When adding cache to my longest increasing subsequence solution, answer is not correct?

I am working on https://leetcode.com/problems/longest-increasing-subsequence

For test case

input: [5, 4, 19, 5, 7, 12]; output: 4

Without the cache, I am able to output 4 correctly.

var lengthOfLIS = function (ns) {
  if (ns.length === 1) return 1;

  const prev_i = -1;
  const curr_i = 0;
  const res = 0;
  return recur(ns, prev_i, curr_i, res);
};

// good
var recur = function (ns, prev_i, curr_i, res) {
  //
  if (curr_i >= ns.length) {
    return res;
  }

  // take
  let out_1 = 0;
  if (prev_i === -1) {
    out_1 = recur(ns, curr_i, curr_i + 1, res + 1);
  } else if (ns[curr_i] > ns[prev_i]) {
    out_1 = recur(ns, curr_i, curr_i + 1, res + 1);
  } else if (ns[curr_i] <= ns[prev_i]) {
    out_1 = res;
  }

  // !take
  let out_2 = 0;
  out_2 = recur(ns, prev_i, curr_i + 1, res);

  const max = Math.max(out_1, out_2);
  return max;
};

When I adding cache to it, my output is 3.

var recur = function (dp, ns, prev_i, curr_i, res) {
  //
  if (curr_i >= ns.length) {
    return res;
  }

  if (prev_i !== -1 && dp[prev_i + 1][curr_i + 1] !== undefined) {
    return dp[prev_i + 1][curr_i + 1];
  }

  // take
  let out_1 = 0;
  if (prev_i === -1) {
    out_1 = recur(dp, ns, curr_i, curr_i + 1, res + 1);
  } else if (ns[curr_i] > ns[prev_i]) {
    out_1 = recur(dp, ns, curr_i, curr_i + 1, res + 1);
  } else if (ns[curr_i] <= ns[prev_i]) {
    out_1 = res;
  }

  // !take
  let out_2 = 0;
  out_2 = recur(dp, ns, prev_i, curr_i + 1, res);

  const max = Math.max(out_1, out_2);
  dp[prev_i + 1][curr_i + 1] = max;
  return max;
};

var lengthOfLIS = function (ns) {
  if (ns.length === 1) return 1;

  const dp = Array(ns.length + 1)
    .fill(undefined)
    .map((_, i) => {
      return Array(ns.length + 1).fill(undefined);
    });

  const prev_i = -1;
  const curr_i = 0;
  const res = 0;
  return recur(dp, ns, prev_i, curr_i, res);
};

Could someone point out any mistake I made? I am trying to use debugger to see what is happening.