Simple elevator algorithm (return total stops of elevator)

Elevator.js

// A [] queue of ppl and their weights
// B [] floors each person is going to
// M maximum floors of building
// X maximum people on elv
// Y maximum wight on elv

function solution(A, B, M, X, Y) {
	// if queue had one person and is going to a reachable floor
	if (A.length === 1 && B[0] <= M) {
		return 2;
		// elif one person but heading to a floor that is out of reach
	} else if (A.length === 1 && B[0] > M ) {
		return 0;
		// if more than one person, then we have a queue [array]
	} else if (A.length > 1) {
		// Filter out (remove) those who are heading to floor > M 
		for (i=0; i<B.length; i++){
			if (B[i] > M) {
				B.splice(i,1);
				A.splice(i,1);
			}
		}
	}
	// if queue is not empty after filteration
	if (A.length > 0) {
		// onElv obj representing those who are on-elevator at any given time:
		// ttlP {p: how many people on board, d: [where each p is heading -- unique]}
		// 			ttlW: total weight of ppl on board
		// 			ttlS: will hold the amount of stops the elevator will make (will be used later on)
		var onElv = {
			ttlP: {
				p:1, 
				d:[B[0]]
			}, 
			ttlW: A[0],
			ttlS: 0
		};
		// extracting first el of array into the obj above
		A.shift(); B.shift();

		if (X > 1) { // elevator capacity allows > 1 person
			
			for (i = 0; i < A.length; i++ ) {
				// if taking on next person doesn't violate the wieght limit(Y) nor the capacity limit (X)
				if (((onElv.ttlW + A[i]) <= Y) &&
						(onElv.ttlP.p < X) ) { 
				
					// check if this new person's destination floor exists in d[]
					var exists = false, p = onElv.ttlP.p;
					onElv.ttlP.d.forEach(function(el) {
						if (el === B[i]) exists = true;
					});

					if (!exists) { // if destination floor is new (no one  on elevator is heading there)
						onElv.ttlP.d.push(B[i]); // add it to d[]
					}

					onElv.ttlP.p ++; // + person, on-boarding person
					onElv.ttlW += A[i]; // + weight, increasing total weight

				} else { //  Y or X doens't allow it, evelvator should go up, then come back down to take new ones
					onElv.ttlP.p = 1; // new number of ppl on elv *starting point*
					onElv.ttlS += onElv.ttlP.d.length + 1; // adding amount of destations(stops) of last batch/group of ppl
					onElv.ttlP.d = [B[i]]; // new dest. 
					onElv.ttlW = A[i]; // new total weight
				}
			}// end for
		}

		// adding last group's dests + 1 for coming back down
		onElv.ttlS += onElv.ttlP.d.length + 1;

		return onElv.ttlS; // return result
	
	} else { // if filtered queue/array is 0

		return 0;
	}

}

// example data below shoul return 6
// elev will take first 3 persons (ttlW: 180, d[3,2])
// elevator will go up stops at 2 (#1), 3 (#2) then comes back down (#3)
// takes last two persons goes up drops them at 2 (#4), 3 (#5) comes back down (#6)
var A = [40,40,100,80,20], 
		B = [3,3,2,2,3], 
		M = 7, X = 5, Y = 200;

console.log(solution(A,B,M,X,Y));