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Problem solution · Python

Number of Beautiful Integers in the Range

Number of Beautiful Integers in the Range: 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
139 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 Number of Beautiful Integers in the Range, 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

  • 139 lines of Python from the credited upstream file number-of-beautiful-integers-in-the-range.py.
  • The implementation visibly relies on ordered lookup, 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 codeNumber of Beautiful Integers in the Range · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(n^2 * k), n = len(str(high))# Space: O(n^2 * k) # memoization (faster but more space)class Solution(object):    def numberOfBeautifulIntegers(self, low, high, k):        """        :type low: int        :type high: int        :type k: int        :rtype: int        """        TIGHT, UNTIGHT, UNBOUND = range(3)        def f(x):            digits = map(int, str(x))            lookup = [[[[-1]*k for _ in xrange(2*len(digits)+1)] for _ in xrange(3)] for _ in xrange(len(digits))]            def memoization(i, state, diff, total):                if i == len(digits):                    return int(state != UNBOUND and diff == total == 0)                if lookup[i][state][diff][total] == -1:                    result = int(i != 0 and diff == total == 0)  # count if the beautiful integer x s.t. len(str(x)) < len(digits)                    for d in xrange(1 if i == 0 else 0, 10):                        new_state = state                        if state == TIGHT and d != digits[i]:                            new_state = UNTIGHT if d < digits[i] else UNBOUND                        new_diff = diff+(1 if d%2 == 0 else -1)                        new_total = (total*10+d)%k                        result += memoization(i+1, new_state, new_diff, new_total)                    lookup[i][state][diff][total] = result                return lookup[i][state][diff][total]                return memoization(0, TIGHT, 0, 0)         return f(high)-f(low-1)  # Time:  O(n^2 * k), n = len(str(high))# Space: O(n * k)# dp (slower but less space)class Solution2(object):    def numberOfBeautifulIntegers(self, low, high, k):        """        :type low: int        :type high: int        :type k: int        :rtype: int        """        TIGHT, UNTIGHT, UNBOUND = range(3)        def f(x):            digits = map(int, str(x))            dp = [[[0]*k for _ in xrange(2*len(digits)+1)] for _ in xrange(3)]            for tight in xrange(2):                for state in (TIGHT, UNTIGHT):                    dp[state][0][0] = 1            for i in reversed(xrange(len(digits))):                new_dp = [[[0]*k for _ in xrange(2*len(digits)+1)] for _ in xrange(3)]                for state in (TIGHT, UNTIGHT, UNBOUND):                    new_dp[state][0][0] = int(i != 0)  # count if the beautiful integer x s.t. len(str(x)) < len(digits)                    for d in xrange(1 if i == 0 else 0, 10):                        new_state = state                        if state == TIGHT and d != digits[i]:                            new_state = UNTIGHT if d < digits[i] else UNBOUND                        for diff in xrange(-len(digits), len(digits)+1):                            new_diff = diff+(1 if d%2 == 0 else -1)                            for total in xrange(k):                                new_total = (total*10+d)%k                                new_dp[state][diff][total] += dp[new_state][new_diff][new_total]                dp = new_dp            return dp[TIGHT][0][0]         return f(high)-f(low-1)  # Time:  O(n^2 * k), n = len(str(high))# Space: O(n^2 * k)# memoization (faster but more space)class Solution3(object):    def numberOfBeautifulIntegers(self, low, high, k):        """        :type low: int        :type high: int        :type k: int        :rtype: int        """        def f(x):            digits = map(int, str(x))            lookup = [[[[[-1]*k for _ in xrange(2*len(digits)+1)] for _ in xrange(2)] for _ in xrange(2)] for _ in xrange(len(digits))]            def memoization(i, zero, tight, diff, total):                if i == len(digits):                    return int(zero == diff == total == 0)                if lookup[i][zero][tight][diff][total] == -1:                    result = 0                    for d in xrange((digits[i] if tight else 9)+1):                        new_zero = int(zero and d == 0)                        new_tight = int(tight and d == digits[i])                        new_diff = diff+((1 if d%2 == 0 else -1) if new_zero == 0 else 0)                        new_total = (total*10+d)%k                        result += memoization(i+1, new_zero, new_tight, new_diff, new_total)                    lookup[i][zero][tight][diff][total] = result                return lookup[i][zero][tight][diff][total]                return memoization(0, 1, 1, 0, 0)         return f(high)-f(low-1)  # Time:  O(n^2 * k), n = len(str(high))# Space: O(n * k)# dp (slower but less space)class Solution4(object):    def numberOfBeautifulIntegers(self, low, high, k):        """        :type low: int        :type high: int        :type k: int        :rtype: int        """        def f(x):            digits = map(int, str(x))            dp = [[[[0]*k for _ in xrange(2*len(digits)+1)] for _ in xrange(2)] for _ in xrange(2)]            for tight in xrange(2):                dp[0][tight][0][0] = 1            for i in reversed(xrange(len(digits))):                new_dp = [[[[0]*k for _ in xrange(2*len(digits)+1)] for _ in xrange(2)] for _ in xrange(2)]                for zero in xrange(2):                    for tight in xrange(2):                        for d in xrange((digits[i] if tight else 9)+1):                            new_zero = int(zero and d == 0)                            new_tight = int(tight and d == digits[i])                            for diff in xrange(-len(digits), len(digits)+1):                                new_diff = diff+((1 if d%2 == 0 else -1) if new_zero == 0 else 0)                                for total in xrange(k):                                    new_total = (total*10+d)%k                                    new_dp[zero][tight][diff][total] += dp[new_zero][new_tight][new_diff][new_total]                dp = new_dp            return dp[1][1][0][0]         return f(high)-f(low-1)

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