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

Larry's Array

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  • sainipihu1235
     + 0 comments

    string larrysArray(vector A) { int count = 0; int n = A.size(); for(int i=0; iA[j]){ count++; } } }

    if(count%2 == 0){ return "YES"; } return "NO"; }

  • sainipihu1235
     + 0 comments

    c++ code string larrysArray(vector A) { int count = 0; int n = A.size(); for(int i=0; iA[j]){ count++; } } }

    if(count%2 == 0){ return "YES"; } return "NO"; }

  • cscys_2310807
     + 0 comments

    //code develop by POLY in C

    #include<stdio.h>
int main() {
    int t;
    scanf("%d",&t);
    while (t--)
    {
        int n;
        int arr[1000];
        scanf("%d",&n);
        for (int i=0;i<n;i++)
            scanf("%d",&arr[i]);
        int inv = 0;
        for (int i=0;i<n;i++)
            for (int j=i+1;j<n;j++)
                if (arr[i] > arr[j])
                    inv ++;
        if (inv%2==0)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}

  • CamaroTheBoss
     + 0 comments

    Solution O(N^3) in C++ with sorting, cause I think it's the purpose of these challenges.

    int rotate(vector<int>& A, int pos) {
    int validPos = std::min(std::max(1, pos), (int)A.size() - 2);
    int tmp = A[validPos - 1];
    A[validPos - 1] = A[validPos];
    A[validPos] = A[validPos + 1];
    A[validPos + 1] = tmp;
    return pos - 1;
}

 void move(vector<int>& A, int currPos, int dstPos) {
     while (currPos != dstPos) {
         currPos = rotate(A, currPos);
     }
 }

string larrysArray(vector<int> A) {
    int base = 1;
    int N = A.size();
    for (int i = 0; i < N - 2; i++) {
        for (int j = i; j < N; j++) {
            if (A[j] == base) {
                move(A, j, i);
                break;
            }
        }
        base++;
    }
    if (A[N - 3] <= A[N - 2] && A[N - 2] <= A[N - 1]) {
        return "YES";
    }
    return "NO";
}

  • leoisaac_lameda
     + 1 comment
    def larrysArray(A):
    count = 0
    
    for i in range(len(A)):
        for j in range(i+1, len(A)):
            if A[i] > A[j]:
                count += 1
    
    return 'YES' if count % 2 == 0 else 'NO'