Longest Common Substring (Java)

In computer science, the longest common substring problem is to find the longest string that is a substring of two or more strings.

Analysis

Given two strings a and b, let dp[i][j] be the length of the common substring ending at a[i] and b[j].

longest-common-substring-java

The dp table looks like the following given a="abc" and b="abcd".

longest-common-substring-java

Java Solution

public static int getLongestCommonSubstring(String a, String b){
	int m = a.length();
	int n = b.length();
 
	int max = 0;
 
	int[][] dp = new int[m][n];
 
	for(int i=0; i<m; i++){
		for(int j=0; j<n; j++){
			if(a.charAt(i) == b.charAt(j)){
				if(i==0 || j==0){
					dp[i][j]=1;
				}else{
					dp[i][j] = dp[i-1][j-1]+1;
				}
 
				if(max < dp[i][j])
					max = dp[i][j];
			}
 
		}
	}
 
	return max;
}

This is a similar problem like longest common subsequence. The difference of the solution is that for this problem when a[i]!=b[j], dp[i][j] are all zeros by default. However, in the longest common subsequence problem, dp[i][j] values are carried from the previous values, i.e., dp[i-1][j] and dp[i][j-1].

Category >> Algorithms  
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  • Coding freek

    There are four different ways to solve this problem –

    // Method1()- recursive solution(Top- down approach)
    // time complexity - O(3^(m+n))
    // space complexity - O(m+n)
    public static int LCSubStrM1(char[] X, char[] Y, int m, int n, int lcsCount) {
    if (m <= 0 || n <= 0)
    return lcsCount;

    int lcsCount1=lcsCount;
    if (X[m - 1] == Y[n - 1])
    lcsCount1 = LCSubStrM1(X, Y, m - 1, n - 1, lcsCount + 1);

    int lcsCount2 = LCSubStrM1(X, Y, m, n - 1, 0);
    int lcsCount3 = LCSubStrM1(X, Y, m - 1, n, 0);

    return Math.max(lcsCount1, Math.max(lcsCount2, lcsCount3));
    }

    // Method2A1()- recursive solution with memoization (Top-down approach caching on method level)
    public static int LCSubStrM2A1(char[] X, char[] Y, int m, int n, int lcsCount, Integer[][][] dp) {
    if (m <= 0 || n <= 0)
    return lcsCount;

    if (dp[m][n][lcsCount] != null)
    return dp[m][n][lcsCount];

    int lcsCount1=lcsCount;
    if (X[m - 1] == Y[n - 1])
    lcsCount1 = LCSubStrM2A1(X, Y, m - 1, n - 1, lcsCount + 1, dp);

    int lcsCount2 = LCSubStrM2A1(X, Y, m, n - 1, 0, dp);
    int lcsCount3 = LCSubStrM2A1(X, Y, m - 1, n, 0, dp);

    return dp[m][n][lcsCount] = Math.max(lcsCount1, Math.max(lcsCount2, lcsCount3));
    }

    // Method2A2()- recursive solution with memoization (Top-down approach caching on global level)
    public static int LCSubStrM2A2(char[] X, char[] Y, int m, int n, int lcsCount) {
    if (m <= 0 || n <= 0)
    return lcsCount;

    if (dp[m][n][lcsCount] != null)
    return dp[m][n][lcsCount];

    int lcsCount1=lcsCount;
    if (X[m - 1] == Y[n - 1])
    lcsCount1 = LCSubStrM2A2(X, Y, m - 1, n - 1, lcsCount + 1);

    int lcsCount2 = LCSubStrM2A2(X, Y, m, n - 1, 0);
    int lcsCount3 = LCSubStrM2A2(X, Y, m - 1, n, 0);

    return dp[m][n][lcsCount] = Math.max(lcsCount1, Math.max(lcsCount2, lcsCount3));
    }

    // Method3()- DP solution(Bottom up approach)
    // time complexity - O(m*n)
    // space complexity - O(m*n)
    public static int LCSubStrA3(char[] X, char[] Y, int m, int n) {
    int memo[][] = new int[m + 1][n + 1];
    int result = 0;

    for (int i = 0; i <= m; i++) {
    for (int j = 0; j <= n; j++) {
    if (i == 0 || j == 0) {
    memo[i][j] = 0;
    } else if (X[i - 1] == Y[j - 1]) {
    memo[i][j] = memo[i - 1][j - 1] + 1;
    result = Math.max(result, memo[i][j]);
    } else {
    memo[i][j] = 0;
    }
    }
    }
    cache = memo;
    return result;
    }

    for detailed explanation refer–https://www.youtube.com/watch?v=Lj90FqNCIJE&list=PLSIpQf0NbcClDpWE58Y-oSJro_W3LO8Nb

  • Rishabh Dubey


    public static int substrLength(String str1, String str2){ //photograph,tomography
    int count=0,mCount= 0;
    for(int i=0;i<str1.length();i++){
    for(int j=0;j<str2.length();j++){
    if(str1.charAt(i)== str2.charAt(j)){
    count =0;
    for(int k=0;(k+i)<str1.length()&&(k+j)mCount)
    mCount =count;
    }
    }
    return mCount;
    }