718. Maximum Length of Repeated Subarray LeetCode Solution

In this guide, you will get 718. Maximum Length of Repeated Subarray LeetCode Solution with the best time and space complexity. The solution to Maximum Length of Repeated Subarray problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Maximum Length of Repeated Subarray solution in C++
Maximum Length of Repeated Subarray solution in Java
Maximum Length of Repeated Subarray solution in Python
Additional Resources

718. Maximum Length of Repeated Subarray LeetCode Solution image

Problem Statement of 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

Complexity Analysis

Time Complexity: O(mn)
Space Complexity: O(mn)

718. Maximum Length of Repeated Subarray LeetCode Solution in C++

class Solution {
 public:
  int findLength(vector<int>& nums1, vector<int>& nums2) {
    const int m = nums1.size();
    const int n = nums2.size();
    int ans = 0;
    // dp[i][j] := the maximum length of a subarray that appears in both
    // nums1[i..m) and nums2[j..n)
    vector<vector<int>> dp(m + 1, vector<int>(n + 1));

    for (int i = m - 1; i >= 0; --i)
      for (int j = n - 1; j >= 0; --j)
        if (nums1[i] == nums2[j]) {
          dp[i][j] = dp[i + 1][j + 1] + 1;
          ans = max(ans, dp[i][j]);
        }

    return ans;
  }
};
/* code provided by PROGIEZ */

718. Maximum Length of Repeated Subarray LeetCode Solution in Java

class Solution {
  public int findLength(int[] nums1, int[] nums2) {
    final int m = nums1.length;
    final int n = nums2.length;
    int ans = 0;
    // dp[i][j] := the maximum length of a subarray that appears in both
    // nums1[i..m) and nums2[j..n)
    int[][] dp = new int[m + 1][n + 1];

    for (int i = m - 1; i >= 0; --i)
      for (int j = n - 1; j >= 0; --j)
        if (nums1[i] == nums2[j]) {
          dp[i][j] = dp[i + 1][j + 1] + 1;
          ans = Math.max(ans, dp[i][j]);
        }

    return ans;
  }
}
// code provided by PROGIEZ

718. Maximum Length of Repeated Subarray LeetCode Solution in Python

class Solution:
  def findLength(self, nums1: list[int], nums2: list[int]) -> int:
    m = len(nums1)
    n = len(nums2)
    ans = 0
    # dp[i][j] := the maximum length of a subarray that appears in both
    # nums1[i..m) and nums2[j..n)
    dp = [[0] * (n + 1) for _ in range(m + 1)]

    for i in reversed(range(m)):
      for j in reversed(range(n)):
        if nums1[i] == nums2[j]:
          dp[i][j] = dp[i + 1][j + 1] + 1
          ans = max(ans, dp[i][j])

    return ans
 # code by PROGIEZ