Minimum Index Sum of Two Lists

HashMap, String

Easy

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 theircommon interestwith theleast 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.

Solution

HashMap - Time: O(n1 + n2), Space - O(n1 * x)

Note about clear(): The time complexity ofArrayList.clear()isO(n)and ofremoveAllisO(n^2).

class Solution {
    public String[] findRestaurant(String[] list1, String[] list2) {
        ArrayList<String> res = new ArrayList<>();
        HashMap<String, Integer> map = new HashMap<>();
        int minIndexSum = Integer.MAX_VALUE;

        for (int i = 0; i < list1.length; i++) {
            map.put(list1[i], i);
        }

        for (int j = 0; j < list2.length; j++) {
            if (map.containsKey(list2[j])) {
                int sum = j + map.get(list2[j]);
                if (sum < minIndexSum) {
                    res.clear();
                    res.add(list2[j]);
                    minIndexSum = sum; 
                } else if (sum == minIndexSum) {
                    res.add(list2[j]);
                }
            }
        }
        return res.toArray(new String[res.size()]);
    }
}

HashMap - Time: O(n1 + n2), Space Optimized: O(min(n1, n2) * x)

Here, x refers to the average string length.

Always use the shorter list for building hashmap:

if (list1.length > list2.length) {
    return findRestaurant(list2, list1);
}

--

class Solution {
    public String[] findRestaurant(String[] list1, String[] list2) {
        if (list1.length > list2.length) {
            return findRestaurant(list2, list1);
        }
        ArrayList<String> res = new ArrayList<>();
        HashMap<String, Integer> map = new HashMap<>();
        int minIndexSum = Integer.MAX_VALUE;

        for (int i = 0; i < list1.length; i++) {
            if (!map.containsKey(list1[i])) {
                map.put(list1[i], i);
            }
        }

        for (int j = 0; j < list2.length; j++) {
            if (map.containsKey(list2[j])) {
                int sum = j + map.get(list2[j]);
                if (sum < minIndexSum) {
                    res.clear();
                    res.add(list2[j]);
                    minIndexSum = sum; 
                } else if (sum == minIndexSum) {
                    res.add(list2[j]);
                }
            }
        }
        return res.toArray(new String[res.size()]);
    }
}

Reference

https://leetcode.com/problems/minimum-index-sum-of-two-lists/solution/

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