LintCode "Majority Number III"

Based on Bucketing and "Majority Number I".

class Solution {
    pair<int,int> majorityNumber0(vector<int> &num) {
        int count = 0;
        int ret = 0;
        for(int i = 0; i < num.size(); i ++)
        {
            if (count == 0)
            {
                ret = num[i];
                count = 1;
                continue;
            }
            if(ret != num[i]) count --;
            else if(ret == num[i]) count ++;
        }
        // find count
        int cnt = 0;
        for(auto v: num)
            if(v == ret) cnt ++;
        return make_pair(ret, cnt);
    }
public:
    /**
     * @param nums: A list of integers
     * @param k: As described
     * @return: The majority number
     */
    int majorityNumber(vector<int> nums, int k) 
    {
        int n = nums.size();
        
        auto mm = minmax_element(nums.begin(), nums.end());
        int minv = *mm.first, maxv = *mm.second;

        int dist = ((maxv - minv) / k) + 1;

        vector<pair<int, vector<int>>> bkt(k + 1, make_pair(0, vector<int>()));
        for(auto v: nums)
        {
            int inx = (v - minv) / dist;            
            bkt[inx].first ++;
            bkt[inx].second.push_back(v);
        }

        int tgt = n / k;
        for(auto &p : bkt)
        {
            if(p.first > tgt)
            {
                auto rp = majorityNumber0(p.second);
                if(rp.second > tgt)
                {
                    return rp.first;
                }
            }
        }
        return 0;
    }
};

And yes, the majorityNumber0() call can be inlined in the pass 1. That will make it O(k) space. One solution online with hashmap is very similar with the bucketing idea here.