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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Larry's Array
  5. Discussions

Larry's Array

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  • debrupchatterje1
     + 0 comments

    Using Python :

    def larrysArray(A): # Write your code here c=0 for i in range(0,len(A)-1): for j in range(i,len(A)): if(A[i]>A[j]): c+=1 if(c%2==0): return "YES" else: return "NO"

  • vuhuutuan0
     + 3 comments

    my answer in Typescript

    const NO = 'NO', YES = 'YES';
function larrysArray(A: number[]): string {
    /**
     * idea: create a [num] increament from 1, find index that contains [num] index and 
     *      always larger than 2 be cause i will rotate [index,index-2,index-1] to move
     *      [num] toward start of array. once [num] is at start of array, [num] increate
     *      and shift array. re do this till array have 2 element, if 2 element is not
     *      incrementing by 1, return NO else YES.
     */

    const rotate = (A: number[], index: number): number[] => {
        const a = [...A]
        return [...a.slice(0, index - 2), ...[a[index], a[index - 2], a[index - 1]], ...A.slice(index + 1)]
    }

    let is_larry_array = NO
    let num = 1

    while (true) {
        if (A.length < 3) {
            if (A[0] == A[1] - 1) is_larry_array = YES
            break
        }

        if (A[0] == num) { A.shift(); num++; continue }

        let index = A.indexOf(num); while (index < 2) index++

        A = rotate(A, index)
    }

    return is_larry_array
}

  • wifivop476
     + 0 comments

    string larrysArray(vector A) { int counter =0; for(int i=0;iA[j]){ counter++; } } } return (counter%2==0) ? "YES" : "NO"; } jinx manga

  • wifivop476
     + 0 comments

    string larrysArray(vector A) { int counter =0; for(int i=0;iA[j]){ counter++; } } } return (counter%2==0) ? "YES" : "NO"; }

  • alban_tyrex
     + 0 comments

    Here is my c++ solution 

    string larrysArray(vector<int> A) {
    if(A.size() == 2 && A[0] > A[1]) return "NO";
    else if(A.size() == 1 || A.size() == 0) return "YES";
    else {
        auto min_it = std::min_element(A.begin(), A.end());
        int tempIndex = distance(A.begin(), min_it);
        A.erase(A.begin() +  tempIndex);
        if(tempIndex % 2 != 0){
            swap(A[0], A[1]);
        }
        return larrysArray(A);
    }
}