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

Absolute Permutation

  • wanyunze
     + 0 comments

    O(n) greedy approach that dont need to consider corner cases. We always take the (i-k) choice if possible, if not try (i+k), if that fails return -1.

        # Write your code here
    
    '''
    pos[i] = i+k or i-k
    pos[i] = i
    '''
    
    visited = set()
    res = []
    
    for i in range(1, n+1):
        if i-k in visited and i+k in visited:
            return [-1]
            
        if i-k > 0 and i-k not in visited:
            res.append(i-k)
            visited.add(i-k)
        else:
            if i + k > n:
                return [-1]
                
            res.append(i+k)
            visited.add(i+k)
    
    return res