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599. Minimum Index Sum of Two Lists

Description

Given two arrays of strings list1 and list2, find the common strings with the least index sum.

A common string is a string that appeared in both list1 and list2.

A common string with the least index sum is a common string such that if it appeared at list1[i] and list2[j] then i + j should be the minimum value among all the other common strings.

Return all the common strings with the least index sum. Return the answer in any order.

 

Example 1:

Input: list1 = ["Shogun","Tapioca Express","Burger King","KFC"], list2 = ["Piatti","The Grill at Torrey Pines","Hungry Hunter Steakhouse","Shogun"]
Output: ["Shogun"]
Explanation: The only common string is "Shogun".

Example 2:

Input: list1 = ["Shogun","Tapioca Express","Burger King","KFC"], list2 = ["KFC","Shogun","Burger King"]
Output: ["Shogun"]
Explanation: The common string with the least index sum is "Shogun" with index sum = (0 + 1) = 1.

Example 3:

Input: list1 = ["happy","sad","good"], list2 = ["sad","happy","good"]
Output: ["sad","happy"]
Explanation: There are three common strings:
"happy" with index sum = (0 + 1) = 1.
"sad" with index sum = (1 + 0) = 1.
"good" with index sum = (2 + 2) = 4.
The strings with the least index sum are "sad" and "happy".

 

Constraints:

• 1 <= list1.length, list2.length <= 1000
• 1 <= list1[i].length, list2[i].length <= 30
• list1[i] and list2[i] consist of spaces ' ' and English letters.
• All the strings of list1 are unique.
• All the strings of list2 are unique.
• There is at least a common string between list1 and list2.

Solutions

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• class Solution {
      public String[] findRestaurant(String[] list1, String[] list2) {
          Map<String, Integer> mp = new HashMap<>();
          for (int i = 0; i < list2.length; ++i) {
              mp.put(list2[i], i);
          }
          List<String> ans = new ArrayList<>();
          int mi = 2000;
          for (int i = 0; i < list1.length; ++i) {
              if (mp.containsKey(list1[i])) {
                  int t = i + mp.get(list1[i]);
                  if (t < mi) {
                      ans = new ArrayList<>();
                      ans.add(list1[i]);
                      mi = t;
                  } else if (t == mi) {
                      ans.add(list1[i]);
                  }
              }
          }
          return ans.toArray(new String[0]);
      }
  }
  

• class Solution {
  public:
      vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) {
          unordered_map<string, int> mp;
          for (int i = 0; i < list2.size(); ++i) mp[list2[i]] = i;
          int mi = 2000;
          vector<string> ans;
          for (int i = 0; i < list1.size(); ++i) {
              if (mp.count(list1[i])) {
                  int t = i + mp[list1[i]];
                  if (t < mi) {
                      ans.clear();
                      ans.push_back(list1[i]);
                      mi = t;
                  } else if (t == mi) {
                      ans.push_back(list1[i]);
                  }
              }
          }
          return ans;
      }
  };
  

• class Solution:
      def findRestaurant(self, list1: List[str], list2: List[str]) -> List[str]:
          ans = []
          mp = {v: i for i, v in enumerate(list2)}
          mi = 2000
          for i, v in enumerate(list1):
              if v in mp:
                  t = i + mp[v]
                  if t < mi:
                      mi = t
                      ans = [v]
                  elif t == mi:
                      ans.append(v)
          return ans
  
  

• func findRestaurant(list1 []string, list2 []string) []string {
  	mp := make(map[string]int)
  	for i, v := range list2 {
  		mp[v] = i
  	}
  	mi := 2000
  	var ans []string
  	for i, v := range list1 {
  		if _, ok := mp[v]; ok {
  			t := i + mp[v]
  			if t < mi {
  				ans = []string{v}
  				mi = t
  			} else if t == mi {
  				ans = append(ans, v)
  			}
  		}
  	}
  	return ans
  }
  

• function findRestaurant(list1: string[], list2: string[]): string[] {
      let minI = Infinity;
      const res = [];
      const map = new Map<string, number>(list1.map((s, i) => [s, i]));
      list2.forEach((s, i) => {
          if (map.has(s)) {
              const sumI = i + map.get(s);
              if (sumI <= minI) {
                  if (sumI < minI) {
                      minI = sumI;
                      res.length = 0;
                  }
                  res.push(s);
              }
          }
      });
      return res;
  }
  
  

• use std::collections::HashMap;
  use std::iter::FromIterator;
  
  impl Solution {
      pub fn find_restaurant(list1: Vec<String>, list2: Vec<String>) -> Vec<String> {
          let map: HashMap<String, usize> = HashMap::from_iter(list1.into_iter().zip(0..));
          let mut res = vec![];
          let mut min_i = usize::MAX;
          list2
              .into_iter()
              .enumerate()
              .for_each(|(i, key)| {
                  if map.contains_key(&key) {
                      let sum_i = map.get(&key).unwrap() + i;
                      if sum_i <= min_i {
                          if sum_i < min_i {
                              min_i = sum_i;
                              res.clear();
                          }
                          res.push(key);
                      }
                  }
              });
          res
      }
  }
  
  

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