public class PalindromeNumber {
public static void main(String[] args) {
int x = -121;
if(isPalindrome(x)){
System.out.println(x + " is a Palindrome number");
}
else{
System.out.println(x + " is not a Palindrome number");
}
}

public static boolean isPalindrome(int x) {
if(x < 0 || (x % 10 == 0 && x != 0)) {
return false;
}

int revertedNumber = 0;
while(x > revertedNumber) {
revertedNumber = revertedNumber * 10 + x % 10;
x /= 10;
}

// When the length is an odd number, we can get rid of the middle digit by revertedNumber/10
// For example when the input is 12321, at the end of the while loop we get x = 12, revertedNumber = 123,
// since the middle digit doesn't matter in palidrome(it will always equal to itself), we can simply get rid of it.
return x == revertedNumber || x == revertedNumber/10;
}
}

Given a valid (IPv4) IP address, return a defanged version of that IP address.

A defanged IP address replaces every period "." with "[.]".



Example 1:

Input: address = "1.1.1.1"
Output: "1[.]1[.]1[.]1"
Example 2:

Input: address = "255.100.50.0"
Output: "255[.]100[.]50[.]0"


Constraints:

The given address is a valid IPv4 address.

There is a programming language with only four operations and one variable X:

++X and X++ increments the value of the variable X by 1.
--X and X-- decrements the value of the variable X by 1.
Initially, the value of X is 0.

Given an array of strings operations containing a list of operations, return the final value of X after performing all the operations.



Example 1:

Input: operations = ["--X","X++","X++"]
Output: 1
Explanation: The operations are performed as follows:
Initially, X = 0.
--X: X is decremented by 1, X = 0 - 1 = -1.
X++: X is incremented by 1, X = -1 + 1 = 0.
X++: X is incremented by 1, X = 0 + 1 = 1.
Example 2:

Input: operations = ["++X","++X","X++"]
Output: 3
Explanation: The operations are performed as follows:
Initially, X = 0.
++X: X is incremented by 1, X = 0 + 1 = 1.
++X: X is incremented by 1, X = 1 + 1 = 2.
X++: X is incremented by 1, X = 2 + 1 = 3.
Example 3:

Input: operations = ["X++","++X","--X","X--"]
Output: 0
Explanation: The operations are performed as follows:
Initially, X = 0.
X++: X is incremented by 1, X = 0 + 1 = 1.
++X: X is incremented by 1, X = 1 + 1 = 2.
--X: X is decremented by 1, X = 2 - 1 = 1.
X--: X is decremented by 1, X = 1 - 1 = 0.


Constraints:

1 <= operations.length <= 100
operations[i] will be either "++X", "X++", "--X", or "X--".

1480. Running Sum of 1d Array
Easy
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Given an array nums. We define a running sum of an array as runningSum[i] = sum(nums[0]…nums[i]).

Return the running sum of nums.



Example 1:

Input: nums = [1,2,3,4]
Output: [1,3,6,10]
Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4].
Example 2:

Input: nums = [1,1,1,1,1]
Output: [1,2,3,4,5]
Explanation: Running sum is obtained as follows: [1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1].
Example 3:

Input: nums = [3,1,2,10,1]
Output: [3,4,6,16,17]


Constraints:

1 <= nums.length <= 1000
-10^6 <= nums[i] <= 10^6

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