Before seeing the solution make sure that you tried enough. Don’t paste the whole code, just find out the logic. If you stuck in trouble, just inform me on comment.
/**Bismillahir Rahmanir Rahim.**/ #include <stdio.h> int main() { int i, j=0, lnth; char x[60], y[60]; scanf("%s", x); for(i=0; x[i]; i++) { if(x[i]=='a' || x[i]=='e' || x[i]=='i' || x[i]=='o' || x[i]=='u') { y[j] = x[i]; j++; } } y[j] = '\0'; lnth = strlen(y); for(i=0, j=lnth-1; i < lnth; i++, j--) { if(y[i] != y[j]) { printf("N\n"); return 0; } } printf("S\n"); return 0; }
Hello . Do you know how to solve by Java?
ReplyDeletei do
Deleteim not an expert but i do it:
DeleteString in = ler.nextLine();
char A[] = in.toCharArray();
char Y[] = new char[A.length];
char X[] = new char[A.length];
int j = Y.length - 1;
for (int i = 0; i < A.length; i++) {
if (A[i] == 'a' || A[i] == 'e' || A[i] == 'i' || A[i] == 'o' || A[i] == 'u') {
Y[j] = A[i];
j--;
}
}
j = 0;
for (int i = 0; i < A.length; i++) {
if (A[i] == 'a' || A[i] == 'e' || A[i] == 'i' || A[i] == 'o' || A[i] == 'u') {
X[j] = A[i];
j++;
}
}
String y = new String(Y);
String x = new String(X);
x = x.trim();
y = y.trim();
if (x.equals(y)) {
System.out.println("S");
} else {
System.out.println("N");
}