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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Absolute Permutation
  5. Discussions

Absolute Permutation

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545 Discussions

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  • naufalfh19
     + 0 comments

    Python3 Solution

    def absolutePermutation(n, k):
    if k == 0:
        return [i+1 for i in range(n)]
        
    lexMap = {}
    for i in range(n):
        lexMap[i+1] = True
    
    res = []
    for i in range(1, n+1):
        if i-k in lexMap:
            res.append(i-k)
            del lexMap[i-k]
        elif i+k in lexMap:
            res.append(i+k)
            del lexMap[i+k]
    
    return res if len(lexMap) == 0 else [-1]

  • wifivop476
     + 0 comments

    The task can only be solved under the condition that the array is divided into blocks of size multiples of 2*k, where in each block, result[i] = input[i] + k for "k" times and result[i] = input[i] - k for "k" times. Example for k=2, (k+1,k+2,3-k,4-k) (k+5,k+6,7-k,8-k) ... result=[3,4,1,2 7,8,5,6, ...]

    Sorry if I didn't explain it very clearly. fore more visit https://www.firenzerentals.com/

  • petru_braha
     + 0 comments

    c++ solution:

    vector<int> absolutePermutation(int n, int k) {
    if (k == 0) {
        vector<int> permutation(n, 0);
        for (int index = 0; index < n; index++) {
            permutation[index] = index + 1;
        }
        return permutation;
    }
    
    // we need an even count of k-chucks in order to find an absolute permutation
    if (k > n / 2 || n % k != 0 || (n / k) % 2 != 0) {
        return { -1 };
    }
    
    vector<int> permutation(n, 0);
    // process two chucks in a single iteration
    for (int indexChucks = 0; indexChucks < n; indexChucks += 2 * k) {
        for (int indexElement = 0; indexElement < k; indexElement++) {
            int index = indexChucks + indexElement;
            permutation[index] = k + index + 1;
            permutation[k + index] = index + 1;
        }
    }
    
    return permutation;
}

  • anarod_val25
     + 0 comments

    c++

    vector<int> absolutePermutation(int n, int k) {    
    vector<int> permuted(n,0);
    vector<bool> ussage(n,false);
    
    for (int i = 1; i<=n; i++){
        
        if(i-k>0 && ussage[i-k-1]==false){
            permuted[i-k-1] = i;
            ussage[i-k-1] = true;
        }else if(i+k<=n && ussage[i+k-1]==false){
            permuted[i+k-1] = i;
            ussage[i+k-1] = true;
        }else{
            return {-1};
        }       
    }
    
    
    return permuted;
}

  • aliraza413063321
     + 0 comments

    Locksmith explains the concept of absolute permutation, a mathematical problem often explored in algorithm design and competitive programming. It involves arranging numbers in a sequence where the absolute difference between each element and its position matches a given value. Solving absolute permutation requires logical reasoning, efficient coding, and problem-solving skills. This challenge tests algorithmic thinking and is widely used for learning optimization techniques, making it both educational and practical for programmers and students.