## 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. 1 <= len(A), len(B) <= 1000
2. 0 <= A[i], B[i] < 100

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#### Approach #1: Brute Force with Initial Character Map [Time Limit Exceeded]

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Intuition and Algorithm

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In a typical brute force, for all starting indices i of A and j of B, we will check for the longest matching subarray A[i:i+k] == B[j:j+k] of length k. This would look roughly like the following psuedocode:

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ans = 0\nfor i in [0 .. A.length - 1]:\n    for j in [0 .. B.length - 1]:\n        k = 0\n        while (A[i+k] == B[j+k]): k += 1 #and i+k < A.length etc.\n        ans = max(ans, k)\n
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Our insight is that in typical cases, most of the time A[i] != B[j]. We could instead keep a hashmap Bstarts[A[i]] = all j such that B[j] == A[i], and only loop through those in our j loop.

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Python

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class Solution(object):\n    def findLength(self, A, B):\n        ans = 0\n        Bstarts = collections.defaultdict(list)\n        for j, y in enumerate(B):\n            Bstarts[y].append(j)\n\n        for i, x in enumerate(A):\n            for j in Bstarts[x]:\n                k = 0\n                while i+k < len(A) and j+k < len(B) and A[i+k] == B[j+k]:\n                    k += 1\n                ans = max(ans, k)\n        return ans\n
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Java

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class Solution {\n    public int findLength(int[] A, int[] B) {\n        int ans = 0;\n        Map<Integer, ArrayList<Integer>> Bstarts = new HashMap();\n        for (int j = 0; j < B.length; j++) {\n            Bstarts.computeIfAbsent(B[j], x -> new ArrayList()).add(j);\n        }\n\n        for (int i = 0; i < A.length; i++) if (Bstarts.containsKey(A[i])) {\n            for (int j: Bstarts.get(A[i])) {\n                int k = 0;\n                while (i+k < A.length && j+k < B.length && A[i+k] == B[j+k]) {\n                    k++\n                }\n                ans = Math.max(ans, k);\n            }\n        }\n        return ans;\n    }\n}\n
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Complexity Analysis

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Time Complexity: , where are the lengths of A, B. The worst case is when all the elements are equal.

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Space Complexity: , the space used by Bstarts. (Of course, we could amend our algorithm to make this .)

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#### Approach #2: Binary Search with Naive Check [Time Limit Exceeded]

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Intuition

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If there is a length k subarray common to A and B, then there is a length j <= k subarray as well.

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Let check(length) be the answer to the question "Is there a subarray with length length, common to A and B?" This is a function with range that must take the form [True, True, ..., True, False, False, ..., False] with at least one True. We can binary search on this function.

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Algorithm

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Focusing on the binary search, our invariant is that check(hi) will always be False. We\'ll start with hi = min(len(A), len(B)) + 1; clearly check(hi) is False.

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Now we perform our check in the midpoint mi of lo and hi. When it is possible, then lo = mi + 1, and when it isn\'t, hi = mi. This maintains the invariant. At the end of our binary search, hi == lo and lo is the lowest value such that check(lo) is False, so we want lo - 1.

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As for the check itself, we can naively check whether any A[i:i+k] == B[j:j+k] using set structures.

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Python

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class Solution(object):\n    def findLength(self, A, B):\n        def check(length):\n            seen = set(tuple(A[i:i+length])\n                       for i in xrange(len(A) - length + 1))\n            return any(tuple(B[j:j+length]) in seen\n                       for j in xrange(len(B) - length + 1))\n\n        lo, hi = 0, min(len(A), len(B)) + 1\n        while lo < hi:\n            mi = (lo + hi) / 2\n            if check(mi):\n                lo = mi + 1\n            else:\n                hi = mi\n        return lo - 1\n
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Java

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class Solution {\n    public boolean check(int length, int[] A, int[] B) {\n        Set<String> seen = new HashSet();\n        for (int i = 0; i + length <= A.length; ++i) {\n            seen.add(Arrays.toString(Arrays.copyOfRange(A, i, i+length)));\n        }\n        for (int j = 0; j + length <= B.length; ++j) {\n            if (seen.contains(Arrays.toString(Arrays.copyOfRange(B, j, j+length)))) {\n                return true;\n            }\n        }\n        return false;\n    }\n\n    public int findLength(int[] A, int[] B) {\n        int lo = 0, hi = Math.min(A.length, B.length) + 1;\n        while (lo < hi) {\n            int mi = (lo + hi) / 2;\n            if (check(mi, A, B)) lo = mi + 1;\n            else hi = mi;\n        }\n        return lo - 1;\n    }\n}\n
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Complexity Analysis

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Time Complexity: , where are the lengths of A, B. The log factor comes from the binary search. The complexity of our naive check of a given is , as we will create the seen strings with complexity , then search for them with complexity , and our total complexity when performing our check is the addition of these two.

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Space Complexity: , the space used by seen.

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#### Approach #3: Dynamic Programming [Accepted]

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Intuition and Algorithm

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Since a common subarray of A and B must start at some A[i] and B[j], let dp[i][j] be the longest common prefix of A[i:] and B[j:]. Whenever A[i] == B[j], we know dp[i][j] = dp[i+1][j+1] + 1. Also, the answer is max(dp[i][j]) over all i, j.

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We can perform bottom-up dynamic programming to find the answer based on this recurrence. Our loop invariant is that the answer is already calculated correctly and stored in dp for any larger i, j.

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Python

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class Solution(object):\n    def findLength(self, A, B):\n        memo = [ * (len(B) + 1) for _ in range(len(A) + 1)]\n        for i in range(len(A) - 1, -1, -1):\n            for j in range(len(B) - 1, -1, -1):\n                if A[i] == B[j]:\n                    memo[i][j] = memo[i+1][j+1]+1\n        return max(max(row) for row in memo)\n
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Java

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class Solution {\n    public int findLength(int[] A, int[] B) {\n        int ans = 0;\n        int[][] memo = new int[A.length + 1][B.length + 1];\n        for (int i = A.length - 1; i >= 0; --i) {\n            for (int j = B.length - 1; j >= 0; --j) {\n                if (A[i] == B[j]) {\n                    memo[i][j] = memo[i+1][j+1] + 1;\n                    if (ans < memo[i][j]) ans = memo[i][j];\n                }\n            }\n        }\n        return ans;\n    }\n}\n
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Complexity Analysis

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Time Complexity: , where are the lengths of A, B.

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Space Complexity: , the space used by dp.

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#### Approach #4: Binary Search with Rolling Hash [Accepted]

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Intuition

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As in Approach #2, we will binary search for the answer. However, we will use a rolling hash (Rabin-Karp algorithm) to store hashes in our set structure.

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Algorithm

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For some prime , consider the following function modulo some prime modulus :

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Notably, . This shows we can get the hash of all in linear time. We will also use the fact that .

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For every i >= length - 1, we will want to record the hash of A[i-length+1], A[i-length+2], ..., A[i]. After, we will truncate the first element by h = (h - A[i - (length - 1)]) * Pinv % MOD to get ready to add the next element.

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To make our algorithm air tight, we also make a naive check when our work with rolling hashes says that we have found a match.

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class Solution(object):\n    def findLength(self, A, B):\n        P, MOD = 113, 10**9 + 7\n        Pinv = pow(P, MOD-2, MOD)\n        def check(guess):\n            def rolling(A, length):\n                if length == 0:\n                    yield 0, 0\n                    return\n\n                h, power = 0, 1\n                for i, x in enumerate(A):\n                    h = (h + x * power) % MOD\n                    if i < length - 1:\n                        power = (power * P) % MOD\n                    else:\n                        yield h, i - (length - 1)\n                        h = (h - A[i - (length - 1)]) * Pinv % MOD\n\n            hashes = collections.defaultdict(list)\n            for ha, start in rolling(A, guess):\n                hashes[ha].append(start)\n            for ha, start in rolling(B, guess):\n                iarr = hashes.get(ha, [])\n                if any(A[i:i+guess] == B[start:start+guess] for i in iarr):\n                    return True\n            return False\n\n        lo, hi = 0, min(len(A), len(B)) + 1\n        while lo < hi:\n            mi = (lo + hi) / 2\n            if check(mi):\n                lo = mi + 1\n            else:\n                hi = mi\n        return lo - 1\n
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Java

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import java.math.BigInteger;\n\nclass Solution {\n    int P = 113;\n    int MOD = 1_000_000_007;\n    int Pinv = BigInteger.valueOf(P).modInverse(BigInteger.valueOf(MOD)).intValue();\n\n    private int[] rolling(int[] source, int length) {\n        int[] ans = new int[source.length - length + 1];\n        long h = 0, power = 1;\n        if (length == 0) return ans;\n        for (int i = 0; i < source.length; ++i) {\n            h = (h + source[i] * power) % MOD;\n            if (i < length - 1) {\n                power = (power * P) % MOD;\n            } else {\n                ans[i - (length - 1)] = (int) h;\n                h = (h - source[i - (length - 1)]) * Pinv % MOD;\n                if (h < 0) h += MOD;\n            }\n        }\n        return ans;\n    }\n\n    private boolean check(int guess, int[] A, int[] B) {\n        Map<Integer, List<Integer>> hashes = new HashMap();\n        int k = 0;\n        for (int x: rolling(A, guess)) {\n            hashes.computeIfAbsent(x, z -> new ArrayList()).add(k++);\n        }\n        int j = 0;\n        for (int x: rolling(B, guess)) {\n            for (int i: hashes.getOrDefault(x, new ArrayList<Integer>()))\n                if (Arrays.equals(Arrays.copyOfRange(A, i, i+guess),\n                                  Arrays.copyOfRange(B, j, j+guess))) {\n                    return true;\n                }\n            j++;\n        }\n        return false;\n    }\n\n    public int findLength(int[] A, int[] B) {\n        int lo = 0, hi = Math.min(A.length, B.length) + 1;\n        while (lo < hi) {\n            int mi = (lo + hi) / 2;\n            if (check(mi, A, B)) lo = mi + 1;\n            else hi = mi;\n        }\n        return lo - 1;\n    }\n}\n
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Complexity Analysis

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Time Complexity: , where are the lengths of A, B. The log factor is contributed by the binary search, while creating the rolling hashes is . The checks for duplicate hashes are . If we perform a naive check to make sure our answer is correct, it adds a factor of to our cost of check, which keeps the complexity the same.

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Space Complexity: , the space used to store hashes and the subarrays in our final naive check.

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Analysis written by: @awice. Idea for Solution #2 by @StefanPochmann.

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