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

Solution 1

Python3JavaC++GoTypeScriptRust

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

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;
    }
};

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
    }
}

Solution 2

C++

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
}

Comments