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Problem solution · Python

Choose Numbers From Two Arrays in Range

Choose Numbers From Two Arrays in Range: a Python solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Dynamic programming
Source
walkccc LeetCode Solutions
Length
26 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 Choose Numbers From Two Arrays in 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

  • 26 lines of Python from the credited upstream file 2143.py.
  • The implementation visibly relies on sequence storage, hash 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 walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeChoose Numbers From Two Arrays in Range · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def countSubranges(self, nums1: list[int], nums2: list[int]) -> int:    MOD = 1_000_000_007    ans = 0    # {sum, count}, add if choose from nums1, minus if choose from nums2    dp = collections.Counter()     for a, b in zip(nums1, nums2):      newDp = collections.Counter()      newDp[a] += 1      newDp[-b] += 1       for prevSum, count in dp.items():        # Choose nums1[i]        newDp[prevSum + a] += count        newDp[prevSum + a] %= MOD        # Choose nums2[i]        newDp[prevSum - b] += count        newDp[prevSum - b] %= MOD       dp = newDp      ans += dp[0]      ans %= MOD     return ans

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