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[其他] Check If Given Array Can be Arranged In Left or Right Positioned Array

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讨厌那个你ヾ 发表于 2016-10-17 20:31:28
127 3
Given an array arr[] of size n>4, the task is to check whether the given array can be      arrangedin the form of Left or Right positioned array?   
          Left or Right Positioned Arraymeans each element in the array is equal to the number of elements to its left or number of elements to its right.   
    Examples:
  1. Input  : arr[] = {1, 3, 3, 2}
  2. Output : "YES"  
  3. This array has one such arrangement {3, 1, 2, 3}.
  4. In this arrangement, first element '3' indicates
  5. that three numbers are after it, the 2nd element
  6. '1' indicates that one number is before it, the
  7. 3rd element '2' indicates that two elements are
  8. before it.

  9. Input : arr[] = {1, 6, 5, 4, 3, 2, 1}
  10. Output: "NO"
  11. // No such arrangement is possible

  12. Input : arr[] = {2, 0, 1, 3}
  13. Output: "YES"
  14. // Possible arrangement is {0, 1, 2, 3}

  15. Input : arr[] = {2, 1, 5, 2, 1, 5}
  16. Output: "YES"
  17. // Possible arrangement is {5, 1, 2, 2, 1, 5}
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A    simple solutionis to generate all possible arrangements (seethis article) and check for the Left or Right Positioned Array condition, if each element in the array satisfies the condition then “YES” else “NO”. Time complexity for this approach is O(n*n! + n), n*n! to generate all arrangements and n for checking the condition using temporary array.  
  An    efficient solutionfor this problem needs little bit observation and pen-paper work. To satisfy the Left or Right Positioned Array condition all the numbers in the array should either be equal to index, i or (n-1-i) and arr < n. So we create an visited[] array of size n and initialize its element with 0. Then we traverse array and follow given steps :  
  
       
  • If visited[arr] = 0 then make it 1, which checks for the condition that number of elements on the left side of array arr[0]…arr[i-1] is equal to arr.   
  • Else make visited[n-arr-1] = 1, which checks for the condition that number of elements on the right side of array arr[i+1]…arr[n-1] is equal to arr.   
  • Now traverse visited[] array and if all the elements of visited[] array become 1 that means arrangement is possible “YES” else “NO”.  
  1. // C++ program to check if an array can be arranged
  2. // to left or right positioned array.
  3. #include<bits/stdc++.h>
  4. using namespace std;

  5. // Function to check Left or Right Positioned
  6. // Array.
  7. // arr[] is array of n elements
  8. // visited[] is boolean array of size n
  9. bool leftRight(int arr[],int n)
  10. {
  11.     // Initially no element is placed at any position
  12.     int visited[n] = {0};

  13.     // Traverse each element of array
  14.     for (int i=0; i<n; i++)
  15.     {
  16.         // Element must be smaller than n.
  17.         if (arr[i] < n)
  18.         {
  19.             // Place "arr[i]" at position "i"
  20.             // or at position n-arr[i]-1
  21.             if (visited[arr[i]] == 0)
  22.                 visited[arr[i]] = 1;
  23.             else
  24.                 visited[n-arr[i]-1] = 1;
  25.         }
  26.     }

  27.     // All positions must be occupied
  28.     for (int i=0; i<n; i++)
  29.         if (visited[i] == 0)
  30.             return false;

  31.     return true;
  32. }

  33. // Driver program to test the case
  34. int main()
  35. {
  36.     int arr[] = {2, 1, 5, 2, 1, 5};
  37.     int n = sizeof(arr)/sizeof(arr[0]);
  38.     if (leftRight(arr, n) == true)
  39.         cout << "YES";
  40.     else
  41.         cout << "NO";
  42.     return 0;
  43. }
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Output:
  1. "YES"
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   Time Complexity : O(n)
    Auxiliary Space : O(n)
    This article is contributed by          Shashank Mishra ( Gullu )    . If you like GeeksforGeeks and would like to contribute, you can also write an article using    contribute.geeksforgeeks.orgor mail your article to [email protected] See your article appearing on the GeeksforGeeks main page and help other Geeks.  
  Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
xdagl 发表于 2016-10-18 10:01:28
我小学十年,中学十二年,我被评为全校最熟悉的面孔,新老师来了都跟我打听学校内幕……
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邓妮 发表于 2016-10-19 04:09:28
元芳你怎么看?
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meitime 发表于 2016-10-25 10:53:42
有谁会在时过境迁之后还在那里等你。
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