g0701_0800.s0718_maximum_length_of_repeated_subarray.Solution Maven / Gradle / Ivy

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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0701_0800.s0718_maximum_length_of_repeated_subarray;

// #Medium #Array #Dynamic_Programming #Binary_Search #Sliding_Window #Hash_Function #Rolling_Hash
// #2022_03_24_Time_58_ms_(77.21%)_Space_78.8_MB_(32.62%)

/**
 * 718 - Maximum Length of Repeated Subarray\.
 *
 * Medium
 *
 * 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
 *
 * **Constraints:**
 *
 * *   `1 <= nums1.length, nums2.length <= 1000`
 * *   `0 <= nums1[i], nums2[i] <= 100`
**/
public class Solution {
    public int findLength(int[] nums1, int[] nums2) {
        int m = nums1.length;
        int n = nums2.length;
        int max = 0;
        int[][] dp = new int[m + 1][n + 1];
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (nums1[i - 1] == nums2[j - 1]) {
                    dp[i][j] = 1 + dp[i - 1][j - 1];
                    max = Math.max(max, dp[i][j]);
                }
            }
        }
        return max;
    }
}