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.
Move from brute-force thinking to an efficient approach using array strategy.
You are given two string arrays creators and ids, and an integer array views, all of length n. The ith video on a platform was created by creators[i], has an id of ids[i], and has views[i] views.
The popularity of a creator is the sum of the number of views on all of the creator's videos. Find the creator with the highest popularity and the id of their most viewed video.
Note: It is possible for different videos to have the same id, meaning that ids do not uniquely identify a video. For example, two videos with the same ID are considered as distinct videos with their own viewcount.
Return a 2D array of strings answer where answer[i] = [creatorsi, idi] means that creatorsi has the highest popularity and idi is the id of their most popular video. The answer can be returned in any order.
Example 1:
Input: creators = ["alice","bob","alice","chris"], ids = ["one","two","three","four"], views = [5,10,5,4]
Output: [["alice","one"],["bob","two"]]
Explanation:
The popularity of alice is 5 + 5 = 10.
The popularity of bob is 10.
The popularity of chris is 4.
alice and bob are the most popular creators.
For bob, the video with the highest view count is "two".
For alice, the videos with the highest view count are "one" and "three". Since "one" is lexicographically smaller than "three", it is included in the answer.
Example 2:
Input: creators = ["alice","alice","alice"], ids = ["a","b","c"], views = [1,2,2]
Output: [["alice","b"]]
Explanation:
The videos with id "b" and "c" have the highest view count.
Since "b" is lexicographically smaller than "c", it is included in the answer.
Constraints:
n == creators.length == ids.length == views.length1 <= n <= 1051 <= creators[i].length, ids[i].length <= 5creators[i] and ids[i] consist only of lowercase English letters.0 <= views[i] <= 105Problem summary: You are given two string arrays creators and ids, and an integer array views, all of length n. The ith video on a platform was created by creators[i], has an id of ids[i], and has views[i] views. The popularity of a creator is the sum of the number of views on all of the creator's videos. Find the creator with the highest popularity and the id of their most viewed video. If multiple creators have the highest popularity, find all of them. If multiple videos have the highest view count for a creator, find the lexicographically smallest id. Note: It is possible for different videos to have the same id, meaning that ids do not uniquely identify a video. For example, two videos with the same ID are considered as distinct videos with their own viewcount. Return a 2D array of strings answer where answer[i] = [creatorsi, idi] means that creatorsi has the highest popularity and idi is the id of
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Hash Map
["alice","bob","alice","chris"] ["one","two","three","four"] [5,10,5,4]
["alice","alice","alice"] ["a","b","c"] [1,2,2]
design-video-sharing-platform)design-a-food-rating-system)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2456: Most Popular Video Creator
class Solution {
public List<List<String>> mostPopularCreator(String[] creators, String[] ids, int[] views) {
int n = ids.length;
Map<String, Long> cnt = new HashMap<>(n);
Map<String, Integer> d = new HashMap<>(n);
for (int k = 0; k < n; ++k) {
String c = creators[k], i = ids[k];
long v = views[k];
cnt.merge(c, v, Long::sum);
if (!d.containsKey(c) || views[d.get(c)] < v
|| (views[d.get(c)] == v && ids[d.get(c)].compareTo(i) > 0)) {
d.put(c, k);
}
}
long mx = 0;
for (long x : cnt.values()) {
mx = Math.max(mx, x);
}
List<List<String>> ans = new ArrayList<>();
for (var e : cnt.entrySet()) {
if (e.getValue() == mx) {
String c = e.getKey();
ans.add(List.of(c, ids[d.get(c)]));
}
}
return ans;
}
}
// Accepted solution for LeetCode #2456: Most Popular Video Creator
func mostPopularCreator(creators []string, ids []string, views []int) (ans [][]string) {
cnt := map[string]int{}
d := map[string]int{}
for k, c := range creators {
i, v := ids[k], views[k]
cnt[c] += v
if j, ok := d[c]; !ok || views[j] < v || (views[j] == v && ids[j] > i) {
d[c] = k
}
}
mx := 0
for _, x := range cnt {
if mx < x {
mx = x
}
}
for c, x := range cnt {
if x == mx {
ans = append(ans, []string{c, ids[d[c]]})
}
}
return
}
# Accepted solution for LeetCode #2456: Most Popular Video Creator
class Solution:
def mostPopularCreator(
self, creators: List[str], ids: List[str], views: List[int]
) -> List[List[str]]:
cnt = defaultdict(int)
d = defaultdict(int)
for k, (c, i, v) in enumerate(zip(creators, ids, views)):
cnt[c] += v
if c not in d or views[d[c]] < v or (views[d[c]] == v and ids[d[c]] > i):
d[c] = k
mx = max(cnt.values())
return [[c, ids[d[c]]] for c, x in cnt.items() if x == mx]
// Accepted solution for LeetCode #2456: Most Popular Video Creator
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
pub fn rust_example() {
// Port the logic from the reference block below.
}
}
// Reference (java):
// // Accepted solution for LeetCode #2456: Most Popular Video Creator
// class Solution {
// public List<List<String>> mostPopularCreator(String[] creators, String[] ids, int[] views) {
// int n = ids.length;
// Map<String, Long> cnt = new HashMap<>(n);
// Map<String, Integer> d = new HashMap<>(n);
// for (int k = 0; k < n; ++k) {
// String c = creators[k], i = ids[k];
// long v = views[k];
// cnt.merge(c, v, Long::sum);
// if (!d.containsKey(c) || views[d.get(c)] < v
// || (views[d.get(c)] == v && ids[d.get(c)].compareTo(i) > 0)) {
// d.put(c, k);
// }
// }
// long mx = 0;
// for (long x : cnt.values()) {
// mx = Math.max(mx, x);
// }
// List<List<String>> ans = new ArrayList<>();
// for (var e : cnt.entrySet()) {
// if (e.getValue() == mx) {
// String c = e.getKey();
// ans.add(List.of(c, ids[d.get(c)]));
// }
// }
// return ans;
// }
// }
// Accepted solution for LeetCode #2456: Most Popular Video Creator
function mostPopularCreator(creators: string[], ids: string[], views: number[]): string[][] {
const cnt: Map<string, number> = new Map();
const d: Map<string, number> = new Map();
const n = ids.length;
for (let k = 0; k < n; ++k) {
const [c, i, v] = [creators[k], ids[k], views[k]];
cnt.set(c, (cnt.get(c) ?? 0) + v);
if (!d.has(c) || views[d.get(c)!] < v || (views[d.get(c)!] === v && ids[d.get(c)!] > i)) {
d.set(c, k);
}
}
const mx = Math.max(...cnt.values());
const ans: string[][] = [];
for (const [c, x] of cnt) {
if (x === mx) {
ans.push([c, ids[d.get(c)!]]);
}
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
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.
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.
Review these before coding to avoid predictable interview regressions.
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.
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.