Given two integer arrays A and B, return the maximum length of an subarray that appears in both arrays.
Example 1:
Input:
A: [1,2,3,2,1]
B: [3,2,1,4,7]
Output: 3
Explanation:
The repeated subarray with maximum length is [3, 2, 1].
Note:
1 <= len(A), len(B) <= 1000
0 <= A[i], B[i] < 100
Solutions
My solution
class Solution {
public:
int findLength(vector<int>& A, vector<int>& B) {
int n = A.size();
int m = B.size();
if(n == 0 || m == 0) return 0;
// dp[i][j]: length of the longest common subarray of A[0..i] and B[0..j] ending with A[i] and B[j]
vector<vector<int>> dp(n, vector<int>(m));
int len = 0;
for(int j = 0; j < m; ++j){
if(A[0] == B[j]){
len = dp[0][j] = 1;
}
}
for(int i = 0; i < n; ++i){
if(A[i] == B[0]){
len = dp[i][0] = 1;
}
}
for(int i = 1; i < n; ++i){
for(int j = 1; j < m; ++j){
if(A[i] == B[j]){
dp[i][j] = dp[i - 1][j - 1] + 1;
len = max(len, dp[i][j]);
}
}
}
return len;
}
};
More concise implementation
class Solution {
public:
int findLength(vector<int>& A, vector<int>& B) {
int n = A.size();
int m = B.size();
// dp[i][j]: length of longest common subarray of A[0..i-1] and B[0..j-1] ending with A[i-1] and B[j-1]
vector<vector<int>> dp(n + 1, vector<int>(m + 1));
int len = 0;
for(int i = 1; i <= n; ++i){
for(int j = 1; j <= m; ++j){
if(A[i - 1] == B[j - 1]){
dp[i][j] = dp[i - 1][j - 1] + 1;
len = max(len, dp[i][j]);
}
}
}
return len;
}
};
