69 lines
1.7 KiB
Java
69 lines
1.7 KiB
Java
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// Java implementation of
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// finding length of longest
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// Common substring using
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// Dynamic Programming
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class GFG {
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/*
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Returns length of longest common substring
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of X[0..m-1] and Y[0..n-1]
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*/
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static int LCSubStr(char X[], char Y[], int m, int n)
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{
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// Create a table to store
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// lengths of longest common
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// suffixes of substrings.
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// Note that LCSuff[i][j]
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// contains length of longest
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// common suffix of
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// X[0..i-1] and Y[0..j-1].
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// The first row and first
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// column entries have no
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// logical meaning, they are
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// used only for simplicity of program
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int LCStuff[][] = new int[m + 1][n + 1];
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// To store length of the longest
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// common substring
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int result = 0;
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// Following steps build
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// LCSuff[m+1][n+1] in bottom up fashion
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for (int i = 0; i <= m; i++)
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{
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for (int j = 0; j <= n; j++)
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{
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if (i == 0 || j == 0)
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LCStuff[i][j] = 0;
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else if (X[i - 1] == Y[j - 1])
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{
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LCStuff[i][j] = LCStuff[i - 1][j - 1] + 1;
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result = Integer.max(result, LCStuff[i][j]);
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}
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else
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LCStuff[i][j] = 0;
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}
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}
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return result;
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}
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// Driver Code
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public static void main(String[] args)
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{
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// Test 1:
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String X = "abcdabaabbacd";
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String Y = "abbabaabcabcd";
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int m = X.length();
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int n = Y.length();
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System.out.println(LCSubStr(X.toCharArray(), Y.toCharArray(), m, n));
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// Test 2:
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String X = "abcdefgh";
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String Y = "abaabcfg";
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int m = X.length();
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int n = Y.length();
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System.out.println(LCSubStr(X.toCharArray(), Y.toCharArray(), m, n));
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}
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}
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// Time Complexity -> O(m*n)
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// where m and n are the lengths of the strings X and Y.
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