I am using a divide and conquer strategy to solve the majority problem. An element in said to be in the majority if repeats more than n/2 times where n is the number of elements given in the input. Return 1 if a majority element in present, return 0 otherwise.
This is the algorithm that I am using:- ->We will keep dividing the array into half until we reach an array of size two and we will compare the two elements of each array. ->If they are the same, they are the majority element of that array and we will return their value. ->If they are not the same, we will return a special character to signify that there is no majority element. ->Moving recursively up the arrays, we will check the values we get from the two child-arrays/halves. As with the base case above, if the elements are the same, we return them, otherwise we return the special character(special character is -1 in the code below).
//Function to count the number of occurences of an element in the input in linear time
ll simple_count(vector<ll>&v,ll p){
ll count = 0;
for(ll i=0;i<v.size();i++){
if(v[i]==p)
count++;
}
return count;
}
//Function to find the majority element using the above algorithm
ll majorityElement(vector<ll>&v,ll left, ll right){
if(left==right)
return v[left];
if(left+1==right){
if(v[left]==v[right])
return v[left];
return -1;
}
ll mid = left+(right-left)/2;
ll p = majorityElement(v,left,mid);
ll q = majorityElement(v,mid+1,right);
if(p!=-1&&q==-1)
{
if(simple_count(v,p)>((ll)v.size()/2))
return p;
return -1;
}
else if(p==-1&&q!=-1)
{
if(simple_count(v,q)>((ll)v.size()/2))
{
return q;
}
return -1;
}
else if(p!=-1&&q!=-1){
ll p_count = simple_count(v,p);
ll q_count = simple_count(v,q);
if(p_count>q_count&&p_count>v.size()/2)
return p;
else if(q_count>p_count&&q_count>v.size()/2)
return q;
return -1;
}
else if(p==q&&p!=-1)
{
if(simple_count(v,p)>(ll)v.size()/2)
return p;
return -1;
}
else if(p==q)
return -1;
}
int main(){
ll n;
cin>>n;
vector<ll>v(n);
for(ll i=0;i<n;i++)
cin>>v[i];
ll k = majorityElement(v,0,v.size()-1);
if(k==-1)
cout<<0;
else
cout<<1;
return 0;
}
The problem I am facing is that this code is giving me Time Limit Exceeded(TLE) error for a few inputs(Coursera grader does not share the inputs for which we get an error) I believe the time complexity of my code is O(nlogn). Please help me optimize the code so that this program does not give me TLE.
input format : the first line contains an integer n, followed by n non negative integers (a1,a2....an) . 1<=n<=10^5 and 0<=ai<=10^9. The array is not sorted by default
