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This approach uses a 2D DP array where dp[i][j] represents the length of the longest common subarray ending at nums1[i-1] and nums2[j-1]. Initialize the DP table with zeros. If nums1[i-1] == nums2[j-1], set dp[i][j] = dp[i-1][j-1] + 1, otherwise set it to 0. Track the maximum length found throughout the process.
Time Complexity: O(n * m), where n and m are the lengths of nums1 and nums2, respectively. Space Complexity: O(n * m) due to the DP table.
1class Solution:
2 def findLength(self, nums1: List[int], nums2: List[int]) -> int:
3 max_len = 0
4 dp = [[0] * (len(nums2) + 1) for _ in range(len(nums1) + 1)]
5 for i in range(1, len(nums1) + 1):
6 for j in range(1, len(nums2) + 1):
7 if nums1[i - 1] == nums2[j - 1]:
8 dp[i][j] = dp[i - 1][j - 1] + 1
9 max_len = max(max_len, dp[i][j])
10 return max_lenPython solution uses a list of lists to implement the DP table. For each pair of indices, it checks for matching elements and updates the max length accordingly.
This approach involves using a sliding window over the two arrays with the help of a hashing method to check potential matches of subarrays. By varying the window size, you can find the length of the longest matching subarray without needing a full DP table.
Time Complexity: O(n * m), Space Complexity: O(1).
1 public int FindLength(int[] nums1, int[] nums2) {
int maxLen = 0;
for (int k = 0; k < nums1.Length; k++) {
int len = 0;
for (int i = k, j = 0; i < nums1.Length && j < nums2.Length; i++, j++) {
if (nums1[i] == nums2[j]) {
len++;
maxLen = Math.Max(maxLen, len);
} else {
len = 0;
}
}
}
for (int k = 0; k < nums2.Length; k++) {
int len = 0;
for (int i = 0, j = k; i < nums1.Length && j < nums2.Length; i++, j++) {
if (nums1[i] == nums2[j]) {
len++;
maxLen = Math.Max(maxLen, len);
} else {
len = 0;
}
}
}
return maxLen;
}
}By initializing two sliding match-checks, one per offset array, the C# solution ensures overlap is tracked with max len updates and handled directly via looping conditions.