¿En cuales situaciones concretas se puede aplicar las técnicas de conteo(particularmente las permutuaciones y las combinaciones)
1 answer:
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Answer:
Program.java
import java.util.Scanner;
public class Program {
public static boolean isPalindrome(String str){
int start = 0, end = str.length()-1;
while (start < end){
if(str.charAt(start) != str.charAt(end)){
return false;
}
start++;
end--;
}
return true;
}
public static int countPalindromePairs(String[] inarr){
int count = 0;
for(int i=0; i<inarr.length; i++){
for(int j=i+1; j<inarr.length; j++){
StringBuilder sb = new StringBuilder();
sb.append(inarr[i]).append(inarr[j]);
if(isPalindrome(sb.toString())){
count++;
}
}
}
return count == 0 ? -1 : count;
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String line = sc.next();
String[] inarr = line.split(",");
int count = countPalindromePairs(inarr);
System.out.println("RESULT: "+count);
}
}
Explanation:
OUTPUT:
Answer:
The given problem is solved using language - Python.
def f(x):
new_list=[]
for i in x:
if i%2==0:
new_list.append(i//2)
else:
new_list.append(i*2)
return new_list
my_list=list(range(1,6))
print('Original List:',my_list)
my_list=f(my_list)
print('Modified List:',my_list)
- Create a new list.
- Iterate over the list passed into the function.
- Check if the element is even or not. If true, append half the value of element in the list.
- If false, append twice the value of the element in the list.
- At last, return the new list.
There is another way of doing this - By using map() function.
—————————————————————————————
def f(x):
return list(map(lambda x:x//2 if x%2==0 else 2*x,x))
my_list=list(range(1,6))
print('Original List:',my_list)
my_list=f(2my_list)
print('Modified List:',my_list)
—————————————————————————————
Original List: [1, 2, 3, 4, 5]
Modified List: [2, 1, 6, 2, 10]