Choose Numbers From Two Arrays in Range
<p>You are given two <strong>0-indexed</strong> integer arrays <code>nums1</code> and <code>nums2</code> of length <code>n</code>.</p>

<p>A range <code>[l, r]</code> (<strong>inclusive</strong>) where <code>0 &lt;= l &lt;= r &lt; n</code> is <strong>balanced</strong> if:</p>

<ul>
	<li>For every <code>i</code> in the range <code>[l, r]</code>, you pick either <code>nums1[i]</code> or <code>nums2[i]</code>.</li>
	<li>The sum of the numbers you pick from <code>nums1</code> equals to the sum of the numbers you pick from <code>nums2</code> (the sum is considered to be <code>0</code> if you pick no numbers from an array).</li>
</ul>

<p>Two <strong>balanced</strong> ranges from <code>[l<sub>1</sub>, r<sub>1</sub>]</code> and <code>[l<sub>2</sub>, r<sub>2</sub>]</code> are considered to be <strong>different</strong> if at least one of the following is true:</p>

<ul>
	<li><code>l<sub>1</sub> != l<sub>2</sub></code></li>
	<li><code>r<sub>1</sub> != r<sub>2</sub></code></li>
	<li><code>nums1[i]</code> is picked in the first range, and <code>nums2[i]</code> is picked in the second range or <strong>vice versa</strong> for at least one <code>i</code>.</li>
</ul>

<p>Return <em>the number of <strong>different</strong> ranges that are balanced. </em>Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code><em>.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [1,2,5], nums2 = [2,6,3]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The balanced ranges are:
- [0, 1] where we choose nums2[0], and nums1[1].
  The sum of the numbers chosen from nums1 equals the sum of the numbers chosen from nums2: 2 = 2.
- [0, 2] where we choose nums1[0], nums2[1], and nums1[2].
  The sum of the numbers chosen from nums1 equals the sum of the numbers chosen from nums2: 1 + 5 = 6.
- [0, 2] where we choose nums1[0], nums1[1], and nums2[2].
  The sum of the numbers chosen from nums1 equals the sum of the numbers chosen from nums2: 1 + 2 = 3.
Note that the second and third balanced ranges are different.
In the second balanced range, we choose nums2[1] and in the third balanced range, we choose nums1[1].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [0,1], nums2 = [1,0]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The balanced ranges are:
- [0, 0] where we choose nums1[0].
  The sum of the numbers chosen from nums1 equals the sum of the numbers chosen from nums2: 0 = 0.
- [1, 1] where we choose nums2[1].
  The sum of the numbers chosen from nums1 equals the sum of the numbers chosen from nums2: 0 = 0.
- [0, 1] where we choose nums1[0] and nums2[1].
  The sum of the numbers chosen from nums1 equals the sum of the numbers chosen from nums2: 0 = 0.
- [0, 1] where we choose nums2[0] and nums1[1].
  The sum of the numbers chosen from nums1 equals the sum of the numbers chosen from nums2: 1 = 1.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == nums1.length == nums2.length</code></li>
	<li><code>1 &lt;= n &lt;= 100</code></li>
	<li><code>0 &lt;= nums1[i], nums2[i] &lt;= 100</code></li>
</ul>
Array; Dynamic Programming
class Solution {
    public int countSubranges(int[] nums1, int[] nums2) {
        int n = nums1.length;
        int s1 = Arrays.stream(nums1).sum();
        int s2 = Arrays.stream(nums2).sum();
        int[][] f = new int[n][s1 + s2 + 1];
        int ans = 0;
        final int mod = (int) 1e9 + 7;
        for (int i = 0; i < n; ++i) {
            int a = nums1[i], b = nums2[i];
            f[i][a + s2]++;
            f[i][-b + s2]++;
            if (i > 0) {
                for (int j = 0; j <= s1 + s2; ++j) {
                    if (j >= a) {
                        f[i][j] = (f[i][j] + f[i - 1][j - a]) % mod;
                    }
                    if (j + b <= s1 + s2) {
                        f[i][j] = (f[i][j] + f[i - 1][j + b]) % mod;
                    }
                }
            }
            ans = (ans + f[i][s2]) % mod;
        }
        return ans;
    }
}