LeetCode in Kotlin
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
Explanation: The repeated subarray with maximum length is [0,0,0,0,0].
Constraints:
1 <= nums1.length, nums2.length <= 10000 <= nums1[i], nums2[i] <= 100
Solution
class Solution {
fun findLength(nums1: IntArray, nums2: IntArray): Int {
val m = nums1.size
val n = nums2.size
var max = 0
val dp = Array(m + 1) { IntArray(n + 1) }
for (i in 1..m) {
for (j in 1..n) {
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
}
}