'Hackerrank: Climbing the Leaderboard
i have a deal with a hackerrank algorithm problem.
It works at all cases, except 6-7-8-9. It gives timeout error. I had spent so much time at this level. Someone saw where is problem?
static long[] climbingLeaderboard(long[] scores, long[] alice)
{
//long[] ranks = new long[scores.Length];
long[] aliceRanks = new long[alice.Length]; // same length with alice length
long lastPoint = 0;
long lastRank;
for (long i = 0; i < alice.Length; i++)
{
lastPoint = scores[0];
lastRank = 1;
bool isIn = false; // if never drop in if statement
for (long j = 0; j < scores.Length; j++)
{
if (lastPoint != scores[j]) //if score is not same, raise the variable
{
lastPoint = scores[j];
lastRank++;
}
if (alice[i] >= scores[j])
{
aliceRanks[i] = lastRank;
isIn = true;
break;
}
aliceRanks[i] = !isIn & j + 1 == scores.Length ? ++lastRank : aliceRanks[i]; //drop in here
}
}
return aliceRanks;
}
Solution 1:[1]
This problem can be solved in O(n) time, no binary search needed at all. First, we need to extract the most useful piece of data given in the problem statement, which is,
The existing leaderboard, scores, is in descending order.
Alice's scores, alice, are in ascending order.
An approach that makes this useful is to create two pointers, one at the start of alice array, let's call it "i", and the second is at the end of scores array, let's call it "j". We then loop until i reaches the end of alice array and at each iteration, we check for three main conditions. We increment i by one if alice[i] is less than scores[j] because the next element of alice may be also less than the current element of scores, or we decrement j if alice[i] is greater than scores[j] because we are sure that the next elements of alice are also greater than those elements discarded in scores. The last condition is that if alice[i] == scores[j], we only increment i.
I solved this question in C++, my goal here is to make you understand the algorithm, I think you can easily convert it to C# if you understand it. If there are any confusions, please tell me. Here is the code:
// Complete the climbingLeaderboard function below.
vector<int> climbingLeaderboard(vector<int> scores, vector<int> alice) {
int j = 1, i = 1;
// this is to remove duplicates from the scores vector
for(i =1; i < scores.size(); i++){
if(scores[i] != scores[i-1]){
scores[j++] = scores[i];
}
}
int size = scores.size();
for(i = 0; i < size-j; i++){
scores.pop_back();
}
vector<int> ranks;
i = 0;
j = scores.size()-1;
while(i < alice.size()){
if(j < 0){
ranks.push_back(1);
i++;
continue;
}
if(alice[i] < scores[j]){
ranks.push_back(j+2);
i++;
} else if(alice[i] > scores[j]){
j--;
} else {
ranks.push_back(j+1);
i++;
}
}
return ranks;
}
I think this may help you too:
vector is like an array list that resizes itself.
push_back() is inserting at the end of the vector.
pop_back() is removing from the end of the vector.
Solution 2:[2]
Here is a solution that utilizes BinarySearch. This method returns the index of the searched number in the array, or if the number is not found then it returns a negative number that is the bitwise complement of the index of the next element in the array. Binary search only works in sorted arrays.
public static int[] GetRanks(long[] scores, long[] person)
{
var defaultComparer = Comparer<long>.Default;
var reverseComparer = Comparer<long>.Create((x, y) => -defaultComparer.Compare(x, y));
var distinctOrderedScores = scores.Distinct().OrderBy(i => i, reverseComparer).ToArray();
return person
.Select(i => Array.BinarySearch(distinctOrderedScores, i, reverseComparer))
.Select(pos => (pos >= 0 ? pos : ~pos) + 1)
.ToArray();
}
Usage example:
var scores = new long[] { 100, 100, 50, 40, 40, 20, 10 };
var alice = new long[] { 5, 25, 50, 120 };
var ranks = GetRanks(scores, alice);
Console.WriteLine($"Ranks: {String.Join(", ", ranks)}");
Output:
Ranks: 6, 4, 2, 1
Solution 3:[3]
Here is my solution with c#
public static List<int> climbingLeaderboard(List<int> ranked, List<int> player)
{
List<int> result = new List<int>();
ranked = ranked.Distinct().ToList();
var pLength = player.Count;
var rLength = ranked.Count-1;
int j = rLength;
for (int i = 0; i < pLength; i++)
{
for (; j >= 0; j--)
{
if (player[i] == ranked[j])
{
result.Add(j + 1);
break;
}
else if(player[i] < ranked[j])
{
result.Add(j + 2);
break;
}
else if(player[i] > ranked[j]&&j==0)
{
result.Add(1);
break;
}enter code here
}
}
return result;
}
Solution 4:[4]
I was bored so i gave this a go with Linq and heavily commented it for you,
Given
public static IEnumerable<int> GetRanks(long[] scores, long[] person)
// Convert scores to a tuple
=> scores.Select(s => (scores: s, isPerson: false))
// convert persons score to a tuple and concat
.Concat(person.Select(s => (scores: s, isPerson: true)))
// Group by scores
.GroupBy(x => x.scores)
// order by score
.OrderBy(x => x.Key)
// select into an indexable tuple so we know everyones rank
.Select((groups, i) => (rank: i, groups))
// Filter the person
.Where(x => x.groups.Any(y => y.isPerson))
// select the rank
.Select(x => x.rank);
Usage
static void Main(string[] args)
{
var scores = new long[]{1, 34, 565, 43, 44, 56, 67};
var alice = new long[]{578, 40, 50, 67, 6};
var ranks = GetRanks(scores, alice);
foreach (var rank in ranks)
Console.WriteLine(rank);
}
Output
1
3
6
8
10
Solution 5:[5]
Based on the given constraint brute-force solution will not be efficient for the problem. you have to optimize your code and the key part here is to look up for exact place which can be effectively done by using binary search.
Here is the solution using binary search:-
static int[] climbingLeaderboard(int[] scores, int[] alice) {
int n = scores.length;
int m = alice.length;
int res[] = new int[m];
int[] rank = new int[n];
rank[0] = 1;
for (int i = 1; i < n; i++) {
if (scores[i] == scores[i - 1]) {
rank[i] = rank[i - 1];
} else {
rank[i] = rank[i - 1] + 1;
}
}
for (int i = 0; i < m; i++) {
int aliceScore = alice[i];
if (aliceScore > scores[0]) {
res[i] = 1;
} else if (aliceScore < scores[n - 1]) {
res[i] = rank[n - 1] + 1;
} else {
int index = binarySearch(scores, aliceScore);
res[i] = rank[index];
}
}
return res;
}
private static int binarySearch(int[] a, int key) {
int lo = 0;
int hi = a.length - 1;
while (lo <= hi) {
int mid = lo + (hi - lo) / 2;
if (a[mid] == key) {
return mid;
} else if (a[mid] < key && key < a[mid - 1]) {
return mid;
} else if (a[mid] > key && key >= a[mid + 1]) {
return mid + 1;
} else if (a[mid] < key) {
hi = mid - 1;
} else if (a[mid] > key) {
lo = mid + 1;
}
}
return -1;
}
You can refer to this link for a more detailed video explanation.
Solution 6:[6]
static int[] climbingLeaderboard(int[] scores, int[] alice) {
int[] uniqueScores = IntStream.of(scores).distinct().toArray();
int [] rank = new int [alice.length];
int startIndex=0;
for(int j=alice.length-1; j>=0;j--) {
for(int i=startIndex; i<=uniqueScores.length-1;i++) {
if (alice[j]<uniqueScores[uniqueScores.length-1]){
rank[j]=uniqueScores.length+1;
break;
}
else if(alice[j]>=uniqueScores[i]) {
rank[j]=i+1;
startIndex=i;
break;
}
else{continue;}
}
}
return rank;
}
Solution 7:[7]
My solution in javascript for climbing the Leaderboard Hackerrank problem. The time complexity of the problem can be O(i+j), i is the length of scores and j is the length of alice. The space complexity is O(1).
// Complete the climbingLeaderboard function below.
function climbingLeaderboard(scores, alice) {
const ans = [];
let count = 0;
// the alice array is arranged in ascending order
let j = alice.length - 1;
for (let i = 0 ; i < scores.length ; i++) {
const score = scores[i];
for (; j >= 0 ; j--) {
if (alice[j] >= score) {
// if higher than score
ans.unshift(count+1);
} else if (i === scores.length - 1) {
// if smallest
ans.unshift(count+2);
} else {
break;
}
}
// actual rank of the score in leaderboard
if (score !== scores[i-1]) {
count++;
}
}
return ans;
}
Solution 8:[8]
My solution in Java for climbing the Leaderboard Hackerrank problem.
// Complete the climbingLeaderboard function below.
static int[] climbingLeaderboard(int[] scores, int[] alice) {
Arrays.sort(scores);
HashSet<Integer> set = new HashSet<Integer>();
int[] ar = new int[alice.length];
int sc = 0;
for(int i=0; i<alice.length; i++){
sc = 1;
set.clear();
for(int j=0; j<scores.length; j++){
if(alice[i] < scores[j] && !set.contains(scores[j])){
sc++;
set.add(scores[j]);
}
}
ar[i] = sc;
}return ar;
}
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Community |
| Solution 2 | Theodor Zoulias |
| Solution 3 | Aslam |
| Solution 4 | TheGeneral |
| Solution 5 | Kanahaiya |
| Solution 6 | Yunus Temurlenk |
| Solution 7 | user3058251 |
| Solution 8 | halfer |