## Google Interview Question for Software Engineers

Country: United States

Comment hidden because of low score. Click to expand.
3
of 3 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

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0
of 0 vote

ss

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0
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 ):
while(1):
l = i*2
r = i*2 +1
if l >= last:
return
if r >= last:
m = l
else:
if heap[l] <= heap[r]:
m = l
else:
m = r
if heap[m] < heap[i]:
heap[m],heap[i] = heap[i],heap[m]
i = m
else:
return

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 )
else:
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 ]
print index
if index > start and index < n - k+ c:
res+= heap
c+=1
start = index
del h[heap]
heap = heap[last]
last-=1
heapify( heap, 0, last )
return res

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

Comment hidden because of low score. Click to expand.
0
of 0 vote

O(N), sliding window approach, returns start and end indices of subsequence.

``````#include <bits/stdc++.h>
using namespace std;

pair<int, int> lexographically_smallest(vector<int>& v, int k) {
assert (v.size() >= k );
int mul = pow(10, k - 1);
int cur = 0, ans = 0;
pair<int, int> result_pair = {0, k-1};
for (int i = k - 1; i >= 0; i --) ans += v[i] * pow(10, k - i - 1);
cur = ans;
// cout << "cur " << cur << endl;
for (int i = k; i < v.size(); i ++) {
int temp_num = (cur % mul) * pow(10, k - 2) + v[i];

cout << "temp_num " << temp_num << endl;
if (ans > temp_num) {
ans = temp_num;
result_pair = {i - k + 1, i};
}
cur = temp_num;
}

return result_pair;
}

int main() {
vector<int> v = {6,3,2,4,2,6,8,9};
auto result = lexographically_smallest(v, 3) ;
cout << "indices " << result.first << " " << result.second << endl;
return 0;
}``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

The problem can be solved using a deque or a stack. I am taking a deque here. The idea is to maintain a deque and elements in the deque will hold lexicographically smallest, subsequence till that point of time. The incoming element ll keep popping the elements from the back of the until it is greater than the last element of deque and number of remaining elements + number of elements currently in the deque are >= k

``````vector<int> smallestLexo(vector<int> s, int k) {
deque<int> dq;
for(int i = 0; i < s.size(); i++) {
while(!dq.empty() && s[i] < dq.back() && (dq.size() + (s.size() - i - 1)) >= k) {
dq.pop_back();
}
if(dq.size() < k) {
dq.push_back(s[i]);
}
}
return vector<int> (dq.begin(), dq.end());
}``````

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