599 Minimum Index Sum of Two Lists

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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Output: ["Shogun"]
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Explanation: The only restaurant they both like is "Shogun".
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Example 2:
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Input:
2
["Shogun", "Tapioca Express", "Burger King", "KFC"]
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["KFC", "Shogun", "Burger King"]
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Output: ["Shogun"]
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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. 1.
    The length of both lists will be in the range of [1, 1000].
  2. 2.
    The length of strings in both lists will be in the range of [1, 30].
  3. 3.
    The index is starting from 0 to the list length minus 1.
  4. 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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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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int minSum = Integer.MAX_VALUE, sum = 0;
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List<String> list = new ArrayList<>();
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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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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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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)