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

Absolute Permutation

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542 Discussions

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  • 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.

  • kh_dang_nguyen
     + 0 comments

    Solution in Java using a HashSet

    public static List<Integer> absolutePermutation(int n, int k) {
    // Write your code here
    var set = new HashSet<Integer>();
    for (var i = 1; i <= n; i++) {
        set.add(i);
    }
    var result = new ArrayList<Integer>();
    for (var i = 1; i <= n; i++) {
        var lowerNumber = i - k;
        var higherNumber = i + k;
        if (set.remove(lowerNumber)) {
            result.add(lowerNumber);
        } else if (set.remove(higherNumber)) {
            result.add(higherNumber);
        } else {
            return List.of(-1);
        }
    }
    return result;
}

  • luka_robajac
     + 0 comments
    function absolutePermutation($n, $k) {
    $index = [];
    
    for($i = 1; $i <= $n; $i++) {
        $index[$i] = true;
    }

    $found = 0;
    $p = [];
    for($number = 1; $number <= $n; $number++) {
        $need1 = $k + $number;
        $need2 = abs($k - $number);
        if(abs($need2 - $number) === $k && ($index[$need2] ?? false)) {
            $index[$need2] = false;
            $p[] = $need2;
            $found++;
        } else if (abs($need1 - $number) === $k && ($index[$need1] ?? false)) {
            $index[$need1] = false;
            $p[] = $need1;
            $found++;
        }
    }
    
    return $found === $n ? $p : [-1];
}

  • kpprajwal54
     + 0 comments

    solution in javascript:

    function absolutePermutation(n, k) {
   const useSet = new Set();
   const res = [];
   
   for(let i = 1; i<=n; i++){
    if((i-k)>=1 && !useSet.has(i-k)){
        res.push(i-k);
        useSet.add(i-k);
    } else if((i+k) <= n && !useSet.has(i+k)){
        res.push(i+k);
        useSet.add(i+k);
    } else{
        return [-1];
    }
   }
   
   return res;
}