Minimum, Maximum, Order Statistics

Objectives

Notes

• Given an unorderd array, what is the expected time for finding the minimum or maximum value?

  • MINIMUM(A)
       A: An unordered array
    
     min ← A[0]
     for i ← 1 to A.size-1 do
        if min > A[i]
           min = A[i]
     return min
    
           

  • This clearly works
  • And reversing the comparison would work for max
  • CLRS argues that

    This could be considered a tournment among the values
    And each value but the min must lose once
    So we need n-1 comparisons to determine the min/max in the worst case
    So this is the best we can do.
           

• The Selection Problem: given an array A with n elements, find the element that would be in position i, 0 ≤ i < n
• An obvious solution is

  • SELECT(A,i)
       A:  an unordered array
       i:  an integer 0 ≤ i < A.size
    
     B = sort(A)
     return B[i]
     

  • This is clearly O(n log₂ n)
• We could use quicksort's pivot function
  • Pivot the array, if the pivot position is i, return, otherwise pivot the sub array i would be in.

    PARTITION-SELECT(A, i, low, high)
        A: an unordered array
        i:  an integer 0 ≤ i < A.size
        low and high, valid positions in the array such that low ≤ i ≤ high
    
     if low = high 
        return A[low]
     part ←  Partition(A, low, high)
     if i == part
        return A[part]
     if i < part
        return PARTITION_SELECT(A, i, low, part-1)
     else 
        return PARTITION_SELECT(A, i, part+1, high)
     

  • This is O(n²)
  • But it does either take extra space or it scrambles the array.
  • Complex but average time will be O(n)