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by Sumit Chourasia | Sep 29, 2020 | Category :coding | Tags : algorithmडेटा-संरचनाआसानhash-tableleetcode

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++){
}

for(int i=0;i<list2.Length;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();
}
else if(sum==min){
}
}
}

return result.ToArray();
}
}``````

Time Complexity: O(n)

Space Complexity: O(n)

Contributed By: Sumit Chourasia
Contributed By: Sumit Chourasia
Contributed By: Sumit Chourasia
Contributed By: Sumit Chourasia