String matching time complexity problem (Asked in Castle Global)

castle-global.js

const problemSolution = function(str){
	let val = str.split(''),
		stack = [],
		count = 0;
	stack.push(val[0]);
	for(let i=1; i< val.length; i++) {
		if((/[a-z]/.test(stack[stack.length-1])) && (/[a-z]/.test(val[i]))) {
			console.log('1'+ val[i]);
			return count;
		}
		else if((/[A-Z]/.test(stack[stack.length-1])) && (/[A-Z]/.test(val[i]))) {
			console.log('2'+ val[i]);
			stack.push(val[i]);
			++count;
		}
		else if(stack[stack.length-1] && (/[a-z]/.test(val[i])) && stack[stack.length-1].toLowerCase() === val[i]) {
			console.log('3'+val[i]);
			++count;
			stack.pop();
		}
		else {
			console.log('4'+ val[i]);
			stack.push(val[i]);
		}
	}
	return count;
}

String Matching problem with s set of rules
i). Count 1 if an Upper case letter is followed by any Upper case letter

ii). Count 1 if an Upper case letter is followed by its same Lower case letter and dissolve the pair. (After dissolving again check condition 2 if it needs to be dissolved again)

iii). if two lower case letters are present without/with a match stop and return count

Example:

ABba
count= 3

gGdLLgCDBbdcf
count= 6