LeetCode 169 - Majority Number

Majority Number I

Given an array of integers, the majority number is the number that occurs more than half of the size of the array. Find it.

 

Example

Given [1, 1, 1, 1, 2, 2, 2], return 1 

Challenge

O(n) time and O(1) extra space 

public int majorityNumber(List<Integer> nums) {
    int candidate = 0, count = 0;
    for(int num : nums) {
        if(num == candidate) {
            count++;
        } else if(count != 0) {
            count--;
        } else {
            candidate = num;
            count = 1;
        }
    }
    return candidate;
}

Majority Number II

Given an array of integers, the majority number is the number that occursmore than 1/3 of the size of the array. Find it.

 

Example

Given [1, 2, 1, 2, 1, 3, 3], return 1.

public List<Integer> majorityElement(int[] nums) {
    int cnt1=0, cnt2=0;
    int a = 0, b = 0;
    for(int n: nums){
        if (cnt1 == 0 || n == a) {
            cnt1++;
            a = n;
        } else if (cnt2 == 0 || n==b) {
            cnt2++;
            b = n;
        } else {
            cnt1--;
            cnt2--;
        }
    }
    cnt1 = cnt2 = 0;
    for(int n: nums){
        if (n==a) cnt1++;
        else if (n==b) cnt2++;
    }

    List<Integer> result = new ArrayList<>();
    if (cnt1 > nums.length/3) result.add(a);
    if (cnt2 > nums.length/3) result.add(b);
    return result;
}

Majority Number III

Given an array of integers and a number k, the majority number is the number that occurs more than 1/k of the size of the array. Find it.

Example

Given [3,1,2,3,2,3,3,4,4,4] and k=3, return 3.

Note

There is only one majority number in the array.

Challenge

O(n) time and O(k) extra space

public int majorityNumber(List<Integer> nums, int k) {
    Map<Integer, Integer> map = new HashMap<>();
    for(int num:nums) {
        if(map.containsKey(num)) {
            map.put(num, map.get(num)+1);
        } else {
            if(map.size() == k-1) {
                List<Integer> trash = new ArrayList<>();
                for(int key:map.keySet()) {
                    int cnt = map.get(key);
                    if(--cnt == 0) {
                        trash.add(key);
                    } else {
                        map.put(key, cnt);
                    }
                }
                for(int key:trash) map.remove(key);
            } else {
                map.put(num, 1);
            }
        }
    }
    for(int key:map.keySet()) {
        int cnt = 0;
        for(int num:nums) {
            if(num==key) cnt++;
        }
        if(cnt > nums.size()/k) return key;
    }
    return -1;
}

C++的代码： 

vector<int> majorityElement(vector<int>& nums) {
    unordered_map<int,int> map;
    for(int num:nums) {
        if(map.size() < 2 || map.count(num))  {
            map[num]++;
        } else {
            vector<int> keys;
            for(auto& p:map) {
                if(--p.second == 0) {
                    keys.push_back(p.first);
                }
            }
            for(int key:keys) map.erase(key);
        }
    }
    vector<int> result;
    for(auto& p : map) {
        int cnt = 0;
        for(int num:nums) {
            if(num == p.first && ++cnt > nums.size()/3) {
                result.push_back(num);
                break;
            }
        }
    }
    return result;
}

 

Reference:

http://www.geeksforgeeks.org/given-an-array-of-of-size-n-finds-all-the-elements-that-appear-more-than-nk-times/