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

Larry's Array

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351 Discussions

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  • 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);
    }
}

  • 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";
}