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I used a HashMap where the the key is the price and the value is the index and I only hold the higher index in the hashmap. When searching the hashmap for the leftover value (M-price[i]), if the value saved in the hashmap != i, then that is the answer. Remember that each test case has a unique answer so there would only be at most 2x M/2 prices if it is the answer
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Ice Cream Parlor
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I used a HashMap where the the key is the price and the value is the index and I only hold the higher index in the hashmap. When searching the hashmap for the leftover value (M-price[i]), if the value saved in the hashmap != i, then that is the answer. Remember that each test case has a unique answer so there would only be at most 2x M/2 prices if it is the answer