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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.
1#include <vector>
2#include <algorithm>
3
4using namespace std;
5
6int findLength(vector<int>& nums1, vector<int>& nums2) {
7 int maxLen = 0;
8 vector<vector<int>> dp(nums1.size() + 1, vector<int>(nums2.size() + 1, 0));
9 for (int i = 1; i <= nums1.size(); i++) {
10 for (int j = 1; j <= nums2.size(); j++) {
11 if (nums1[i - 1] == nums2[j - 1]) {
12 dp[i][j] = dp[i - 1][j - 1] + 1;
13 maxLen = max(maxLen, dp[i][j]);
14 }
15 }
16 }
17 return maxLen;
18}This C++ implementation uses a vector of vectors for the DP table. Similar logic is applied to update the table and track the maximum length as in the C solution.
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
The C solution selectively shifts and iterates over subarrays, checking integer matches directly as the two pointers overlap, recalculating lengths and ensuring a memory-efficient solution.