# Does anyone have a solution that does not sort and does not use a map?

• All the solutions I'm seeing here are either "sort then count" (satisfying the O(1) space constraint but not the O(n) runtime constraint) or "count by hashmap" (satisfying the O(n) runtime constraint but not the O(1) space constraint).

Does anyone have a solution that actually satisfies the requirements?

• Solution in O(n) runtime and O(1) space

``````vector<int> majorityElement(vector<int>& nums) {
vector<int> ans;
if (nums.size() == 0) return ans;
int qnt = 2;
int x = nums[0];
for (int i = 1; i < nums.size(); i++) {
if (nums[i] == x) {
qnt += 2;
} else {
qnt--;
if (!qnt) {
x = nums[i];
qnt = 1;
}
}
}
qnt = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] == x) qnt++;
}
if (qnt > nums.size() / 3) {
ans.push_back(x);
}
qnt = 1;
int y = x;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] == x) continue;
if (nums[i] == y) {
qnt++;
} else {
qnt--;
if (!qnt) {
y = nums[i];
qnt = 1;
}
}
}
if (y != x) {
qnt = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] == y) qnt++;
}
if (qnt > nums.size() / 3) {
ans.push_back(y);
}
}
return ans;
}``````

• ``````class Solution {
public:
vector<int> majorityElement(vector<int>& nums) {
vector<int> res;
int size = nums.size();
int temp1=INT_MAX-1,temp2=INT_MAX-1,counter1=0,counter2=0;
for(int i=0;i<size;i++){
if(nums[i]==temp1) counter1++;
else if(nums[i]==temp2) counter2++;
else {
if(counter1==0) {
temp1=nums[i];
counter1++;
}
else if(counter2==0){
temp2=nums[i];
counter2++;
}
else {
counter1--;
counter2--;
}
}
}
counter1=0;
counter2=0;
for(int i=0;i<size;i++){
if(nums[i]==temp1) counter1++;
if(nums[i]==temp2) counter2++;
}
if(counter1>size/3) res.push_back(temp1);
if(counter2>size/3) res.push_back(temp2);
return res;

}
};``````

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