CapitalIQ Interview Question for Consultants


Country: India
Interview Type: Phone Interview




Comment hidden because of low score. Click to expand.
5
of 5 vote

map mymap;
int givenarr[n];
j = 0
for i = 0 to n - 1
	if not mymap.containskey(givenarr[i])
		givenarr[j++] = givenarr[i]
		mymap[givenarr[i]] = something
while j < n
	givenarr[j++] = Nothing

- coding.arya June 19, 2013 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

Very compact code with O(n) time and space complexity! Good job @coding_arya +1

- oOZz June 19, 2013 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 votes

Good one. +1

- Murali Mohan June 19, 2013 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 votes

Nice Solution.

- Prashant June 20, 2013 | Flag
Comment hidden because of low score. Click to expand.
-1
of 1 vote

Yep, looks right. I recommend using a set instead of a map because you're not using the value portion of the key-value pairs in the map.

- eugene.yarovoi June 21, 2013 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

We can use LinkedHashSet, where repeated values are removed and it maintains the order.

- Prashanth June 22, 2013 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

#include<iostream>
#include<set>
#include<vector>
using namespace std;
void main() {
	set<int> rmvDuplicates;
	int n;
	cin>>n;
	vector<int> a(n), res;
	for (int i = 0; i < n; i++) {
		cin>>a[i];
	}
	for (int i = 0; i < n; i++) {
		int oldsize = rmvDuplicates.size();
		rmvDuplicates.insert(a[i]);
		if(rmvDuplicates.size() != oldsize) 
			res.push_back(a[i]);
	}
}

- tamojit chatterjee June 27, 2013 | Flag Reply


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