This is a classic computer science problem. It is the basis of data comparison programs like the diff utility, and for version control systems like Git. The idea is to find the longest subsequence to all the sequences, most often involving only two. This is a problem that can be solved using dynamic programming. Dynamic programming is a method for solving complex problems by breaking the problem down into smaller sub problems. Each sub problem is solved just once, it’s solution is stored so that the next time the same subproblem occurs, instead of recomputing it’s solution, it is returned from memory. This technique of storing solutions to sub problems to be retrieved later when needed is called ‘memoization.’

The naive solution for this problem would be to generate all possible subsequences and find the longest matching subsequence. This solution is exponential in terms of time complexity. Because this problem has the following two properties it can be approached with a dynamic programming technique. This problem has an optimal substructure. This means that it can be broken down into smaller and simpler subproblems. This problem also has overlapping subproblems which means that the sub problems repeat themselves. These two properties make this problem solvable by a dynamic programming technique.

The first property, the optimal substructure, is that it is composed of repeating sub problems. Suppose that two strings end with the same letter. Removing that letter and adding it to the resulting substring, you are left with two shorter strings. The solution can be found by repeating this process on the shorter strings until there are no letters left. If the last letters are not the same, the answer can be found by finding the max of either substring A with the last letter removed compared with substring B or substring A compared with substring B with the last letter removed.

The second property, the overlapping subproblems, means that in the above described recursive method the same problem is repeatedly solved. This recomputation can be avoided by either using memoization or tabulation. In this solution we will use tabulation.

The function LCS_calc() which creates the table of the length of the longest common subsequence (LCS). The function LCS_find() goes through that table, retrieves and prints the LCS.

Time Complexity:

  • O(mn)
              
  function LCS(x, y) {
    x = x.split('');
    y = y.split('');
    var lcs = [];
    return LCS_calc(lcs, x, y);
  }
  
  function LCS_calc(lcs, x, y) {
    var i, j, m = x.length, n = y.length;
    for(i = 0; i < m+1; i++) {
      for(j = 0; j < n+1; j++) {
        if (j===0) {
          lcs[i] = [];
        }
        if (i===0 || j===0) {
          lcs[i][j] = 0;
        } else if (x[i-1] === y[j-1]) {
          lcs[i][j] = lcs[i-1][j-1]+1;
        } else {
          lcs[i][j] = Math.max(lcs[i-1][j], lcs[i][j-1]);
        }
      }
    }
    return LCS_find(lcs, x, y, m, n);
  }
  
  function LCS_find(lcs, x, y, m, n) {
    var current = lcs[m][n],
        result = [];
    while (current > 0 && m > 0 && n > 0) {
      if (lcs[m][n-1] === lcs[m][n]-1 && lcs[m-1][n] === lcs[m][n]-1) {
        m--;
        n--;
        current--;
        result.unshift(x[m]);
      } else if (lcs[m][n-1] === lcs[m][n] && lcs[m-1][n] === lcs[m][n] && lcs[m-1][n-1] === lcs[m][n]) {
        m--;
        n--;
      } else if (lcs[m-1][n] === lcs[m][n]-1 && lcs[m][n-1] === lcs[m][n]) {
        n--;
      } else {
        m--;
      }
    }
    return result.join('');
  }

  LCS( "132535365" , "123456789" );// => returns "12356"
              
            

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