LeetCode #599 — EASY

Minimum Index Sum of Two Lists

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

["Shogun","Tapioca Express","Burger King","KFC"]
["Piatti","The Grill at Torrey Pines","Hungry Hunter Steakhouse","Shogun"]

Example 2

["Shogun","Tapioca Express","Burger King","KFC"]
["KFC","Shogun","Burger King"]

Example 3

["happy","sad","good"]
["sad","happy","good"]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #599: Minimum Index Sum of Two Lists
class Solution {
    public String[] findRestaurant(String[] list1, String[] list2) {
        Map<String, Integer> d = new HashMap<>();
        for (int i = 0; i < list2.length; ++i) {
            d.put(list2[i], i);
        }
        List<String> ans = new ArrayList<>();
        int mi = 1 << 30;
        for (int i = 0; i < list1.length; ++i) {
            if (d.containsKey(list1[i])) {
                int j = d.get(list1[i]);
                if (i + j < mi) {
                    mi = i + j;
                    ans.clear();
                    ans.add(list1[i]);
                } else if (i + j == mi) {
                    ans.add(list1[i]);
                }
            }
        }
        return ans.toArray(new String[0]);
    }
}

// Accepted solution for LeetCode #599: Minimum Index Sum of Two Lists
func findRestaurant(list1 []string, list2 []string) []string {
	d := map[string]int{}
	for i, s := range list2 {
		d[s] = i
	}
	ans := []string{}
	mi := 1 << 30
	for i, s := range list1 {
		if j, ok := d[s]; ok {
			if i+j < mi {
				mi = i + j
				ans = []string{s}
			} else if i+j == mi {
				ans = append(ans, s)
			}
		}
	}
	return ans
}

# Accepted solution for LeetCode #599: Minimum Index Sum of Two Lists
class Solution:
    def findRestaurant(self, list1: List[str], list2: List[str]) -> List[str]:
        d = {s: i for i, s in enumerate(list2)}
        ans = []
        mi = inf
        for i, s in enumerate(list1):
            if s in d:
                j = d[s]
                if i + j < mi:
                    mi = i + j
                    ans = [s]
                elif i + j == mi:
                    ans.append(s)
        return ans

// Accepted solution for LeetCode #599: Minimum Index Sum of Two Lists
use std::collections::HashMap;

impl Solution {
    pub fn find_restaurant(list1: Vec<String>, list2: Vec<String>) -> Vec<String> {
        let mut d = HashMap::new();
        for (i, s) in list2.iter().enumerate() {
            d.insert(s, i);
        }

        let mut ans = Vec::new();
        let mut mi = std::i32::MAX;

        for (i, s) in list1.iter().enumerate() {
            if let Some(&j) = d.get(s) {
                if (i as i32 + j as i32) < mi {
                    mi = i as i32 + j as i32;
                    ans = vec![s.clone()];
                } else if (i as i32 + j as i32) == mi {
                    ans.push(s.clone());
                }
            }
        }

        ans
    }
}

// Accepted solution for LeetCode #599: Minimum Index Sum of Two Lists
function findRestaurant(list1: string[], list2: string[]): string[] {
    const d = new Map<string, number>(list2.map((s, i) => [s, i]));
    let mi = Infinity;
    const ans: string[] = [];
    list1.forEach((s, i) => {
        if (d.has(s)) {
            const j = d.get(s)!;
            if (i + j < mi) {
                mi = i + j;
                ans.length = 0;
                ans.push(s);
            } else if (i + j === mi) {
                ans.push(s);
            }
        }
    });
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.