zoukankan      html  css  js  c++  java
  • [Algorithms] Using Dynamic Programming to Solve longest common subsequence problem

    Let's say we have two strings:

    str1 = 'ACDEB'

    str2 = 'AEBC'

    We need to find the longest common subsequence, which in this case should be 'AEB'.

    Using dynamic programming, we want to compare by char not by whole words.

    • we need memo to keep tracking the result which have already been calculated
      •   memo is 2d array, in this case is 5 * 4 array.
    • It devided problem into two parts
      •   If the char at the given indexs for both strings are the same, for example, 'A' for str1 & str2, then we consider 
    'A' + LSC(str1, str2, i1 + 1, i2 + 1)
      • If the char at the given indexs are not the same, we pick max length between LCB('DEB', 'EBC') & LCB('CDEB', 'BC'),  we pick
    Max {
       LCS('DEB', 'EBC'),
       LCS('CDEB', 'BC')
    }

    Bacislly for the str1 = 'CDEB' str2 = 'EBC', the first char is not the same, one is 'C', another is 'E', then we devide into tow cases and get the longer one. The way to devide is cutting 'C' from str1 get LCS('DEB', 'EBC'), and cutting 'E' from str2 get LCS('CDEB', 'BC').

     /**
     * FIND THE LONGEST COMMON SEQUENCES BY USING DYNAMICE PROGRAMMING
     *
     * @params:
     * str1: string
     * str2: string
     * i1: number
     * i2: number
     * memo: array []
     *
     * TC: O(L*M) << O(2^(L*M))
     */
    
    function LCS(str1, str2) {
        const memo = [...Array(str1.length)].map(e => Array(str2.length));
      
        /**
         * @return longest common sequence string
         */
        function helper(str1, str2, i1, i2, memo) {
          console.log(`str1, str2, ${i1}, ${i2}`);
          // if the input string is empty
          if (str1.length === i1 || str2.length === i2) {
            return "";
          }
          // check the memo, whether it contians the value
          if (memo[i1][i2] !== undefined) {
            return memo[i1][i2];
          }
          // if the first latter is the same
          // "A" + LCS(CDEB, EBC)
          if (str1[i1] === str2[i2]) {
            memo[i1][i2] = str1[i1] + helper(str1, str2, i1 + 1, i2 + 1, memo);
            return memo[i1][i2];
          }
      
          // Max { "C" + LCS(DEB, EBC), "E" + LCB(CDEB, BC) }
          let result;
          const resultA = helper(str1, str2, i1 + 1, i2, memo); // L
          const resultB = helper(str1, str2, i1, i2 + 1, memo); // M
      
          if (resultA.length > resultB.length) {
            result = resultA;
          } else {
            result = resultB;
          }
      
          memo[i1][i2] = result;
          return result;
        }
      
        return {
          result: helper(str1, str2, 0, 0, memo),
          memo
        };
      }
      
      //const str1 = "I am current working in Finland @Nordea",
      //str2 = "I am currently working in Finland at Nordea";
      
      const str1 = "ACDEB",
        str2 = "GAEBC";
      
      const { result, memo } = LCS(str1, str2);
      console.log(
        `
         ${str1}  
         ${str2}
         's longest common sequence is 
         "${result === "" ? "Empty!!!" : result}"
        `
      );
      
      console.log(memo);
      

    ----

    Bottom up solution can be:

    1. Init first row and first col value to zero

    2. Then loop thought the data, If row latter and col latter is not the same, then take which is larger Max {the previous row same col value data[row-1][col], same row but previous col data[row][col-1]}

    3. If they are the same, take data[row-1][col-1] + 1.

    Source, Code

  • 相关阅读:
    解决跨域POST登录中IE不能正常工作的bug
    设置一个严格的SESSION过期时间
    一次不成功的脚本Hack[捕鱼达人游戏]
    页面高度定位
    简单实用的跨域表单POST提交
    最简单的记录程序运行时间的方法:[记录PHP程序运行消耗时间]
    火狐下的GreaseMonkey和Chrome下的tampermonkey使用手记
    console.log
    记录最近工作使用javascript对select[option]的操作
    ubuntu16.04让内核编译一次过的方法
  • 原文地址:https://www.cnblogs.com/Answer1215/p/10206497.html
Copyright © 2011-2022 走看看