599. Minimum Index Sum of Two Lists

Posted 2020-12-19 wentiliangkaihua

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:

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.

class Solution {
    public String[] findRestaurant(String[] list1, String[] list2) {
        List<String> res = new ArrayList();
        List<String> l2 = Arrays.asList(list2);
        int min = 2000;
        
        for(int i = 0; i < list1.length; i++) {
            int ind = l2.indexOf(list1[i]);
            if(ind >= 0) {
                if(ind + i < min) {
                    res.clear();
                    res.add(list1[i]);
                    min = ind + i;
                }
                else if(ind + i == min) res.add(list1[i]);
            }
        }
        String[] r = new String[res.size()];
        int i = 0;
        for(String s: res) r[i++] = s;
        return r;
    }
}

brute force

public class Solution {
    public String[] findRestaurant(String[] list1, String[] list2) {
        HashMap < String, Integer > map = new HashMap < String, Integer > ();
        for (int i = 0; i < list1.length; i++) map.put(list1[i], i);        
        List < String > res = new ArrayList < > ();
        int min_sum = 4000；
        int sum = 0;
        for (int j = 0; j < list2.length; j++) {
            if (map.containsKey(list2[j])) {
                sum = j + map.get(list2[j]);
                if (sum < min_sum) {
                    res.clear();
                    res.add(list2[j]);
                    min_sum = sum;
                } else if (sum == min_sum)
                    res.add(list2[j]);
            }
        }
        return res.toArray(new String[res.size()]);
    }
}

用hashmap，O(l1 + l2)