Description Link to heading

718.maximum-length-of-repeated-subarray

Solution Link to heading

dp[i][j] denotes the length of longest public subarray of two subarray ending with nums1[i - 1], nums2[j - 1];

Recurrence formula

if (nums1[i - 1] == nums2[j - 1])
    dp[i][j] = dp[i - 1][j - 1] + 1;

Code Link to heading 

class Solution {
  public:
    int findLength(vector<int> &nums1, vector<int> &nums2) {
        vector<vector<int>> dp(nums1.size() + 1, vector<int>(nums2.size() + 1, 0));
        int m = 0;
        for (int i = 1; i <= nums1.size(); i++) {
            for (int j = 1; j <= nums2.size(); j++) {
                if (nums1[i - 1] == nums2[j - 1])
                    // dp[i][j] = max(dp[i - 1][j - 1] + 1, dp[i][j]);
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                if (dp[i][j] > m)
                    m = dp[i][j];
            }
        }
        return m;
    }
};