funnyString.ts

/*
In this challenge, you will determine whether a string is funny or not. To determine whether a string is funny, create a copy of the string in reverse e.g. .
Iterating through each string, compare the absolute difference in the ascii values of the characters at positions 0 and 1, 1 and 2 and so on to the end.
If the list of absolute differences is the same for both strings, they are funny.

Determine whether a give string is funny. If it is, return Funny, otherwise return Not Funny.

s = 'lmnop'

The ordinal values of the charcters are [108, 109, 110, 111, 112]. s' = 'ponml' and the ordinals are [112, 111, 110, 109, 108].
The absolute differences of the adjacent elements for both strings are [1, 1, 1, 1], so the answer is Funny.

Sample Input

STDIN   Function
-----   --------
2       q = 2
acxz    s = 'acxz'
bcxz    s = 'bcxz'
Sample Output

Funny
Not Funny
*/

// First, we will create a list to save the absolute different of adjacent elements for the original string
// Ex: [108, 109, 110, 111, 112]
// After that, we will run a for loop to the haft of the list and compare follow the rule
// If all of the values has l[i] === l[length of list - i] => It is funny string
// Otherwise, it is not a funny string
function funnyString(s: string) {
	// Approach 1: Time: O(n), Space: O(n)
	// const differenceList = [];
	// for (let i = 0; i < s.length - 1; i++) {
	//   differenceList.push(Math.abs(s.charCodeAt(i + 1) - s.charCodeAt(i)));
	// }
	// for (let i = 0; i < differenceList.length / 2; i++) {
	//   if (differenceList[i] !== differenceList[differenceList.length - i - 1]) {
	//     return 'Not Funny';
	//   }
	// }
	// return 'Funny';

	// Approach 2: Time O(n), Space: O(1)
	// In this approach, we will need 2 pointer
	// The first pointer (left) for the loop from the index 0 to s.length / 2
	// The second pointer (right) for the loop from the index s.length - 2 to s.length / 2
	// We will use the while loop, with condition is left <= right
	// And compare the absolute of the different of the adjacent elements between the haft left and haft right

	let left = 0;
	let right = s.length - 2;
	let leftDifferent: number;
	let rightDifferent: number;

	while (left <= right) {
		leftDifferent = Math.abs(s.charCodeAt(left) - s.charCodeAt(left + 1));
		rightDifferent = Math.abs(s.charCodeAt(right) - s.charCodeAt(right + 1));

		if (leftDifferent !== rightDifferent) {
			return 'Not Funny';
		}

		left++;
		right--;
	}

	return 'Funny';
}

console.log(funnyString('bcxz'));