Write add_counts(a, b) that returns a new dictionary where each key's value is its count in a plus its count in b (0 when missing).
add_counts({'UPI': 3, 'COD': 1}, {'UPI': 2, 'Card': 4})
→ {'UPI': 5, 'COD': 1, 'Card': 4}add_counts({}, {})
→ {}Submit runs 4 tests, including hidden edge cases.
Each hint gives a little more help. Try on your own first.
Run your code to see the output here.