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Yahoo Interview Question for Software Engineer / Developers


Team: Ad
Country: United States
Interview Type: In-Person
More Questions from This Interview



Comment hidden because of low score. Click to expand.
1
of 1 vote

This can be done using a Set. We will put one element from the array at a time in the Set. If the add operation returns false then it means that element was already present in the Set and thus is a duplicate.

- SumitGaur March 20, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

o(n) runtime and O(1) space complexity

- duskan March 20, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
1
of 1 vote

Is there actually a solution can be done by O(N) time and O(1) space?

- hellNoWord March 20, 2014 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 votes

With the question as it is, no. You would need to know more about the data in the array to be able to do it in O(n) time and O(1) space.

- Anonymous March 23, 2014 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

Either a set or a Hash Table. Someone already mentioned how to use a set, but you could create a hash table, w/ boolean value true during inserts. If there's a collision -> return true. There are a bunch of ways to simplify the hash table.

- Joshua March 20, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

a = [1,2,34,545,334,53,534,56,7,4,3,1]
def findDup():
    
    d = {}
    for i in a:
        print d
        try:
            d[i] = d[i] + 1
            if d[i] > 1:
                return True
            else:
                continue
        except:
            d[i] = 1
            continue
        else:
            continue
    return False

print findDup()

- sudevs March 20, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes
If you are doing it in Python, might as well do it in Pythonic way :P (Use collections.Counter) {{{ import collections def findDup(a): d = collections.Counter(a) for key, value in d.items(): if value > 1: return True return False a = [1, 2, 34, 545, 334, 53, 534, 56, 7, 4, 3, 1] print findDup(a) - akshaycj47 November 15, 2015 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 votes

If you are doing it in Python, might as well do it in Pythonic way :P
(Use collections.Counter)

import collections 

def findDup(a): 
	d = collections.Counter(a)
	for key, value in d.items():
		if value > 1:
			return True
	return False

a = [1, 2, 34, 545, 334, 53, 534, 56, 7, 4, 3, 1] 
print findDup(a)

- akshaycj47 November 15, 2015 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

a = [1,2,34,545,334,53,534,56,7,4,3,1]
def findDup():
    
    d = {}
    for i in a:
        print d
        try:
            d[i] = d[i] + 1
            if d[i] > 1:
                return True
            else:
                continue
        except:
            d[i] = 1
            continue
        else:
            continue
    return False

print findDup()

- sudevs March 20, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

var findDuplicatedItems = (function() {
    var obj = {}, result = [],
        len = 0;
    
    return {
        getDup: function(arrData) {
            len = arrData.length;
            for(var i = 0; i < len; i++) {
                if (obj[arrData[i]]) result.push(arrData[i]);
                else obj[arrData[i]] = true;
            }
            console.log(result);
        }
    };
})();

var arr = [1, 2, 3, 4, 77, 54, 3, 57, 2, 99, 54];
findDuplicatedItems.getDup(arr);

- Mason March 22, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

For a Set or HashMap there is space complexity and without them it is impossible to do it in O(n) time complexity and O(1) space complexity unless you are exploiting any data specific properties.

- Sreekar October 29, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

If the array has exactly 0 or 1 duplicate then it can be found in O(N) time and O(1) space complexity if it is an array of char or int.

Step1 : Sum the array .If it is char array convert to ascii no and sum the array up.
Step 2: XOR the array and compare this value with sum from step 1.
if (sum from step1 == sum from step2) {
return no duplicates
}
else {
return ascii (step1 - step2)/2;
}

- sundarrajanvyas June 22, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

How about using a BloomFilter..

- Madhur Kapoor October 14, 2015 | Flag Reply


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