Since both arrays are sorted in a non-increasing way this means that for each value in the first array. We can find the farthest value smaller than it using binary search.

There is another solution using a two pointers approach since the first array is non-increasing the farthest j such that nums2[j] ≥ nums1[i] is at least as far as the farthest j such that nums2[j] ≥ nums1[i-1]