Given two strings ‘X’ and ‘Y’, find the length of longest common substring. Expected space complexity is linear.

Examples:

Input : X = "GeeksforGeeks", Y = "GeeksQuiz" Output : 5 The longest common substring is "Geeks" and is of length 5. Input : X = "abcdxyz", Y = "xyzabcd" Output : 4 The longest common substring is "abcd" and is of length 4.

We have discussed Dynamic programming based solution for Longest common substring. The auxiliary space used by the solution is O(m*n), where m and n are lengths of string X and Y. The space used by solution can be reduced to O(2*n).

Suppose we are at position mat[i][j]. Now if X[i-1] == Y[j-1], then we add the value of mat[i-1][j-1] to our result. That is we add value from previous row and value for all other rows below the previous row are never used. So, at a time we are using only two consecutive rows. This observation can be used to reduce the space required to find length of longest common substring.

Instead of creating a matrix of size m*n, we create a matrix of size 2*n. A variable currRow is used to represent that either row 0 or row 1 of this matrix is currently used to find length. Initially row 0 is used as current row for the case when length of string X is zero. At the end of each iteration, current row is made previous row and previous row is made new current row.

// Space optimized CPP implementation of longest // common substring. #include <bits/stdc++.h> using namespace std; // Function to find longest common substring. int LCSubStr(string X, string Y) { // Find length of both the strings. int m = X.length(); int n = Y.length(); // Variable to store length of longest // common substring. int result = 0; // Matrix to store result of two // consecutive rows at a time. int len[2][n]; // Variable to represent which row of // matrix is current row. int currRow = 0; // For a particular value of i and j, // len[currRow][j] stores length of longest // common substring in string X[0..i] and Y[0..j]. for (int i = 0; i <= m; i++) { for (int j = 0; j <= n; j++) { if (i == 0 || j == 0) { len[currRow][j] = 0; } else if (X[i - 1] == Y[j - 1]) { len[currRow][j] = len[1 - currRow][j - 1] + 1; result = max(result, len[currRow][j]); } else { len[currRow][j] = 0; } } // Make current row as previous row and previous // row as new current row. currRow = 1 - currRow; } return result; } int main() { string X = "GeeksforGeeks"; string Y = "GeeksQuiz"; cout << LCSubStr(X, Y); return 0; }

**Output:**

5

Output:5

**Time Complexity: ** O(m*n)

**Auxiliary Space: ** O(n)

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