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Problem solution · Python

Count Good Integers on a Grid Path

Count Good Integers on a Grid Path: a Python solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Dynamic programming
Source
Kamyu LeetCode Solutions
Length
145 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Dynamic programming

For Count Good Integers on a Grid Path, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 145 lines of Python from the credited upstream file count-good-integers-on-a-grid-path.py.
  • The implementation visibly relies on cached states.
  • No explicit loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCount Good Integers on a Grid Path · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(16 * 2 * 10 * 10)# Space: O(16 + 2 * 10) # dpclass 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)# memoizationclass 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)# memoizationclass 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)

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