599 Minimum Index Sum of Two Lists

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599. Minimum Index Sum of Two Lists
1. Question
Suppose Andy and Doris want to choose a restaurant for dinner, and they both have a list of favorite restaurants represented by strings.
You need to help them find out their common interest with the least list index sum. If there is a choice tie between answers, output all of them with no order requirement. You could assume there always exists an answer.
Example 1:
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Input:
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["Shogun", "Tapioca Express", "Burger King", "KFC"]
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["Piatti", "The Grill at Torrey Pines", "Hungry Hunter Steakhouse", "Shogun"]
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​
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Output: ["Shogun"]
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​
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Explanation: The only restaurant they both like is "Shogun".
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Example 2:
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Input:
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["Shogun", "Tapioca Express", "Burger King", "KFC"]
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["KFC", "Shogun", "Burger King"]
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​
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Output: ["Shogun"]
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​
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Explanation: The restaurant they both like and have the least index sum is "Shogun" with index sum 1 (0+1).
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Note:
1.The length of both lists will be in the range of [1, 1000].
2.The length of strings in both lists will be in the range of [1, 30].
3.The index is starting from 0 to the list length minus 1.
4.No duplicates in both lists.
2. Implementation
(1) Hash Table
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class Solution {
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    public String[] findRestaurant(String[] list1, String[] list2) {
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        Map<String, Integer> map = new HashMap<>();
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​
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        for (int i = 0; i < list1.length; i++) {
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            map.put(list1[i], i);
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        }
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​
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        int minSum = Integer.MAX_VALUE, sum = 0;
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        List<String> list = new ArrayList<>();
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​
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        for (int j = 0; j < list2.length; j++) {
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            if (map.containsKey(list2[j])) {
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                sum = j + map.get(list2[j]);
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​
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                if (sum < minSum) {
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                    list.clear();
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                    list.add(list2[j]);
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                    minSum = sum;
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                }
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                else if (sum == minSum) {
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                    list.add(list2[j]);
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                }
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            }
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        }
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        String[] res = new String[list.size()];
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        int index = 0;
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​
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        for (String s : list) {
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            res[index++] = s;
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        }
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        return res;
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    }
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}
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3. Time & Space Complexity
Hash Table: 时间复杂度O(m + n)), m为list1的长度, n为list2的长度，空间复杂度O(m)
Hash Table
697 Degree of an Array
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Contents
599. Minimum Index Sum of Two Lists
1. Question
2. Implementation
3. Time & Space Complexity