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Problem solution · Python

Form Largest Integer with Digits That Add Up to Target

Form Largest Integer with Digits That Add Up to Target: 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
72 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 Form Largest Integer with Digits That Add Up to Target, 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

  • 72 lines of Python from the credited upstream file form-largest-integer-with-digits-that-add-up-to-target.py.
  • The implementation visibly relies on sequence storage, 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 codeForm Largest Integer with Digits That Add Up to Target · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(t)# Space: O(t) class Solution(object):    def largestNumber(self, cost, target):        """        :type cost: List[int]        :type target: int        :rtype: str        """        dp = [0]        for t in xrange(1, target+1):            dp.append(-1)            for i, c in enumerate(cost):                if t-c < 0 or dp[t-c] < 0:                    continue                dp[t] = max(dp[t], dp[t-c]+1)        if dp[target] < 0:            return "0"        result = []        for i in reversed(xrange(9)):            while target >= cost[i] and dp[target] == dp[target-cost[i]]+1:                target -= cost[i]                result.append(i+1)        return "".join(map(str, result))  # Time:  O(t)# Space: O(t)class Solution2(object):    def largestNumber(self, cost, target):        """        :type cost: List[int]        :type target: int        :rtype: str        """        def key(bag):            return sum(bag), bag                dp = [[0]*9]        for t in xrange(1, target+1):            dp.append([])            for d, c in enumerate(cost):                if t < c or not dp[t-c]:                    continue                curr = dp[t-c][:]                curr[~d] += 1                if key(curr) > key(dp[t]):                    dp[-1] = curr                if not dp[-1]:            return "0"        return "".join(str(9-i)*c for i, c in enumerate(dp[-1]))  # Time:  O(t^2)# Space: O(t^2)class Solution3(object):    def largestNumber(self, cost, target):        """        :type cost: List[int]        :type target: int        :rtype: str        """        dp = [0]        for t in xrange(1, target+1):            dp.append(-1)            for i, c in enumerate(cost):                if t-c < 0:                    continue                dp[t] = max(dp[t], dp[t-c]*10 + i+1)        return str(max(dp[t], 0))

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