Solution :
check_palindrome
lower__string
string
_lower()
Let stack = new Stack()
Let queue = new Queue();
for each character c in lower_case_string:
stack.push(c);
queue.enqueue(c);
let isPalindrome = true;
while queue is not empty {
if (queue.remove().equals(stack.pop())) {
continue;
} else {
isPalindrome=false;
break while loop;
}
}
return isPalindrome
Input = aabb
output = true
input =abcd
output = false
Hello, since you did not specify a programming language, I wrote this algorithm in C++. Good luck!
<h2>Code:</h2>
#include <iostream>
#include <vector>
std::vector<int> v;
int main(int argc, char* argv[]) {
while(1) {
int temp;
std::cout << "\nEnter a number: ";std::cin>>temp;
if(temp<0) {
std::cout << "\nEven number(s) is/are:\n---------------------\n";
for(int i=0;i<v.size();i++) {
if(v.at(i)%2==0) std::cout << v[i] << " ";
else continue;
}
std::cout << std::endl;
break;
}else {
v.push_back(temp);
}
}
return 0;
}
