Leetcode Majority Element II problem solution YASH PAL, 31 July 202419 January 2026 In this Leetcode Majority Element II problem solution, you are given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times.Leetcode Majority Element II problem solution in Python.import collections class Solution: def majorityElement(self, nums): length = len(nums) counter = collections.Counter(nums) if length<3: return counter.keys() result = [] words = counter.most_common(3) for i in range(0,min(3,len(counter))): if words[i][-1]>length/3: result.append(words[i][0]) return result Majority Element II problem solution in Java.public List<Integer> majorityElement(int[] nums) { List<Integer> result = new ArrayList<>(); if(nums.length < 3){ int n = nums.length; for(int i=0; i<n; i++){ if(!result.contains(nums[i])) result.add(nums[i]); } return result; } Arrays.sort(nums); int count=1; int gt = nums.length/3; for(int i=0; i<nums.length-1; i++){ if(nums[i] == nums[i+1]){ count++; } if(count > 1 && nums[i] != nums[i+1]){ count = 1; } if(count >= gt+1){ if(!result.contains(nums[i])){ result.add(nums[i]); count = 1; } } } return result; } Problem solution in C++.vector<int> majorityElement(vector<int>& nums) { vector<int> vec; int n= floor(nums.size()/3)+1; map<int,int> mp; map<int,int>::iterator it; for(int i=0;i<nums.size();i++){ it=mp.find(nums[i]); if(it==mp.end()) mp.insert(pair<int,int>(nums[i],1)); else it->second=mp[nums[i]]+1; } for(it=mp.begin();it!=mp.end();it++){ if(it->second>=n) vec.push_back(it->first); } return vec; } Problem solution in C.void add_back(int **ret, int *rS, int num){ *ret = realloc(*ret, sizeof(int) * (*rS + 1)); (*ret)[*rS] = num; *rS += 1; return; } int* majorityElement(int* nums, int numsSize, int* returnSize){ *returnSize = 0; if(nums == NULL || numsSize < 1) return NULL; int cand1, cand2, count1, count2; cand1 = cand2 = count1 = count2 = 0; for (int i = 0; i < numsSize; i++) { int num = nums[i]; if (cand1 == num) { count1++; }else if (cand2 == num) { count2++; }else if (count1 == 0) { cand1 = num; count1++; }else if (count2 == 0) { cand2 = num; count2++; }else { count1--; count2--; } } count1 = count2 = 0; for (int i = 0; i < numsSize; i++) { int num = nums[i]; if (cand1 == num) count1++; else if (cand2 == num) count2++; } int *ret = NULL; if(count1 > numsSize/3) add_back(&ret, returnSize, cand1); if(count2 > numsSize/3) add_back(&ret, returnSize, cand2); return ret; } coding problems solutions Leetcode Problems Solutions Leetcode