CareerCup

def findIndex(arr):
	starting = 0
	ending = len(arr)
	for i in range(len(arr)-1):
		if(arr[i] > arr[i+1]):
			starting = i
			break
			
	for i in range(len(arr)-1,-1,-1):
		if(arr[i] < arr[i-1]):
			ending = i
			break
	print(starting)
	print(ending)

Actual problem can be solved, for all inputs this way , worthy of SDE 3 tag. Kudos Amazon.

/* 
select domain of inversions,
and imagine starting with (0,0) which is sorted ascending.
*/
def find_if_whatever( array ){
  N = size(array)
  if ( N < 2 ) return true // boundary condition
  // now, anything else.
  domains = list()
  domain = { 's' : 0, 'e'  : 0 , 'st' : true }
  for ( i : [ 1 : N ] ){
    sorted = ( array[domain.e] <= array[i] ) 
    if ( sorted == domain.st ){
      // add to domain 
      domain.e = i 
    } else {
       // end domain, create domain 
       domains += domain
       domain = { 's' : i, 'e'  : i , 'st' : sorted }
    }
  } 
  // now am done ...
  domains += domain
  println( jstr( domains ) ) // debug???
  // now, well, well... 
  D = size( domains) 
  if ( D > 3 ) return false // no go... right?
  // this is a clear go 
  if ( D == 1 ) return true 
  // now, when D=2, we have.. problems.. so 
  if ( D == 2 ){
    if ( domains[0].st && !domains[1].st ) return array[domains[0].e] <= array[domains[1].s] 
    if ( !domains[0].st && domains[1].st ) return array[domains[0].s] <= array[domains[1].s] 
    return false 
  }
  // now when D==3, so... larger problems...
   if ( domains[0].st && domains[2].st ) { 
     return array[domains[0].e] <= array[domains[1].e] && 
            array[domains[1].s] <= array[domains[2].s] 
   } 
   return false // done 
}

 
println( find_if_whatever([ 1, 10,-1 ]) )
println( find_if_whatever([ 1, 2,3,4,8,7,6,9  ]) )
println( find_if_whatever([ 1,2,3,50,40,90,10,23,99  ]) )