Subsequence with difference of 1 (Memoization)

Given a list of numbers, find the longest subsequence such that the absolute difference between adjacent elements is 1.

Hint

This algorithm uses memoization to improve performance. It's same as the recursive version of Subsequence with difference of 1, but with a special 'cache' to store results. The 'cache key' is crucial. It's how we identify and store past results. Sometimes we need one key, sometimes multiple keys. We add code to:

• Initialize cache to store results.
• Using the key to check cached result. If the result is not found in the cache, do the calculation and save the result in cache.
• Return the cached result.

• Python
• Javascript
•  Run

cache = None

def count(ls, i, prev):
  if i >= len(ls):
    return 0

  global cache
  if not cache:
    cache = [{} for _ in range(len(ls))]

  if prev not in cache[i]:
    skip = count(ls, i + 1, prev)

    take = 0
    if prev == -1 or abs(ls[i] - ls[prev]) == 1:
      take = 1 + count(ls, i + 1, i)

    cache[i][prev] = max(skip, take)
  
  return cache[i][prev]

ls = [10, 9, 4, 5, 4, 8, 6]

print(count(ls, 0, -1))

let cache = null;

function count(list, i, prev) {
  if (i >= list.length) {
    return 0;
  }

  if (!cache) {
    cache = Array(list.length).fill({});
  }

  if (cache[i][prev] === undefined) {
    let skip = count(list, i + 1, prev);

    let take = 0;
    if (prev === -1 || Math.abs(list[i] - list[prev]) === 1) {
      take = 1 + count(list, i + 1, i);
    }

    cache[i][prev] = Math.max(skip, take);
  }

  return cache[i][prev];
}

const list = [10, 9, 4, 5, 4, 8, 6];

console.log(count(list, 0, -1));