spik
BAN USER
void make_unique(vector<int> & first , vector<int> & second , vector<int> &third)
{
sort(first.begin(),first.end());
sort(second.begin(), second.end());
third.clear();
int f_pos = 0 , s_pos = 0 ; // index on first array and second array
while(f_pos < first.size() && s_pos <second.size())
{
// 2 elements to compare
if(first[f_pos] < second[s_pos])
{
third.push_back(first[f_pos]);
// i want to skip all element in first array which are the same as first[f_pos]
skip_dup(first,f_pos);
}
else if( second[s_pos] < first[f_pos])
{
third.push_back(second[s_pos]);
// // i want to skip all element in second array which are the same as second[f_pos]
skip_dup(second,s_pos);
}
else
{
// 2 elements quals so push any of them and skip dup in both
third.push_back(first[f_pos]);
skip_dup(first,f_pos);
skip_dup(second,s_pos);
}
}
while(f_pos <first.size())
{
third.push_back(first[f_pos]);
skip_dup(first,f_pos);
}
while(s_pos <second.size())
{
third.push_back(second[s_pos]);
skip_dup(second,s_pos);
}
}
Time O(NlogN) , Space O(1)
- spik August 26, 2013void rearrange(vector <int> &nums)
{
vector<int> temp;
int start_pos = 0;
for(int i=0;i<nums.size();++i)
{
if(nums[i] <0 )
nums[start_pos++] = nums[i];
else
temp.push_back(nums[i]);
}
for(int i=0;i<temp.size();++i)
nums[start_pos++] = temp[i];
}
It's O(N) time and (N) memory , i can't find solution with less memory.
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How can you check if the word you added in the heap after remove the min is not already exist in the heap. for example if you have only 2 words {ab,6} , {ac,3} so min heap head will be {ac,3} if you countered ab you will increase ab with 1 to be{ab,7} comparing it with {ac,3} which is > so remove head {ac} and insert {ab} so heap now contains repeated strings
- spik August 30, 2013