Discussion on Circular Array Rotation Challenge

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3 days ago+ 0 comments

PHP

function circularArrayRotation($a, $k, $queries) {
    $result = [];
    
    foreach ($queries as $queryIndex) {
        $lastIterationOffset = $k % count($a); // eliminate iterations over whole array
        $index = $queryIndex - $lastIterationOffset; // calculate initial index before shifting
        if ($index < 0) { // initial index exceeded beggining of array limit
            $index = count($a) + $index; // counting index from the end of array
        }
        $result[] = $a[$index];
    }
    
    return $result;
}

5 days ago+ 0 comments

JAVASCRIPT BIG O(N) SOLUTION:

for(let i=0;i<k;i++) {
			// Remove the last elment and add into the first position and moves remaining element to the right
        a.unshift(a.pop());
    }
    let finalArr = queries.map((query) => a[query]);
    return finalArr;

7 days ago+ 0 comments

Simple Java solution without additional list

   int size = a.size();
    k=k%size;
    List<Integer> res = new ArrayList<>();
    for(int i:queries){
        int idx = i-k;
        if(idx<0) idx+=size;
        res.add(a.get(idx));
    }
    return res;

1 week ago+ 0 comments

Here is my c++ solution, you can watch the explanation here : https://youtu.be/MaT-4dnozJI

Solution 1 O(n + m) :

vector<int> circularArrayRotation(vector<int> a, int k, vector<int> queries) {
    vector<int> newArray(a.size()), result;
    for(int i = 0; i < a.size(); i++){
        newArray[(i + k) % a.size()] = a[i];
    }
    for(int i = 0; i < queries.size(); i++) result.push_back(newArray[queries[i]]);
    return result;
}

Solution 2 O(m) :

vector<int> circularArrayRotation(vector<int> a, int k, vector<int> queries) {
    vector<int> result;
    for(int i = 0; i < queries.size(); i++) {
        int previousIndex = (queries[i] - k + a.size() * 100000 ) % a.size();
        result.push_back(a[previousIndex]);
    }
    return result;
}

3 weeks ago+ 0 comments

Java Streams with better efficiency:

  
Deque<Integer> deque = new ArrayDeque<>(a);

IntStream.rangeClosed(1, k).forEach(value -> {
    Integer last = deque.pollLast();
    deque.addFirst(last);
});

List<Integer> dequeList = new ArrayList<>(deque);

return queries.stream()
        .map(dequeList::get)
        .collect(Collectors.toList());