Problem (in simple words)

• Xiao Kou wants to buy 1 staple food and 1 drink.
• Prices of staples are in array staple.
• Prices of drinks are in array drinks.
• Budget = x.
• Question: How many different (staple + drink) combos can he buy without going over budget?
• Because the answer can be very big, return it modulo 1,000,000,007.

Example 1

staple = [10, 20, 5]
drinks = [5, 5, 2]
x = 15

Valid pairs:

• 10 + 5 = 15 ✅
• 10 + 5 = 15 ✅
• 10 + 2 = 12 ✅
• 5 + 5 = 10 ✅
• 5 + 5 = 10 ✅
• 5 + 2 = 7 ✅

Answer = 6