Another Queue Example

• Given list of numbers, order is important, a span consists of the number of concecutive numbers to the left of a given number that are less than or equal to that number.
• Examples:

  7 4 3 5 8
      The span for 8 is 5
      The span for 5 is 3
      The span for 3 is 1
      The span for 4 is 1
      The span for 7 is 1
     

• Given a list of numbers, compute the span:
• A naive algorithm

  
  Input array A of size n
  Output s an array of spans
  
  for(i=0;i<n;i++) {
     k = 0
     done = false
     repeat
         if (A[i-k] <= A[i]) {
            k = k + 1
         } else {
            done = true
     while (k < i) 
     s[i] = k
  }
      

• For each element, step back through the array and compute the span
• What is the worst and best?
• A better algorithm

  Make an empty queue D 
  for(i=0;i<n-1;i++) {
     done = false
     while (!D.isEmpty()) && (!done) {
         if A[i] >= P[D.top()] 
             D.pop()
         else
             done = true
     }
  
     if D.isEmpty() then
         h = -1
     else
         h = D.top()
     S[i] = i-h
     D.push(i)
  }
      

• We then remove all spans that began with elements smaller than the current element from the stack
• We put the index of the element that began the last span onto the stack. (The current element)
• Consider:

  8 6 2 4 2 5 7 
      Action    Elements     Stack
      push 8    8            0
      push 6    8 6          0 1
      push 2    8 6 2        0 1 2
      pop 2     
      push 4    8 6 4        0 1 3
      push 2    8 6 4 2      0 1 3 4
      pop 2 
      pop 4
      push 5    8 6 5        0 1 5
      pop 5
      pop 6
      push 7    8 7          0 6
      

• Argue that this is O(n)