Solution Explanation

This problem asks to find the number of balanced ranges within two arrays, nums1 and nums2, where a balanced range is defined as a sub-array where the sum of elements chosen from either nums1 or nums2 (for each index within the range) are equal. The solution uses dynamic programming to efficiently count these ranges.

Approach

The dynamic programming approach cleverly leverages the cumulative sums of the two arrays.

1. Preprocessing: Calculate the total sums s1 and s2 of nums1 and nums2 respectively. This is crucial for efficiently tracking the sums of chosen elements.

2. DP Table: A 2D DP table f is created, where f[i][j] represents the number of ways to achieve a sum difference of j considering only the elements up to index i. The size of the second dimension is s1 + s2 + 1 because the maximum possible sum difference will be at most the sum of all elements in nums1 or nums2.

3. Iteration: The code iterates through the arrays. For each index i:

   • It updates f[i][j] by considering two possibilities:

     • Choose nums1[i] : If j >= nums1[i], then f[i][j] += f[i-1][j - nums1[i]]. This adds the number of ways to reach a sum difference of j by choosing nums1[i] at the current index.
     • Choose nums2[i] : If j + nums2[i] <= s1 + s2, then f[i][j] += f[i-1][j + nums2[i]]. This adds the number of ways to reach a sum difference of j by choosing nums2[i] at the current index.

   • The modulo operation % mod is used to avoid integer overflow.

4. Result: The final answer is f[n-1][s2], because a balanced range requires the sum of chosen elements from nums1 to equal the sum of elements chosen from nums2, resulting in a sum difference of 0 (relative to s2).

Time and Space Complexity Analysis

• Time Complexity: O(n * (s1 + s2)), where n is the length of the arrays, and s1 and s2 are the sums of the arrays. The nested loop iterates through each index and each possible sum difference.

• Space Complexity: O(n * (s1 + s2)). This is due to the DP table f that stores the counts of possible sum differences at each index.

Code Explanation (Python)

The Python code directly implements the dynamic programming approach described above. Note the use of accumulate from the itertools library for efficient sum calculation and the modulo operator to handle potential integer overflow.

from itertools import accumulate
 
class Solution:
    def countSubranges(self, nums1: List[int], nums2: List[int]) -> int:
        n = len(nums1)
        s1 = sum(nums1)
        s2 = sum(nums2)
        f = [[0] * (s1 + s2 + 1) for _ in range(n)]
        ans = 0
        mod = 10**9 + 7
        for i, (a, b) in enumerate(zip(nums1, nums2)):
            f[i][a + s2] += 1
            f[i][-b + s2] += 1
            if i:
                for j in range(s1 + s2 + 1):
                    if j >= a:
                        f[i][j] = (f[i][j] + f[i - 1][j - a]) % mod
                    if j + b < s1 + s2 + 1:
                        f[i][j] = (f[i][j] + f[i - 1][j + b]) % mod
            ans = (ans + f[i][s2]) % mod
        return ans
 

The code in other languages (Java, C++, Go, TypeScript) follows the same logic and algorithmic structure, differing only in syntax and specific library functions.