LeetCode: Maximum Length of Repeated Subarray Solution

Approach

recursion(i, j) returns longest common post-fix of nums1.slice(0, i) and nums2.slice(0, j)

Result will be max(recursion(i, j)) all over i, j pairs

Implementation

1/**
2 * @param {number[]} nums1
3 * @param {number[]} nums2
4 * @return {number}
5 */
6var findLength = function (nums1, nums2) {
7  const [m, n] = [nums1.length, nums2.length]
8  const memo = Array.from({ length: m }, _ => Array(n).fill(null))
9
10  const recursion = (i, j) => {
11    if (i < 0 || j < 0) return 0
12    if (memo[i][j] !== null) return memo[i][j]
13    const res = nums1[i] === nums2[j] ? 1 + recursion(i - 1, j - 1) : 0
14    return (memo[i][j] = res)
15  }
16
17  let res = 0
18
19  for (let i = 0; i < m; i++) {
20    for (let j = 0; j < n; j++) {
21      res = Math.max(res, recursion(i, j))
22    }
23  }
24
25  return res
26}