Leetcode | Maximum Length of Repeated Subarray

718. Maximum Length of Repeated Subarray

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

! subarray, not subsequence

I. DP,

Similar to LCS

Time Complexity: O(LAx LB)

Space Complexity: O(LAx LB)

//c++
class Solution {
public:
int findLength(vector<int>& A, vector<int>& B) {
int l = A.size(),ans = 0;
vector<vector<int>> dp(l,vector<int>(l,0));
for (int i=0;i<l;i++){
for (int j=0;j<l;j++){
if(i == 0 || j == 0) dp[i][j] = (A[i]==B[j]);
else dp[i][j] = (A[i]==B[j]) ? dp[i-1][j-1]+1:0;
ans = max(ans,dp[i][j]);
}
}
return ans;
}
};

II. Optimization

DP

Time Complexity: O(LAx LB)

Space Complexity: O(LA)

//c++
class Solution {
public:
int findLength(vector<int>& A, vector<int>& B) {
int m = A.size();
int n = B.size();
vector<int> dp(n + 1);
int result = 0;
for (int i = 1; i <= m; ++i) {
vector<int> ndp(n + 1);
for (int j = 1; j <= n; ++j) {
if (A[i-1] == B[j-1]) {
ndp[j] = dp[j-1] + 1;
result = max(ndp[j], result);
}
}
dp = ndp;
}
return result;
}
};

III. Optimization 2

DP

Time Complexity: O(LAx LB)

Space Complexity: O(1)

since dp[i][j] only depends on dp[i-1][j-1], we only need to keep it and traverse the matrix diagonally.

//Java
public int findLength(int[] A, int[] B) {
int maxLen = 0;
for (int j = 0; j < B.length; j++) {
int maxLenEnding = 0;
for (int i = 0, k = j; i < A.length && k < B.length; i++, k++) {
if (A[i] != B[k]) maxLenEnding = 0;
else {
maxLenEnding++;
maxLen = Math.max(maxLen, maxLenEnding);
}
}
}

for (int i =1; i < A.length; i++) {
int maxLenEnding = 0;
for (int j = 0, k = i; k < A.length && j < B.length; j++, k++) {
if (A[k] != B[j]) maxLenEnding = 0;
else {
maxLenEnding++;
maxLen = Math.max(maxLen, maxLenEnding);
}
}
}
return maxLen;
}

IV. HashMap

//c++
struct Node {
int len;
Node *ch[101], *f;
Node() : len(0), f(NULL) { memset(ch, 0, sizeof ch); }
};
struct SAM {
Node *last, *root;
SAM() { root = last = new Node(); };
void add(int c) {
Node *e = new Node(), *tmp = last;
e->len = last->len + 1;
for (; tmp && !tmp->ch[c]; tmp = tmp->f)
tmp->ch[c] = e;
if (!tmp) {
e->f = root;
} else {
Node *nxt = tmp->ch[c];
if (tmp->len + 1 == nxt->len)
e->f = nxt;
else {
Node *np = new Node();
*np = *nxt;
np->len = tmp->len + 1;
nxt->f = e->f = np;
for (; tmp && tmp->ch[c] == nxt; tmp = tmp->f)
tmp->ch[c] = np;
}
}
last = e;
}
};

class Solution {
public:
int findLength(vector<int> &A, vector<int> &B) {
SAM a = SAM();
for (int x : A) a.add(x);
int ans = 0, cur = 0;
Node *t = a.root;
for (int i = 0; i < B.size() && t; ++i) {
int idx = B[i];
while (t != a.root && !t->ch[idx]) {
t = t->f;
cur = t->len;
}
if (t->ch[idx]) {
t = t->ch[idx];
cur++;
}
if (cur > ans)
ans = cur;
}
return ans;
}
};

相关题目

115. Distinct Subsequences

count the number of distinct subsequences of S which equals T

DP, O(n) space

class Solution {
public:
int numDistinct(string s, string t) {
int l1 = s.length(), l2 = t.length(), tmp, pre;
vector<int> dp(l2+1,0);
dp[0] = 1;
for (int i=1;i<=l1;i++){
pre = 1;
for (int j=1;j<=l2;j++){
tmp = dp[j];
dp[j] = dp[j] + (s[i-1]==t[j-1] ? pre:0);
pre = tmp;
}
cout << endl;
}
return dp[l2];
}
};

最大连续子序列和，乘积，最长递增子串，最长公共子串，子序列等问题(动态规划等)

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