字母缩写。
注意: Pangram号包含每个数字[ 0-9 ]至少一次。
例子:
Input : N = 10239876540022
Output : Yes
Explanation: N contains all the digits from 0 to 9. Therefore, it is a pangram.
Input : N = 234567890
Output : No
Explanation: N doesn’t contain the digit 1. Therefore, it is not a pangram.
基于集合的方法:这个想法是使用集合来存储N中存在的不同数字的计数。请按照以下步骤解决问题:
- 将数字N转换为等效字符串。
- 将字符串转换为集合。
- 如果Set的大小为10 ,则它包含所有可能的不同数字[0 – 9] 。因此,打印“是” 。
- 否则,打印“否”
下面是上述方法的实现:
Java
// Java implementation of above approach
import java.math.BigInteger;
import java.util.HashSet;
class GFG{
// Function to check if N
// is a Pangram or not
static String numberPangram(BigInteger N)
{
// Stores equivalent string
// representation of N
String num = N.toString();
// Convert the string to Character array
char[] arrNum = num.toCharArray();
// Add all characters pf arrNum to set
HashSet setNum = new HashSet();
for(char ch : arrNum)
{
setNum.add(ch);
}
// If the length of set is 10
// The number is a Pangram
if (setNum.size() == 10)
return "True";
else
return "False";
}
// Driver code
public static void main(String[] args)
{
BigInteger N = new BigInteger("10239876540022");
System.out.print(numberPangram(N));
}
}
// This code is contributed by abhinavjain194
Python3
# Python3 implementation of above approach
# Function to check if N
# is a Pangram or not
def numberPangram(N):
# Stores equivalent string
# representation of N
num = str(N)
# Convert the string to set
setnum = set(num)
# If the length of set is 10
if(len(setnum) == 10):
# The number is a Pangram
return True
else:
return False
# Driver Code
N = 10239876540022
print(numberPangram(N))
Python3
# Python implementation of above approach
from collections import Counter
# Function to check if
# N is a Pangram or not
def numberPangram(N):
# Stores equivalent string
# representation of N
num = str(N)
# Count frequencies of
# digits present in N
frequency = Counter(num)
# If the length of the
# dictionary frequency is 10
if(len(frequency) == 10):
# The number is a Pangram
return True
else:
return False
# Driver Code
N =10239876540022
print(numberPangram(N))
输出:
True
时间复杂度: O(log 10 N * log(log 10 N))
辅助空间: O(1)
基于散列的方法:请按照以下步骤解决问题:
- 将N转换为其等效的字符串。
- 计算该字符串中所有字符的频率。 Counter()函数可在Python用于此目的。
- 如果存储频率的Dictionary / HashMap的长度为10 ,则该数字包含所有可能的不同数字。因此,打印“是”。
- 否则,打印“否” 。
下面是上述方法的实现:
Python3
# Python implementation of above approach
from collections import Counter
# Function to check if
# N is a Pangram or not
def numberPangram(N):
# Stores equivalent string
# representation of N
num = str(N)
# Count frequencies of
# digits present in N
frequency = Counter(num)
# If the length of the
# dictionary frequency is 10
if(len(frequency) == 10):
# The number is a Pangram
return True
else:
return False
# Driver Code
N =10239876540022
print(numberPangram(N))
输出:
True
时间复杂度: O(log 10 N * log(log 10 N))
辅助空间: O(1)