Programmer Interview Gold Code-Interview Question 08.14. Boolean Operations (Interval Dynamic Programming)

1. Title

Given a Boolean expression and a Boolean result desired result, a Boolean expression consists of 0 (false)、1 (true)、& (AND)、 | (OR) 和 ^ (XOR)symbols.
Implement a function and work out several parenthetical methods that allow the expression to get the result value.

示例 1:
输入: s = "1^0|0|1", result = 0
输出: 2
解释: 两种可能的括号方法是
1^(0|(0|1))
1^((0|0)|1)

示例 2:
输入: s = "0&0&0&1^1|0", result = 1
输出: 10

提示：
运算符的数量不超过 19 个

Source: LeetCode
Link: https://leetcode-cn.com/problems/boolean-evaluation-lcci
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2. Interval DP problem solving

• dp？[i][j]Interval represents [i,j]the arithmetic value of ？（0 or 1）the number of programs
• Initialize, at each digitdp?[i][i]=1, if s[i]==?
• Then lenincrease by length and solvedp[i][i+len]
• dp[i][i+len]The solution can be obtained according to the product of the left and right sides of its internal solution
• So it is divided into two parts dp[i,j],dp[j+2][i+len], iterates over all j, j+1and the operator is
• Then, according to the three possibilities of the operator, discuss the result of 0,1, and add up

class Solution {
public:
    int countEval(string s, int result) {
        if(s=="")
            return 0;
        int i, j, n = s.size(), len;
        vector<vector<int>> dp0(n,vector<int>(n,0));
        vector<vector<int>> dp1(n,vector<int>(n,0));
        //dp？[i][j] 表示 区间[i,j]内运算值为 ？ 的方案数
        for(i = 0; i < n; i+=2)
        {
            if(s[i]=='1')
                dp1[i][i] = 1;
            else
                dp0[i][i] = 1;
        }
        for(len = 2; len <= n-1; len += 2)
        {	//按长度递增
            for(i = 0; i < n-len; i += 2)
            {	//左端点i
                for(j = i; j <= i+len-2; j+=2)
                {	//中间端点j
                    if(s[j+1]=='&')
                    {
                        dp1[i][i+len] += dp1[i][j]*dp1[j+2][i+len];
                        dp0[i][i+len] += dp0[i][j]*dp0[j+2][i+len]+dp1[i][j]*dp0[j+2][i+len]+dp0[i][j]*dp1[j+2][i+len];
                    }
                    else if(s[j+1]=='|')
                    {
                        dp1[i][i+len] += dp1[i][j]*dp1[j+2][i+len]+dp1[i][j]*dp0[j+2][i+len]+dp0[i][j]*dp1[j+2][i+len];
                        dp0[i][i+len] += dp0[i][j]*dp0[j+2][i+len];
                    }
                    else//^
                    {
                        dp1[i][i+len] += dp1[i][j]*dp0[j+2][i+len]+dp0[i][j]*dp1[j+2][i+len];
                        dp0[i][i+len] += dp0[i][j]*dp0[j+2][i+len]+dp1[i][j]*dp1[j+2][i+len];
                    }
                }
            }
        }
        if(result)
            return dp1[0][n-1];
        return dp0[0][n-1];
    }
};

8 ms 7 MB