Let <code>dp[i]</code> be the maximum number of elements in the increasing sequence after processing the first <code>i</code> elements of the original array.

We have <code>dp[0] = 0</code>. <code>dp[i + 1] >= dp[i]</code> (since if we have the solution for the first <code>i</code> elements, we can always merge the last one of the first <code>i + 1</code> elements which is <code>nums[i]</code> into the solution of the first <code>i</code> elements.

For <code>i > 0</code>, we want to <code>dp[i] = max(dp[j] + 1)</code> where <code>sum(nums[i - 1] + nums[i - 2] +… + nums[j]) >= v[j]</code> and <code>v[j]</code> is the last element of the solution ending with <code>nums[j - 1]</code>.