As the question states, I am trying to solve a leetcode problem. The solutions are available online but, I want to implement my own solution. I have built my logic. Logic is totally fine. However, I am unable to optimize the code as the time limit is exceeding for the large numbers.
Here's my code:
let count = 0;
const climbingStairs = (n, memo = [{stairs: null}]) => {
if(n === memo[n]) {
count = memo[n].stairs;
}
if(n < 0) return;
if(n === 0) return memo[n].stairs = count ;
memo[n] = climbingStairs(n - 1, memo) climbingStairs(n - 2, memo);
return memo[n];
}
climbingStairs(20); //running fine on time
climbingStairs(40); //hangs as the code isn't optimized
console.log(count); //the output for the given number
The code optimization using the memoization object is not working. I have tried multiple ways but still, facing issues. Any help would be appreciated in optimizing the code. Thanks!
CodePudding user response:
Actually, you do not store a value, but NaN
to the array.
You need to return zero to get a numerical value for adding.
Further more, you assign in each call a new value, even if you already have this value in the array.
A good idea is to use only same types (object vs number) in the array and not mixed types, because you need a differen hndling for each type.
const climbingStairs = (n, memo = [1]) => {
if (n < 0) return 0;
return memo[n] ??= climbingStairs(n - 1, memo) climbingStairs(n - 2, memo);
}
console.log(climbingStairs(5));
console.log(climbingStairs(20));
console.log(climbingStairs(40));
CodePudding user response:
no need for count value, you can memoize this way:
const climbStairs = (n, memo = []) => {
if(n <= 2) return n;
if(memo[n]) {
return memo[n];
}
memo[n] = climbStairs(n - 1, memo) climbStairs(n - 2, memo);
return memo[n];
}