Dada una array de enteros positivos, encuentre todos los alfabetos que podamos hacer usando elementos de la array.
Ejemplos:
Input : arr[] = {6, 5}
Output : A
In this example, we can take 6 and 5
as 65 that gives us A.
Input : arr[] = {5, 6, 6}
Output : A B
Here we can take 6 and 5 as 65 to make A.
And 6 and 6 as 66 to make B.
Input : arr[] = {1, 0, 1, 0}
Output: d e n
Here we can take 1, 0 and 0 as 100 to make d.
1, 0 and 1 as 101 to make e. And
1, 1, and 0 as 110 to make n.
Enfoque: el enfoque consiste en iterar un bucle desde el carácter A hasta la Z y verificar si ambos dígitos de sus valores ASCII (el valor ASCII de A es 65 y los dígitos del valor ASCII para A son 6 y 5) existen en la array o no. Para esto, tenemos que calcular la frecuencia de cada elemento de la array. Igual que para los alfabetos pequeños.
A continuación se muestra la implementación del enfoque anterior.
C++
// CPP program to find all the capital characters
// we can make from the given array.
#include <bits/stdc++.h>
using namespace std;
// Function returns all the capital characters
// we can make.
string findCharacters(int arr[], int n)
{
string ans = "";
int brr[10] = { 0 };
// Calculating the frequency of the elements.
for (int i = 0; i < n; i++)
brr[arr[i]]++;
// Checking for capital alphabets.
for (char ch = 'A'; ch <= 'Z'; ch++)
{
// Storing one digit in x and other in y.
int x = ch / 10;
int y = ch % 10;
brr[x]--;
brr[y]--;
// If both the digits exist.
if (brr[x] >= 0 && brr[y] >= 0)
{
// Putting in the result
ans += ch;
ans += ' ';
}
brr[x]++;
brr[y]++;
}
// Checking for alphabets a, b and c.
for (char ch = 'a'; ch <= 'c'; ch++)
{
int x = ch / 10;
int y = ch % 10;
brr[x]--;
brr[y]--;
// If all the digits exist.
if (brr[x] >= 0 && brr[y] >= 0)
{
// Putting in the result
ans += ch;
ans += ' ';
}
brr[x]++;
brr[y]++;
}
// Checking for d to z.
for (char ch = 'd'; ch <= 'z'; ch++)
{
int x = (ch / 10) / 10;
int y = (ch / 10) % 10;
int z = ch % 10;
brr[x]--;
brr[y]--;
brr[z]--;
// If all the digits exist.
if (brr[x] >= 0 && brr[y] >= 0 && brr[z] >= 0)
{
// Putting in the result
ans += ch;
ans += ' ';
}
brr[x]++;
brr[y]++;
brr[z]++;
}
return ans;
}
// Driver function
int main()
{
int arr[] = { 5, 6, 6 }, n;
n = sizeof(arr) / sizeof(arr[0]);
cout << findCharacters(arr, n) << endl;
}
Java
// Java program to find all the capital characters
// we can make from the given array.
import java.util.*;
import java.lang.*;
public class GeeksforGeeks
{
// Function returns all the capital characters we can make.
public static String findCharacters(int arr[], int n)
{
String ans = "";
int[] brr = new int[10];
// Calculating the frequency of the elements.
for (int i = 0; i < n; i++)
brr[arr[i]]++;
for (char ch = 'A'; ch <= 'Z'; ch++)
{
// Storing one digit in x and other in y.
int x = ch / 10;
int y = ch % 10;
brr[x]--;
brr[y]--;
// If both the digits exist.
if (brr[x] >= 0 && brr[y] >= 0)
{
// Putting in the result
ans += ch;
ans += ' ';
}
brr[x]++;
brr[y]++;
}
// Checking for alphabets a, b and c.
for (char ch = 'a'; ch <= 'c'; ch++)
{
int x = ch / 10;
int y = ch % 10;
brr[x]--;
brr[y]--;
// If all the digits exist.
if (brr[x] >= 0 && brr[y] >= 0)
{
// Putting in the result
ans += ch;
ans += ' ';
}
brr[x]++;
brr[y]++;
}
// Checking for d to z.
for (char ch = 'd'; ch <= 'z'; ch++)
{
int x = (ch / 10) / 10;
int y = (ch / 10) % 10;
int z = ch % 10;
brr[x]--;
brr[y]--;
brr[z]--;
// If all the digits exist.
if (brr[x] >= 0 && brr[y] >= 0 && brr[z] >= 0)
{
// Putting in the result
ans += ch;
ans += ' ';
}
brr[x]++;
brr[y]++;
brr[z]++;
}
return ans;
}
// Driver function
public static void main(String argc[])
{
int arr[] = { 5, 6, 6 }, n;
n = 3;
System.out.println(findCharacters(arr, n));
}
}
Python
# Python3 program to find all the capital characters
# we can make from the given array.
# Function returns all the capital characters
# we can make.
def findCharacters(arr, n):
ans = ""
brr = [0] * 10
# Calculating the frequency of the elements.
for i in range(n):
brr[arr[i]] += 1
# Checking for capital alphabets.
for ch in range(ord('A'), ord('Z') + 1):
# Storing one digit in x and other in y.
x = ch // 10
y = ch % 10
brr[x] -= 1
brr[y] -= 1
# If both the digits exist.
if (brr[x] >= 0 and brr[y] >= 0):
# Putting in the result
ans += chr(ch)
ans += ' '
brr[x] += 1
brr[y] += 1
# Checking for alphabets a, b and c.
for ch in range(ord('a'), ord('c') + 1):
x = ch // 10
y = ch % 10
brr[x] -= 1
brr[y] -= 1
# If all the digits exist.
if (brr[x] >= 0 and brr[y] >= 0):
# Putting in the result
ans += chr(ch)
ans += ' '
brr[x] += 1
brr[y] += 1
# Checking for d to z.
for ch in range(ord('d'), ord('z') + 1):
x = (ch // 10) // 10
y = (ch // 10) % 10
z = ch % 10
brr[x] -= 1
brr[y] -= 1
brr[z] -= 1
# If all the digits exist.
if (brr[x] >= 0 and brr[y] >= 0 and brr[z] >= 0):
# Putting in the result
ans += chr(ch)
ans += ' '
brr[x] += 1
brr[y] += 1
brr[z] += 1
return ans
# Driver function
arr = [5, 6, 6]
n = len(arr)
print(findCharacters(arr, n))
# This code is contributed by mohit kumar 29
C#
// C# program to find all the capital
// characters we can make from the given array.
using System;
class GeeksforGeeks
{
// Function returns all the capital
// characters we can make.
public static String findCharacters(int []arr, int n )
{
String ans = "";
int []brr = new int[10];
// Calculating the frequency of the elements.
for (int i = 0; i < n; i++)
brr[arr[i]]++;
for (char ch = 'A'; ch <= 'Z'; ch++)
{
// Storing one digit in x and other in y.
int x = ch / 10;
int y = ch % 10;
brr[x]--;
brr[y]--;
// If both the digits exist.
if (brr[x] >= 0 && brr[y] >= 0)
{
// Putting in the result
ans += ch;
ans += ' ';
}
brr[x]++;
brr[y]++;
}
// Checking for alphabets a, b and c.
for (char ch = 'a'; ch <= 'c'; ch++)
{
int x = ch / 10;
int y = ch % 10;
brr[x]--;
brr[y]--;
// If all the digits exist.
if (brr[x] >= 0 && brr[y] >= 0)
{
// Putting in the result
ans += ch;
ans += ' ';
}
brr[x]++;
brr[y]++;
}
// Checking for d to z.
for (char ch = 'd'; ch <= 'z'; ch++)
{
int x = (ch / 10) / 10;
int y = (ch / 10) % 10;
int z = ch % 10;
brr[x]--;
brr[y]--;
brr[z]--;
// If all the digits exist.
if (brr[x] >= 0 && brr[y] >= 0 && brr[z] >= 0)
{
// Putting in the result
ans += ch;
ans += ' ';
}
brr[x]++;
brr[y]++;
brr[z]++;
}
return ans;
}
// Driver function
public static void Main()
{
int []arr = {5, 6, 6};
int n = 3;
Console.Write(findCharacters(arr, n));
}
}
// This code is contributed by nitin mittal.
Javascript
<script>
// JavaScript program to find
// all the capital characters
// we can make from the given array.
// Function returns all the capital
// characters we can make.
function findCharacters(arr,n)
{
let ans = "";
let brr = new Array(10);
for(let i=0;i<brr.length;i++)
{
brr[i]=0;
}
// Calculating the frequency of the elements.
for (let i = 0; i < n; i++)
brr[arr[i]]++;
for (let ch = 'A'.charCodeAt(0);
ch <= 'Z'.charCodeAt(0); ch++)
{
// Storing one digit in x and other in y.
let x = Math.floor(ch / 10);
let y = ch % 10;
brr[x]--;
brr[y]--;
// If both the digits exist.
if (brr[x] >= 0 && brr[y] >= 0)
{
// Putting in the result
ans += String.fromCharCode(ch);
ans += ' ';
}
brr[x]++;
brr[y]++;
}
// Checking for alphabets a, b and c.
for (let ch = 'a'.charCodeAt(0);
ch <= 'c'.charCodeAt(0); ch++)
{
let x = Math.floor(ch / 10);
let y = ch % 10;
brr[x]--;
brr[y]--;
// If all the digits exist.
if (brr[x] >= 0 && brr[y] >= 0)
{
// Putting in the result
ans += String.fromCharCode(ch);
ans += ' ';
}
brr[x]++;
brr[y]++;
}
// Checking for d to z.
for (let ch = 'd'.charCodeAt(0);
ch <= 'z'.charCodeAt(0); ch++)
{
let x = Math.floor((ch / 10) / 10);
let y = (ch / 10) % 10;
let z = ch % 10;
brr[x]--;
brr[y]--;
brr[z]--;
// If all the digits exist.
if (brr[x] >= 0 && brr[y] >= 0 &&
brr[z] >= 0)
{
// Putting in the result
ans += String.fromCharCode(ch);
ans += ' ';
}
brr[x]++;
brr[y]++;
brr[z]++;
}
return ans;
}
// Driver function
let arr=[5, 6, 6 ];
let n = 3;
document.write(findCharacters(arr, n));
// This code is contributed by patel2127
</script>
Producción
A B
Complejidad de tiempo: O(n)
Publicación traducida automáticamente
Artículo escrito por Sagar Shukla y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA