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Minimum Index Sum of Two Lists - Hash Table - Easy - LeetCode - MiniTV

Minimum Index Sum of Two Lists - Hash Table - Easy - LeetCode - मिनी टीवी

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.

public class Solution {
    public string[] FindRestaurant(string[] list1, string[] list2) {
        var map1 = new Dictionary<string,int>();
        var map2 = new Dictionary<string,int>();
        for(int i=0;i<list1.Length;i++){
            map1.Add(list1[i],i);
        }
        
        for(int i=0;i<list2.Length;i++){
            map2.Add(list2[i],i);
        }
        
        int min = int.MaxValue;
        var result = new List<string>();
        foreach(var item in map1){
            if(map2.ContainsKey(item.Key)){
                var sum = item.Value+map2[item.Key];
                if(sum<min){
                    min = sum;
                    result.Clear();
                    result.Add(item.Key);
                }         
                else if(sum==min){
                    result.Add(item.Key);
                }
            }
        }
        
        return result.ToArray();
    }
}

 


Time Complexity: O(n)

Space Complexity: O(n)