We use a monotonic stack to precompute all next greater elements in nums2. We store these results in a hash map. Then, we simply query the map for each element in nums1.

1. Initialize an empty stack and a dictionary mapping.
2. Iterate through each num in nums2:
   • While stack is not empty and num > top of stack:
     • temp = stack.pop()
     • mapping[temp] = num
   • Push num to stack.
3. For values remaining in stack, set their mapping value to -1.
4. Build the result for nums1 using the mapping.

• Time Complexity: O(N + M), where N, M are lengths of nums1 and nums2. Each element in nums2 is pushed and popped exactly once.
• Space Complexity: O(M) for map and stack.

def next_greater_element(nums1, nums2):
    mapping = {}
    stack = []
    
    for num in nums2:
        while stack and num > stack[-1]:
            mapping[stack.pop()] = num
        stack.append(num)
        
    while stack:
        mapping[stack.pop()] = -1
        
    return [mapping[num] for num in nums1]