Majority Element II - Elements occurring more than ⌊n/3⌋ times
Last Updated :
26 Jul, 2025
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Given an array arr[] consisting of n integers, find all the array elements which occurs more than floor(n/3) times.
Note: The returned array of majority elements should be sorted.
Examples:
Input: arr[] = [2, 2, 3, 1, 3, 2, 1, 1]
Output: [1, 2]
Explanation: The frequency of 1 and 2 is 3, which is more than floor n/3 (8/3 = 2).Input: arr[] = [-5, 3, -5]
Output: [-5]
Explanation: The frequency of -5 is 2, which is more than floor n/3 (3/3 = 1).Input: arr[] = [3, 2, 2, 4, 1, 4]
Output: [ ]
Explanation: There is no majority element.
Table of Content
[Naive Approach] Using Nested Loops - O(n^2) Time and O(1) Space
The idea is to iterate over all elements and count the frequency of the element in the array. If the frequency of the element is greater than floor(n/3), add it to the result. To avoid adding duplicate elements into the result, we can check if the element is already present in the result. We can stop the iteration if we have already found two majority elements.
#include <iostream>
#include <vector>
using namespace std;
vector<int> findMajority(vector<int> &arr) {
int n = arr.size();
vector<int> res;
for (int i = 0; i < n; i++) {
// Count the frequency of arr[i]
int cnt = 0;
for (int j = i; j < n; j++) {
if (arr[j] == arr[i])
cnt += 1;
}
// Check if arr[i] is a majority element
if (cnt > (n / 3)) {
// Add arr[i] only if it is not already
// present in the result
if (res.size() == 0 || arr[i] != res[0]) {
res.push_back(arr[i]);
}
}
// If we have found two majority elements,
// we can stop our search
if (res.size() == 2) {
if(res[0] > res[1])
swap(res[0], res[1]);
break;
}
}
return res;
}
int main() {
vector<int> arr = {2, 2, 3, 1, 3, 2, 1, 1};
vector<int> res = findMajority(arr);
for (int ele : res)
cout << ele << " ";
return 0;
}
#include <stdio.h>
int *findMajority(int *arr, int n, int *resSize) {
int *res = (int *)malloc(2 * sizeof(int));
*resSize = 0;
for (int i = 0; i < n; i++) {
// Count the frequency of arr[i]
int cnt = 0;
for (int j = i; j < n; j++) {
if (arr[j] == arr[i])
cnt += 1;
}
// Check if arr[i] is a majority element
if (cnt > (n / 3)) {
// Add arr[i] only if it is not already
// present in the result
if (*resSize == 0 || arr[i] != res[0]) {
res[*resSize] = arr[i];
(*resSize)++;
}
}
// If we have found two majority elements,
// we can stop our search
if (*resSize == 2) {
if (res[0] > res[1]) {
int temp = res[0];
res[0] = res[1];
res[1] = temp;
}
break;
}
}
return res;
}
int main() {
int arr[] = {2, 2, 3, 1, 3, 2, 1, 1};
int n = sizeof(arr) / sizeof(arr[0]);
int resSize;
int *res = findMajority(arr, n, &resSize);
for (int i = 0; i < resSize; i++)
printf("%d ", res[i]);
return 0;
}
import java.util.ArrayList;
class GfG {
static ArrayList<Integer> findMajority(int[] arr) {
int n = arr.length;
ArrayList<Integer> res = new ArrayList<>();
for (int i = 0; i < n; i++) {
// Count the frequency of arr[i]
int cnt = 0;
for (int j = i; j < n; j++) {
if (arr[j] == arr[i])
cnt += 1;
}
// Check if arr[i] is a majority element
if (cnt > (n / 3)) {
// Add arr[i] only if it is not already present
if (res.size() == 0 || arr[i] != res.get(0)) {
res.add(arr[i]);
}
}
// If we have found two majority elements,
// we can stop our search
if (res.size() == 2) {
if (res.get(0) > res.get(1))
java.util.Collections.swap(res, 0, 1);
break;
}
}
return res;
}
public static void main(String[] args) {
int[] arr = {2, 2, 3, 1, 3, 2, 1, 1};
ArrayList<Integer> res = findMajority(arr);
for (int ele : res)
System.out.print(ele + " ");
}
}
def findMajority(arr):
n = len(arr)
res = []
for i in range(n):
# Count the frequency of arr[i]
cnt = 0
for j in range(i, n):
if arr[j] == arr[i]:
cnt += 1
# Check if arr[i] is a majority element
if cnt > (n // 3):
# Add arr[i] only if it is not already
# present in the result
if len(res) == 0 or arr[i] != res[0]:
res.append(arr[i])
# If we have found two majority elements,
# we can stop our search
if len(res) == 2:
if res[0] > res[1]:
res[0], res[1] = res[1], res[0]
break
return res
if __name__ == "__main__":
arr = [2, 2, 3, 1, 3, 2, 1, 1]
res = findMajority(arr)
for ele in res:
print(ele, end=" ")
using System;
using System.Collections.Generic;
class GfG {
static List<int> findMajority(int[] arr) {
int n = arr.Length;
List<int> res = new List<int>();
for (int i = 0; i < n; i++) {
// Count the frequency of arr[i]
int cnt = 0;
for (int j = i; j < n; j++) {
if (arr[j] == arr[i])
cnt += 1;
}
// Check if arr[i] is a majority element
if (cnt > (n / 3)) {
// Add arr[i] only if it is not already
// present in the result
if (res.Count == 0 || arr[i] != res[0]) {
res.Add(arr[i]);
}
}
// If we have found two majority elements,
// we can stop our search
if (res.Count == 2) {
if (res[0] > res[1]) {
int temp = res[0];
res[0] = res[1];
res[1] = temp;
}
break;
}
}
return res;
}
static void Main(string[] args) {
int[] arr = { 2, 2, 3, 1, 3, 2, 1, 1 };
List<int> res = findMajority(arr);
foreach (int ele in res)
Console.Write(ele + " ");
}
}
function findMajority(arr) {
const n = arr.length;
const res = [];
for (let i = 0; i < n; i++) {
// Count the frequency of arr[i]
let cnt = 0;
for (let j = i; j < n; j++) {
if (arr[j] === arr[i]) {
cnt += 1;
}
}
// Check if arr[i] is a majority element
if (cnt > (n / 3)) {
// Add arr[i] only if it is not already
// present in the result
if (res.length === 0 || arr[i] !== res[0]) {
res.push(arr[i]);
}
}
// If we have found two majority elements,
// we can stop our search
if (res.length === 2) {
if (res[0] > res[1]) {
[res[0], res[1]] = [res[1], res[0]];
}
break;
}
}
return res;
}
// Driver Code
const arr = [2, 2, 3, 1, 3, 2, 1, 1];
const res = findMajority(arr);
console.log(res.join(" "));
Output
1 2
[Better Approach] Using Hash Map or Dictionary - O(n) Time and O(n) Space
The idea is to use a hash map or dictionary to count the frequency of each element in the array. After counting, iterate over the hash map and if the frequency of any element is greater than (n/3), push it into the result. Finally, the majority elements are returned after sorting.
#include <iostream>
#include <unordered_map>
#include <vector>
using namespace std;
vector<int> findMajority(vector<int> &arr) {
int n = arr.size();
unordered_map<int, int> freq;
vector<int> res;
// find frequency of each number
for (int ele : arr)
freq[ele]++;
// Iterate over each key-value pair
// in the hash map
for (auto it : freq) {
int ele = it.first;
int cnt = it.second;
// Add the element to the result, if its frequency
// if greater than floor(n/3)
if (cnt > n / 3)
res.push_back(ele);
}
if (res.size() == 2 && res[0] > res[1])
swap(res[0], res[1]);
return res;
}
int main() {
vector<int> arr = {2, 2, 3, 1, 3, 2, 1, 1};
vector<int> res = findMajority(arr);
for (auto ele : res) {
cout << ele << " ";
}
return 0;
}
import java.util.HashMap;
import java.util.ArrayList;
import java.util.Map;
class GfG {
static ArrayList<Integer> findMajority(int[] arr) {
int n = arr.length;
HashMap<Integer, Integer> freq = new HashMap<>();
ArrayList<Integer> res = new ArrayList<>();
// find frequency of each number
for (int ele : arr)
freq.put(ele, freq.getOrDefault(ele, 0) + 1);
// Iterate over each key-value
// pair in the hash map
for (Map.Entry<Integer, Integer> it : freq.entrySet()) {
int ele = it.getKey();
int cnt = it.getValue();
// Add the element to the result if its
// frequency is greater than n / 3
if (cnt > n / 3)
res.add(ele);
}
// Sort result if there are two elements
// and they are out of order
if (res.size() == 2 && res.get(0) > res.get(1)) {
int temp = res.get(0);
res.set(0, res.get(1));
res.set(1, temp);
}
return res;
}
public static void main(String[] args) {
int[] arr = {2, 2, 3, 1, 3, 2, 1, 1};
ArrayList<Integer> res = findMajority(arr);
for (int ele : res) {
System.out.print(ele + " ");
}
}
}
def findMajority(arr):
n = len(arr)
freq = {}
res = []
# find frequency of each number
for ele in arr:
freq[ele] = freq.get(ele, 0) + 1
# Iterate over each key-value
# pair in the hash map
for ele, cnt in freq.items():
# Add the element to the result, if its frequency
# is greater than floor(n/3)
if cnt > n // 3:
res.append(ele)
if len(res) == 2 and res[0] > res[1]:
res[0], res[1] = res[1], res[0]
return res
if __name__ == "__main__":
arr = [2, 2, 3, 1, 3, 2, 1, 1]
res = findMajority(arr)
for ele in res:
print(ele, end=" ")
using System;
using System.Collections.Generic;
class GfG {
static List<int> findMajority(int[] arr) {
int n = arr.Length;
Dictionary<int, int> freq = new Dictionary<int, int>();
List<int> res = new List<int>();
// find frequency of each number
foreach (int ele in arr) {
if (freq.ContainsKey(ele))
freq[ele]++;
else
freq[ele] = 1;
}
// Iterate over each key-value
// pair in the hash map
foreach (var it in freq) {
int ele = it.Key;
int cnt = it.Value;
// Add the element to the result, if its frequency
// is greater than floor(n/3)
if (cnt > n / 3)
res.Add(ele);
}
if (res.Count == 2 && res[0] > res[1]) {
int temp = res[0];
res[0] = res[1];
res[1] = temp;
}
return res;
}
static void Main() {
int[] arr = { 2, 2, 3, 1, 3, 2, 1, 1 };
List<int> res = findMajority(arr);
foreach (int ele in res) {
Console.Write(ele + " ");
}
}
}
function findMajority(arr) {
const n = arr.length;
const freq = {};
const res = [];
// find frequency of each number
for (const ele of arr) {
freq[ele] = (freq[ele] || 0) + 1;
}
// Iterate over each key-value pair in the hash map
for (const it in freq) {
const ele = Number(it);
const cnt = freq[it];
// Add the element to the result, if its frequency
// is greater than floor(n/3)
if (cnt > Math.floor(n / 3)) {
res.push(ele);
}
}
if (res.length === 2 && res[0] > res[1]) {
[res[0], res[1]] = [res[1], res[0]];
}
return res;
}
// Driver Code
const arr = [2, 2, 3, 1, 3, 2, 1, 1];
const res = findMajority(arr);
console.log(res.join(" "));
Output
1 2
[Expected Approach] Boyer-Moore’s Voting Algorithm - O(n) Time and O(1) Space
The idea is based on the observation that there can be at most two majority elements, which appear more than n/3 times. so we can use Boyer-Moore’s Voting algorithm. As we iterate the array, We identify potential majority elements by keeping track of two candidates and their respective counts.
Steps:
- Initialize two variables ele1 = -1 and ele2 = -1, for candidates and two variables cnt1 = 0 and cnt2 = 0, for counting.
- In each iteration,
- If an element is equal to any candidate, update that candidate's count.
- If count of a candidate reaches zero then replace that candidate with current element.
- If neither candidate matches and both counts are non zero, decrement the counts.
- After this, in second pass we check if the chosen candidates appear more than n/3 times in the array. If they do then include them in result array.
Since any element than appears more than floor(n/3) times, will dominate over elements that appear less frequently. Whenever we encounter a different element, we decrement the count of both the candidates. This maintains at most two candidates in the array.
#include <iostream>
#include <vector>
#include <algorithm>
#include <limits.h>
using namespace std;
vector<int> findMajority(vector<int> &arr) {
int n = arr.size();
// Initialize two candidates and their counts
int ele1 = -1, ele2 = -1, cnt1 = 0, cnt2 = 0;
for (int ele : arr) {
// Increment count for candidate 1
if (ele1 == ele) {
cnt1++;
}
// Increment count for candidate 2
else if (ele2 == ele) {
cnt2++;
}
// New candidate 1 if count is zero
else if (cnt1 == 0) {
ele1 = ele;
cnt1++;
}
// New candidate 2 if count is zero
else if (cnt2 == 0) {
ele2 = ele;
cnt2++;
}
// Decrease counts if neither candidate
else {
cnt1--;
cnt2--;
}
}
vector<int> res;
cnt1 = 0;
cnt2 = 0;
// Count the occurrences of candidates
for (int ele : arr) {
if (ele1 == ele) cnt1++;
if (ele2 == ele) cnt2++;
}
// Add to result if they are majority elements
if (cnt1 > n / 3) res.push_back(ele1);
if (cnt2 > n / 3 && ele1 != ele2) res.push_back(ele2);
if(res.size() == 2 && res[0] > res[1])
swap(res[0], res[1]);
return res;
}
int main() {
vector<int> arr = {2, 2, 3, 1, 3, 2, 1, 1};
vector<int> res = findMajority(arr);
for (int ele : res) {
cout << ele << " ";
}
return 0;
}
#include <stdio.h>
#include <limits.h>
void findMajority(int arr[], int n, int *res, int *resSize) {
// Initialize two candidates and their counts
int ele1 = -1, ele2 = -1, cnt1 = 0, cnt2 = 0;
for (int i = 0; i < n; i++) {
int ele = arr[i];
// Increment count for candidate 1
if (ele1 == ele) {
cnt1++;
}
// Increment count for candidate 2
else if (ele2 == ele) {
cnt2++;
}
// New candidate 1 if count is zero
else if (cnt1 == 0) {
ele1 = ele;
cnt1++;
}
// New candidate 2 if count is zero
else if (cnt2 == 0) {
ele2 = ele;
cnt2++;
}
// Decrease counts if neither candidate
else {
cnt1--;
cnt2--;
}
}
cnt1 = 0;
cnt2 = 0;
// Count the occurrences of candidates
for (int i = 0; i < n; i++) {
int ele = arr[i];
if (ele1 == ele) cnt1++;
if (ele2 == ele) cnt2++;
}
// Add to result if they are majority elements
*resSize = 0;
if (cnt1 > n / 3) res[(*resSize)++] = ele1;
if (cnt2 > n / 3 && ele1 != ele2) res[(*resSize)++] = ele2;
// Sort the result if there are two elements
if (*resSize == 2 && res[0] > res[1]) {
int temp = res[0];
res[0] = res[1];
res[1] = temp;
}
}
int main() {
int arr[] = {2, 2, 3, 1, 3, 2, 1, 1};
int n = sizeof(arr) / sizeof(arr[0]);
int res[2];
int resSize;
findMajority(arr, n, res, &resSize);
for (int i = 0; i < resSize; i++) {
printf("%d ", res[i]);
}
return 0;
}
import java.util.ArrayList;
class GfG {
// Function to find Majority element in an array
static ArrayList<Integer> findMajority(int[] arr) {
int n = arr.length;
// Initialize two candidates and their counts
int ele1 = -1, ele2 = -1;
int cnt1 = 0, cnt2 = 0;
for (int ele : arr) {
// Increment count for candidate 1
if (ele1 == ele) {
cnt1++;
}
// Increment count for candidate 2
else if (ele2 == ele) {
cnt2++;
}
// New candidate 1 if count is zero
else if (cnt1 == 0) {
ele1 = ele;
cnt1++;
}
// New candidate 2 if count is zero
else if (cnt2 == 0) {
ele2 = ele;
cnt2++;
}
// Decrease counts if neither candidate
else {
cnt1--;
cnt2--;
}
}
ArrayList<Integer> res = new ArrayList<>();
cnt1 = 0;
cnt2 = 0;
// Count the occurrences of candidates
for (int ele : arr) {
if (ele1 == ele) cnt1++;
if (ele2 == ele) cnt2++;
}
// Add to result if they are majority elements
if (cnt1 > n / 3) res.add(ele1);
if (cnt2 > n / 3 && ele1 != ele2) res.add(ele2);
// Sort the result if needed
if (res.size() == 2 && res.get(0) > res.get(1)) {
int temp = res.get(0);
res.set(0, res.get(1));
res.set(1, temp);
}
return res;
}
public static void main(String[] args) {
int[] arr = {2, 2, 3, 1, 3, 2, 1, 1};
ArrayList<Integer> res = findMajority(arr);
for (int ele : res) {
System.out.print(ele + " ");
}
}
}
def findMajority(arr):
n = len(arr)
# Initialize two candidates and their counts
ele1, ele2 = -1, -1
cnt1, cnt2 = 0, 0
for ele in arr:
# Increment count for candidate 1
if ele1 == ele:
cnt1 += 1
# Increment count for candidate 2
elif ele2 == ele:
cnt2 += 1
# New candidate 1 if count is zero
elif cnt1 == 0:
ele1 = ele
cnt1 += 1
# New candidate 2 if count is zero
elif cnt2 == 0:
ele2 = ele
cnt2 += 1
# Decrease counts if neither candidate
else:
cnt1 -= 1
cnt2 -= 1
res = []
cnt1, cnt2 = 0, 0
# Count the occurrences of candidates
for ele in arr:
if ele1 == ele:
cnt1 += 1
if ele2 == ele:
cnt2 += 1
# Add to result if they are majority elements
if cnt1 > n / 3:
res.append(ele1)
if cnt2 > n / 3 and ele1 != ele2:
res.append(ele2)
if len(res) == 2 and res[0] > res[1]:
res[0], res[1] = res[1], res[0]
return res
if __name__ == "__main__":
arr = [2, 2, 3, 1, 3, 2, 1, 1]
res = findMajority(arr)
for ele in res:
print(ele, end = " ")
using System;
using System.Collections.Generic;
class GfG {
static List<int> findMajority(int[] arr) {
int n = arr.Length;
// Initialize two candidates and their counts
int ele1 = -1, ele2 = -1, cnt1 = 0, cnt2 = 0;
foreach (int ele in arr) {
// Increment count for candidate 1
if (ele1 == ele) {
cnt1++;
}
// Increment count for candidate 2
else if (ele2 == ele) {
cnt2++;
}
// New candidate 1 if count is zero
else if (cnt1 == 0) {
ele1 = ele;
cnt1++;
}
// New candidate 2 if count is zero
else if (cnt2 == 0) {
ele2 = ele;
cnt2++;
}
// Decrease counts if neither candidate
else {
cnt1--;
cnt2--;
}
}
List<int> res = new List<int>();
cnt1 = 0;
cnt2 = 0;
// Count the occurrences of candidates
foreach (int ele in arr) {
if (ele1 == ele) cnt1++;
if (ele2 == ele) cnt2++;
}
// Add to result if they are majority elements
if (cnt1 > n / 3) res.Add(ele1);
if (cnt2 > n / 3 && ele1 != ele2) res.Add(ele2);
if (res.Count == 2 && res[0] > res[1]) {
int temp = res[0];
res[0] = res[1];
res[1] = temp;
}
return res;
}
static void Main() {
int[] arr = { 2, 2, 3, 1, 3, 2, 1, 1 };
List<int> res = findMajority(arr);
foreach (int ele in res) {
Console.Write(ele + " ");
}
}
}
function findMajority(arr) {
const n = arr.length;
// Initialize two candidates and their counts
let ele1 = -1, ele2 = -1;
let cnt1 = 0, cnt2 = 0;
for (let ele of arr) {
// Increment count for candidate 1
if (ele1 === ele) {
cnt1++;
}
// Increment count for candidate 2
else if (ele2 === ele) {
cnt2++;
}
// New candidate 1 if count is zero
else if (cnt1 === 0) {
ele1 = ele;
cnt1++;
}
// New candidate 2 if count is zero
else if (cnt2 === 0) {
ele2 = ele;
cnt2++;
}
// Decrease counts if neither candidate
else {
cnt1--;
cnt2--;
}
}
const res = [];
cnt1 = 0;
cnt2 = 0;
// Count the occurrences of candidates
for (let ele of arr) {
if (ele1 === ele) cnt1++;
if (ele2 === ele) cnt2++;
}
// Add to result if they are majority elements
if (cnt1 > n / 3) res.push(ele1);
if (cnt2 > n / 3 && ele1 != ele2) res.push(ele2);
if (res.length === 2 && res[0] > res[1]) {
[res[0], res[1]] = [res[1], res[0]];
}
return res;
}
// Driver Code
const arr = [2, 2, 3, 1, 3, 2, 1, 1];
const res = findMajority(arr);
console.log(res.join(" "));
Output
1 2
Majority Element II
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