Google Interview Question for SDE1s


Country: United States




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

Sort the array, then use two pointers to count subsets.

- ak_domi March 13, 2017 | Flag Reply
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1
of 1 vote

public static int countSubSet(int[] nums, int target){ 
         if(nums.length==0)
             return 0;
     Arrays.sort(nums);
        int subsets=0;
        int start=0, end=nums.length-1;
        
        while(start<=end){
            if(2*nums[start]>=target)
                return subsets;
           if(nums[start]+nums[end]<target){
               if(start==end)
                   subsets=subsets+1;
               else 
                subsets= subsets+(int)Math.pow(2,end-start-1);
                 
               start++;
               }
            else
                end--;
        }
  return subsets;       
}

- deep March 14, 2017 | Flag Reply
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0
of 0 vote

For example, if target will be equal 2*Max + 1 (Max - is a maximum element in the array) we have to print all possible subsets N * (N + 1) / 2
So, solution can't be faster than O(N*N) ?

- dmitry.labutin March 13, 2017 | Flag Reply
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0
of 0 vote

the idea is to learn from past subsets, if you increase the lower bound, the upper bound may stay the same or decrease... maybe be neatly expressed in a DP recursion as well, here the hands-on solution in O(n*log(n)) due to sort, O(n) if nums is passed already sorted (and nums.sort() is removed)

# 
# 1) sort the array
# 2) start with i at 0 and j at n-1
# 3) repeat as long as nums[i] < target
# 4) try to find an upper bound which is smaller or equal than last j
# 5) add j - i + 1 or at least 1
# 6) increment goto 3)
# 
# example with sorted input for simplicity:
# nums: [1, 4, 5, 7, 11, 14, 20]
# 1..11: 5 subsets: {1}, {1,4}, {1,4,5}, {1,4,5,7}, {1,4,5,7,11}
# 4..7:  3 subsets: {4}, {4,5}, {4,5,7}
# 5..7:  2 subsets: {5}, {5,7}
# 7:     1 subset: {7}
# 11:    1 subset: {11}
# total: 5+3+2+1+1 = 12 subsets 
#
def countSubSet(nums, target):
    nums.sort()
    count = 0
    j =  len(nums) - 1
    i = 0
    while nums[i] < target:
        while nums[i] + nums[j] >= target: j -= 1      
        count += max(j - i + 1, 1)
        i += 1
    return count

print(countSubSet([1, 4, 5, 7, 11, 14, 20], 14))

- Chris March 13, 2017 | Flag Reply
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0
of 0 vote

If I got this correct then this is basically a combination problem. If we simplify this Lets say I give you 10 numbers then you have to generate possible combination of two numbers [ Because I think max(subset) and min(subset) is the length of subsets] that will sum to a value less than 10.

- Rohit March 14, 2017 | Flag Reply
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0
of 0 vote

If i understand this correct we need to generate K-subsets here. Given a number lets say 10 then we need to generate all possible combination of its subsets. for eg {1},{1,2},{5,6,7,9} having length 1,2,4 etc etc. So basically this boils down to combination problem where we need to find all possible 2 element combination of Target numbers and choose only those numbers whose sum is less than Target Value.

- SO90 March 14, 2017 | Flag Reply
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0
of 0 vote

If i understand this correct we need to generate K-subsets here. Given a number lets say 10 then we need to generate all possible combination of its subsets. for eg {1},{1,2},{5,6,7,9} having length 1,2,4 etc etc. So basically this boils down to combination problem where we need to find all possible 2 element combination of Target numbers and choose only those numbers whose sum is less than Target Value.

- testuser8190 March 14, 2017 | Flag Reply
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0
of 0 vote

dd

- Anonymous March 17, 2017 | Flag Reply
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0
of 0 vote

int a[] = {1,4,5,7,11,14,20};
Arrays.sort(a);
int target = 14;

int i =0;
int j = a.length -1;
int count =0;

while(a[i] < target) {
while (a[i] + a[j] >= target) {
j --;
}
count += Math.max(j - i + 1, 1);
i ++;
}
System.out.println(count);

- Srikanth Raj March 18, 2017 | Flag Reply
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0
of 0 vote

int main() {
	// your code goes here
	int n;
	vector<int> v;
	int x;
	cin>>n;
	for(int i=0;i<n;i++) {
		cin>>x;
		v.push_back(x);
	}
	cin>>x;
	sort(v.begin(), v.end());
	int i = 0;
	int j = n-1;
	while(v[i]+v[j] > x) j--;
	int count = 0;
	while(v[i] <= x - v[j]) i++;
	cout<<i<<" "<<j<<endl;
	count += ((pow(2, i)-1) * (pow(2, j-i+1)));
	
	for(;i<n;i++) {
		if(v[i] <x) count++;
	}
	
	cout<<count<<endl;
	return 0;
}

- Janani March 24, 2017 | Flag Reply
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0
of 0 vote

int countSubSet( const vector<int>& nums , int target ){
	int total = 0;
	sort( nums.begin() , nums.end() );
	
	for( int i = 0 ; i < nums.size(); ++i ){
		auto it = lower_bound( nums.begin() , nums.end() , target - nums[i] );
		if( it - nums.begin() > i )
			continue;
		
		int j = ( nums[i] + *it == target ) ? it - nums.begin() - 1 : it - nums.begin();
		while( j >= 0 ){
			total += pow( 2 , max( 0 , i - j - 1 ) );
			--j;
		}
	}
	return total;
}

- anonymous March 25, 2017 | Flag Reply
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-1
of 1 vote

We don't need to sort the array! First we find the minimum by one pass. If 2*minimum is not less than the target, there is no such subset and the count is zero. Other wise, we do another pass and count all elements that minimum plus such element is less than the target, say we get N. Since any subset with the above property must be a subset of such elements, and vice versa, the answer is 2^N. So the solution is O(n)!

Note that there no repeated elements, otherwise we would need to sort.

- Maryam March 14, 2017 | Flag Reply


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