README.md

718. Maximum Length of Repeated Subarray

Given two integer arrays nums1 and nums2, return the maximum length of a subarray that appears in both arrays.

Example 1:

Input: nums1 = [1,2,3,2,1], nums2 = [3,2,1,4,7]
Output: 3
Explanation: The repeated subarray with maximum length is [3,2,1].

Example 2:

Input: nums1 = [0,0,0,0,0], nums2 = [0,0,0,0,0]
Output: 5
Explanation: The repeated subarray with maximum length is [0,0,0,0,0].

Constraints:

• 1 <= nums1.length, nums2.length <= 1000
• 0 <= nums1[i], nums2[i] <= 100

Solutions (Rust)

1. Solution

impl Solution {
    pub fn find_length(nums1: Vec<i32>, nums2: Vec<i32>) -> i32 {
        let mut dp = vec![vec![0; nums2.len() + 1]; nums1.len() + 1];
        let mut ret = 0;

        for i in 0..nums1.len() {
            for j in 0..nums2.len() {
                if nums1[i] == nums2[j] {
                    dp[i + 1][j + 1] = dp[i][j] + 1;
                    ret = ret.max(dp[i + 1][j + 1]);
                }
            }
        }

        ret
    }
}