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:
Input:
["Shogun", "Tapioca Express", "Burger King", "KFC"]
["Piatti", "The Grill at Torrey Pines", "Hungry Hunter Steakhouse", "Shogun"]
Output: ["Shogun"]
Explanation: The only restaurant they both like is "Shogun".
Example 2:
Input:
["Shogun", "Tapioca Express", "Burger King", "KFC"]
["KFC", "Shogun", "Burger King"]
Output: ["Shogun"]
Explanation: The restaurant they both like and have the least index sum is "Shogun" with index sum 1 (0+1).
Note:
The length of both lists will be in the range of [1, 1000].
The length of strings in both lists will be in the range of [1, 30].
The index is starting from 0 to the list length minus 1.
No duplicates in both lists.
2. Implementation
(1) Hash Table
class Solution {
public String[] findRestaurant(String[] list1, String[] list2) {
Map<String, Integer> map = new HashMap<>();
for (int i = 0; i < list1.length; i++) {
map.put(list1[i], i);
}
int minSum = Integer.MAX_VALUE, sum = 0;
List<String> list = new ArrayList<>();
for (int j = 0; j < list2.length; j++) {
if (map.containsKey(list2[j])) {
sum = j + map.get(list2[j]);
if (sum < minSum) {
list.clear();
list.add(list2[j]);
minSum = sum;
}
else if (sum == minSum) {
list.add(list2[j]);
}
}
}
String[] res = new String[list.size()];
int index = 0;
for (String s : list) {
res[index++] = s;
}
return res;
}
}
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)
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