LeetCode-Solutions/Python/count-good-integers-on-a-grid-path.py at master · kamyu104/LeetCode-Solutions · GitHub

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
# Time:  O(16 * 2 * 10 * 10)
# Space: O(16 + 2 * 10)

# dp
class Solution(object):
    def countGoodIntegersOnPath(self, l, r, directions):
        """
        :type l: int
        :type r: int
        :type directions: str
        :rtype: int
        """
        L = 16
        def count(n):
            digits = [0]*L
            for i in reversed(xrange(len(digits))):
                digits[i] = n%10
                n //= 10
            dp = [[0]*10 for _ in xrange(2)]
            dp[1][0] = 1
            for i in xrange(L):
                new_dp = [[0]*10 for _ in xrange(2)]
                for t in xrange(2):
                    bound = digits[i] if t else 9
                    for k in xrange(10):
                        if not dp[t][k]:
                            continue
                        for d in xrange(bound+1):
                            nk = k
                            if lookup[i]:
                                if d < k:
                                    continue
                                nk = d
                            new_dp[t and d == bound][nk] += dp[t][k]
                dp = new_dp
            return sum(sum(row) for row in dp)

        i = j = 0
        lookup = [False]*L
        lookup[i*4+j] = True
        for x in directions:
            if x == 'D':
                i += 1
            else:
                j += 1
            lookup[i*4+j] = True
        return count(r)-count(l-1)


# Time:  O(16 * 2 * 10 * 10)
# Space: O(16 * 10)
# memoization
class Solution2(object):
    def countGoodIntegersOnPath(self, l, r, directions):
        """
        :type l: int
        :type r: int
        :type directions: str
        :rtype: int
        """
        L = 16
        def count(n):
            def memoization(i, t, k):
                if i == L:
                    return 1
                if not t and memo[i][k] != -1:
                    return memo[i][k]
                result = 0
                bound = digits[i] if t else 9
                for d in xrange(bound+1):
                    nk = k
                    if lookup[i]:
                        if d < k:
                            continue
                        nk = d
                    result += memoization(i+1, t and d == bound, nk)
                if not t:
                    memo[i][k] = result
                return result

            digits = [0]*L
            for i in reversed(xrange(len(digits))):
                digits[i] = n%10
                n //= 10
            memo = [[-1]*10 for _ in xrange(L)]
            return memoization(0, True, 0)

        i = j = 0
        lookup = [False]*L
        lookup[i*4+j] = True
        for x in directions:
            if x == 'D':
                i += 1
            else:
                j += 1
            lookup[i*4+j] = True
        return count(r)-count(l-1)


# Time:  O(16 * 2 * 10 * 10)
# Space: O(16 * 2 * 10)
# memoization
class Solution3(object):
    def countGoodIntegersOnPath(self, l, r, directions):
        """
        :type l: int
        :type r: int
        :type directions: str
        :rtype: int
        """
        L = 16
        def count(n):
            def memoization(i, t, k):
                if i == L:
                    return 1
                if memo[i][t][k] == -1:
                    memo[i][t][k] = 0
                    bound = digits[i] if t else 9
                    for d in xrange(bound+1):
                        nk = k
                        if lookup[i]:
                            if d < k:
                                continue
                            nk = d
                        memo[i][t][k] += memoization(i+1, t and d == bound, nk)
                return memo[i][t][k]

            digits = [0]*L
            for i in reversed(xrange(len(digits))):
                digits[i] = n%10
                n //= 10
            memo = [[[-1]*10 for _ in xrange(2)] for _ in xrange(L)]
            return memoization(0, True, 0)

        i = j = 0
        lookup = [False]*L
        lookup[i*4+j] = True
        for x in directions:
            if x == 'D':
                i += 1
            else:
                j += 1
            lookup[i*4+j] = True
        return count(r)-count(l-1)