Google Interview Question for Software Engineers

Country: United States

Comment hidden because of low score. Click to expand.
of 2 vote

I feel this question can be solved by stack. once we found previous element in the stack is larger than current element, than pop out

- lixx3527 July 09, 2019 | Flag Reply
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of 0 vote


- Anonymous July 09, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
of 0 vote

def makeheap( heap, start, N ):
	for i in reversed( xrange( start, N/2 +1 ) ):
		heapify( heap, i, N )

def heapify( heap, i, last ):
		l = i*2
		r = i*2 +1
		if l >= last:
		if r >= last:
			m = l
			if heap[l] <= heap[r]:
				m = l
				m = r
		if heap[m] < heap[i]:
			heap[m],heap[i] = heap[i],heap[m]
			i = m

def lexi( arr, k):
	n = len(arr )
	heap = [ i for i in arr]
	makeheap( heap, 0, len(arr))
	h = dict()
	for ind,i in enumerate(arr):
		if i in h:
			h[i].append( ind )
			h[i] = [ind]
	i =0
	start = 0
	end = n - k
	res = ""
	last = n-1
	c = 0
	print h
	while c < k:
		print heap
		index = h[ heap[0] ][0] 
		print index
		if index > start and index < n - k+ c:
			res+= heap[0]
			start = index
			del h[heap[0]][0]
		heap[0] = heap[last]
		heapify( heap, 0, last )
	return res

print lexi( ["5","1","2","4","12","2"],2)

- Newman October 07, 2019 | Flag Reply

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