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📜  检查字符串数组是否可以对应于特定数字X

📅  最后修改于: 2021-04-24 16:14:20             🧑  作者: Mango

给定一个整数X和一个字符串数组str ,该数组表示范围为[2,36]的任意基数的数字,任务是检查是否可以通过为每个字符串指定2至36的所需基数来将所有字符串都转换为X,这样该字符串的十进制基数等效为X。

例子:

方法:想法是通过将数组的每个数字分配为2到36的基数,将其转换为十进制基数,然后检查每个转换后的数字是否等于X。

下面介绍了上述方法的分步算法–

  1. 将计数初始化为0,以检查转换后等于X的数字计数。
  2. 运行循环以遍历数组的数字,然后遍历每个数字–
    • 运行另一个从2到36的循环,为数字指定基数,然后找到该数字的十进制等效项。
    • 如果该数字的十进制等效项等于X,则将计数加1并中断循环,以便不为该数字分配任何其他基数。
  3. 如果可转换为X的数字的计数等于数组的长度,则该数组可以对应于数字X。

下面是上述方法的实现:

C++
// C++ implementation to check
// wheather array of strings
// can correspond to a number X
  
#include 
using namespace std;
  
// Function to find the maximum 
// base possible for the number N
int val(char c)
{
    if (c >= '0' && c <= '9')
        return (int)c - '0';
    else
        return (int)c - 'A' + 10;
}
  
// Function to find the decimal
// equivalent of the number
int toDeci(string str, int base)
{
    int len = str.size();
    int power = 1;
    int num = 0;
    int i;
    for (i = len - 1; i >= 0; i--) {
          
        // Condition to check if the
        // number is convertible 
        // to another base
        if (val(str[i]) >= base) {
            return -1;
        }
        num += val(str[i]) * power;
        power = power * base;
    }
    return num;
}
  
// Function to check that the 
// array can correspond to a number X 
void checkCorrespond(vector str,
                                int x){
                                      
    // counter to count the numbers
    // those are convertible to X
    int counter = 0;
    int n = str.size();
  
    // Loop to iterate over the array
    for (int i = 0; i < n; i++) {
        for (int j = 2; j <= 36; j++) {
              
            // Convert the current string
            // to every base for checking
            // whether it will correspond
            // to X from any base
            if (toDeci(str[i], j) == x) {
                counter++;
                break;
            }
        }
    }
      
    // Condition to check if every
    // number of the array can
    // be converted to X
    if (counter == n)
        cout << "YES"
            << "\n";
    else
        cout << "NO"
            << "\n";
}
  
// Driver Code
int main()
{
    int x = 16;
  
    // The set of strings 
    // in base from [2, 36]
    vector str =
         { "10000", "20", "16" };
    checkCorrespond(str, x);
    return 0;
}


Java
// Java implementation to check
// wheather array of Strings
// can correspond to a number X
  
class GFG{
   
// Function to find the maximum 
// base possible for the number N
static int val(char c)
{
    if (c >= '0' && c <= '9')
        return (int)c - '0';
    else
        return (int)c - 'A' + 10;
}
   
// Function to find the decimal
// equivalent of the number
static int toDeci(String str, int base)
{
    int len = str.length();
    int power = 1;
    int num = 0;
    int i;
    for (i = len - 1; i >= 0; i--) {
           
        // Condition to check if the
        // number is convertible 
        // to another base
        if (val(str.charAt(i)) >= base) {
            return -1;
        }
        num += val(str.charAt(i)) * power;
        power = power * base;
    }
    return num;
}
   
// Function to check that the 
// array can correspond to a number X 
static void checkCorrespond(String[] str,
                                int x){
                                       
    // counter to count the numbers
    // those are convertible to X
    int counter = 0;
    int n = str.length;
   
    // Loop to iterate over the array
    for (int i = 0; i < n; i++) {
        for (int j = 2; j <= 36; j++) {
               
            // Convert the current String
            // to every base for checking
            // whether it will correspond
            // to X from any base
            if (toDeci(str[i], j) == x) {
                counter++;
                break;
            }
        }
    }
       
    // Condition to check if every
    // number of the array can
    // be converted to X
    if (counter == n)
        System.out.print("YES"
           + "\n");
    else
        System.out.print("NO"
           + "\n");
}
   
// Driver Code
public static void main(String[] args)
{
    int x = 16;
   
    // The set of Strings 
    // in base from [2, 36]
    String[] str =
         { "10000", "20", "16" };
    checkCorrespond(str, x);
}
}
  
// This code contributed by PrinciRaj1992


Python3
# Python3 implementation to check
# wheather array of strrings
# can correspond to a number X
  
# Function to find the maximum 
# base possible for the number N
def val(c):
    if (c >= '0' and c <= '9'):
        return int(c)
    else:
        return c - 'A' + 10
          
# Function to find the decimal
# equivalennt of the number
def toDeci(strr, base):
      
    lenn = len(strr)
    power = 1
    num = 0
    for i in range(lenn - 1, -1, -1):
          
        # Condition to check if the
        # number is convertible 
        # to another base
        if (val(strr[i]) >= base):
            return -1
          
        num += val(strr[i]) * power
        power = power * base
      
    return num
  
  
# Function to check that the 
# array can correspond to a number X 
def checkCorrespond(strr, x):
      
    # counter to count the numbers
    # those are convertible to X
    counter = 0
    n = len(strr)
      
    # Loop to iterate over the array
    for i in range(n):
        for j in range(2,37):
              
            # Convert the current strring
            # to every base for checking
            # whether it will correspond
            # to X from any base
            if (toDeci(strr[i], j) == x):
                counter += 1
                break
              
    # Condition to check if every
    # number of the array can
    # be converted to X
    if (counter == n):
        print("YES")
    else:
        print("NO")
  
# Driver Code
x = 16
  
# The set of strrings 
# in base from [2, 36]
strr = ["10000", "20", "16"]
checkCorrespond(strr, x)
  
# This code is contributed by shubhamsingh10


C#
// C# implementation to check
// wheather array of Strings
// can correspond to a number X
using System;
  
class GFG{
    
// Function to find the maximum 
// base possible for the number N
static int val(char c)
{
    if (c >= '0' && c <= '9')
        return (int)c - '0';
    else
        return (int)c - 'A' + 10;
}
    
// Function to find the decimal
// equivalent of the number
static int toDeci(String str, int Base)
{
    int len = str.Length;
    int power = 1;
    int num = 0;
    int i;
    for (i = len - 1; i >= 0; i--) {
            
        // Condition to check if the
        // number is convertible 
        // to another base
        if (val(str[i]) >= Base) {
            return -1;
        }
        num += val(str[i]) * power;
        power = power * Base;
    }
    return num;
}
    
// Function to check that the 
// array can correspond to a number X 
static void checkCorrespond(String[] str,
                                int x){
                                        
    // counter to count the numbers
    // those are convertible to X
    int counter = 0;
    int n = str.Length;
    
    // Loop to iterate over the array
    for (int i = 0; i < n; i++) {
        for (int j = 2; j <= 36; j++) {
                
            // Convert the current String
            // to every base for checking
            // whether it will correspond
            // to X from any base
            if (toDeci(str[i], j) == x) {
                counter++;
                break;
            }
        }
    }
        
    // Condition to check if every
    // number of the array can
    // be converted to X
    if (counter == n)
        Console.Write("YES"
           + "\n");
    else
        Console.Write("NO"
           + "\n");
}
    
// Driver Code
public static void Main(String[] args)
{
    int x = 16;
    
    // The set of Strings 
    // in base from [2, 36]
    String[] str =
         { "10000", "20", "16" };
    checkCorrespond(str, x);
}
}
  
// This code is contributed by Princi Singh


输出:
YES

性能分析:

  • 时间复杂度: O(N)。
  • 辅助空间: O(1)。