Given an array arr[] of size n, find the element that appears more than ⌊n/2⌋ times. If no such element exists, return -1.
Examples:
Input: arr[] = [1, 1, 2, 1, 3, 5, 1]
Output: 1
Explanation: Element 1 appears 4 times. Since ⌊7/2⌋ = 3, and 4 > 3, it is the majority element.
Input: arr[] = [7]
Output: 7
Explanation: Element 7 appears once. Since ⌊1/2⌋ = 0, and 1 > 0, it is the majority element.
Input: arr[] = [2, 13]
Output: -1
Explanation: No element appears more than ⌊2/2⌋ = 1 time, so there is no majority element.
[Naive Approach] Using Two Nested Loops - O(n^2) Time and O(1) Space
The idea is to use nested loops to count frequencies. The outer loop selects each element as a candidate, and the inner loop counts how many times it appears. If any element appears more than n / 2 times, it is the majority element.
C++
#include <iostream>
#include <vector>
using namespace std;
int majorityElement(vector<int>& arr) {
int n = arr.size();
// Loop to consider each element as a
// candidate for majority
for (int i = 0; i < n; i++) {
int count = 0;
// Inner loop to count the frequency of arr[i]
for (int j = 0; j < n; j++) {
if (arr[i] == arr[j]) {
count++;
}
}
// Check if count of arr[i] is more
// than half the size of the array
if (count > n / 2) {
return arr[i];
}
}
// If no majority element found, return -1
return -1;
}
int main() {
vector<int> arr = {1, 1, 2, 1, 3, 5, 1};
cout << majorityElement(arr) << endl;
return 0;
}
#include <stdio.h>
int majorityElement(int arr[], int n) {
// Loop to consider each element as
// a candidate for majority
for (int i = 0; i < n; i++) {
int count = 0;
// Inner loop to count the frequency of arr[i]
for (int j = 0; j < n; j++) {
if (arr[i] == arr[j]) {
count++;
}
}
// Check if count of arr[i] is more
// than half the size of the array
if (count > n / 2) {
return arr[i];
}
}
// If no majority element found, return -1
return -1;
}
int main() {
int arr[] = {1, 1, 2, 1, 3, 5, 1};
int n = sizeof(arr) / sizeof(arr[0]);
printf("%d\n", majorityElement(arr, n));
return 0;
}
class GfG {
static int majorityElement(int[] arr) {
int n = arr.length;
// Loop to consider each element as
// a candidate for majority
for (int i = 0; i < n; i++) {
int count = 0;
// Inner loop to count the frequency of arr[i]
for (int j = 0; j < n; j++) {
if (arr[i] == arr[j]) {
count++;
}
}
// Check if count of arr[i] is more
// than half the size of the array
if (count > n / 2) {
return arr[i];
}
}
// If no majority element found, return -1
return -1;
}
public static void main(String[] args) {
int[] arr = {1, 1, 2, 1, 3, 5, 1};
System.out.println(majorityElement(arr));
}
}
def majorityElement(arr):
n = len(arr)
# Loop to consider each element as
# a candidate for majority
for i in range(n):
count = 0
# Inner loop to count the frequency of arr[i]
for j in range(n):
if arr[i] == arr[j]:
count += 1
# Check if count of arr[i] is more
# than half the size of the array
if count > n // 2:
return arr[i]
# If no majority element found, return -1
return -1
if __name__ == "__main__":
arr = [1, 1, 2, 1, 3, 5, 1]
print(majorityElement(arr))
using System;
class GfG {
static int majorityElement(int[] arr) {
int n = arr.Length;
// Loop to consider each element
// as a candidate for majority
for (int i = 0; i < n; i++) {
int count = 0;
// Inner loop to count the frequency of arr[i]
for (int j = 0; j < n; j++) {
if (arr[i] == arr[j]) {
count++;
}
}
// Check if count of arr[i] is more
// than half the size of the array
if (count > n / 2) {
return arr[i];
}
}
// If no majority element found, return -1
return -1;
}
static void Main(string[] args) {
int[] arr = { 1, 1, 2, 1, 3, 5, 1 };
Console.WriteLine(majorityElement(arr));
}
}
function majorityElement(arr) {
let n = arr.length;
// Loop to consider each element as
// a candidate for majority
for (let i = 0; i < n; i++) {
let count = 0;
// Inner loop to count the frequency of arr[i]
for (let j = 0; j < n; j++) {
if (arr[i] === arr[j]) {
count++;
}
}
// Check if count of arr[i] is more than
// half the size of the array
if (count > n / 2) {
return arr[i];
}
}
// If no majority element found, return -1
return -1;
}
// Driver Code
let arr = [1, 1, 2, 1, 3, 5, 1];
console.log(majorityElement(arr));
[Better Approach - 1] Using Sorting - O(nlog(n)) Time and O(1) Space
The idea is to sort the array so that similar elements are next to each other. Once sorted, go through the array and keep track of how many times each element appears. When you encounter a new element, check if the count of the previous element was more than half the total number of elements in the array. If it was, that element is the majority and should be returned. If no element meets this requirement, no majority element exists.
C++
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int majorityElement(vector<int>& arr) {
int n = arr.size();
sort(arr.begin(), arr.end());
// Potential majority element
int candidate = arr[n/2];
// Count how many times candidate appears
int count = 0;
for (int num : arr) {
if (num == candidate) {
count++;
}
}
if (count > n/2) {
return candidate;
}
// No majority element
return -1;
}
int main() {
vector<int> arr = {1, 1, 2, 1, 3, 5, 1};
cout << majorityElement(arr);
return 0;
}
import java.util.Arrays;
class GfG {
static int majorityElement(int[] arr) {
int n = arr.length;
Arrays.sort(arr);
// Potential majority element
int candidate = arr[n/2];
// Count how many times candidate appears
int count = 0;
for (int num : arr) {
if (num == candidate) {
count++;
}
}
if (count > n/2) {
return candidate;
}
// No majority element
return -1;
}
public static void main(String[] args) {
int[] arr = {1, 1, 2, 1, 3, 5, 1};
System.out.println(majorityElement(arr));
}
}
def majorityElement(arr):
n = len(arr)
arr.sort()
# Potential majority element
candidate = arr[n // 2]
# Count how many times candidate appears
count = 0
for num in arr:
if num == candidate:
count += 1
if count > n // 2:
return candidate
else:
return -1
if __name__ == "__main__":
arr = [1, 1, 2, 1, 3, 5, 1]
print(majorityElement(arr))
using System;
class GfG {
static int majorityElement(int[] arr) {
int n = arr.Length;
Array.Sort(arr);
// Potential majority element
int candidate = arr[n / 2];
// Count how many times candidate appears
int count = 0;
foreach (int num in arr) {
if (num == candidate) {
count++;
}
}
if (count > n / 2) {
return candidate;
} else {
return -1;
}
}
static void Main() {
int[] arr = {1, 1, 2, 1, 3, 5, 1};
Console.WriteLine(majorityElement(arr));
}
}
function majorityElement(arr) {
let n = arr.length;
arr.sort((a, b) => a - b);
// Potential majority element
let candidate = arr[Math.floor(n / 2)];
// Count how many times candidate appears
let count = 0;
for (let num of arr) {
if (num === candidate) {
count++;
}
}
if (count > Math.floor(n / 2)) {
return candidate;
} else {
return -1;
}
}
// Driver Code
let arr = [1, 1, 2, 1, 3, 5, 1];
console.log(majorityElement(arr));
[Better Approach - 2] Using Hashing - O(n) Time and O(n) Space
The idea is to use a hash map to track frequencies and identify the majority element in a single pass.
Step By Step Implementations:
- Initialize an empty hash map.
- Traverse the array and update the count of each element.
- After each update, check if the count exceeds n / 2.
- If found, return that element immediately.
- If no such element exists after the loop, return -1.
C++
#include <iostream>
#include<unordered_map>
#include<vector>
using namespace std;
int majorityElement(vector<int> &arr) {
int n = arr.size();
unordered_map<int, int> countMap;
// Traverse the array and count occurrences using the hash map
for (int num : arr) {
countMap[num]++;
// Check if current element count exceeds n / 2
if (countMap[num] > n / 2) {
return num;
}
}
// If no majority element is found, return -1
return -1;
}
int main() {
vector<int> arr = {1, 1, 2, 1, 3, 5, 1};
cout << majorityElement(arr) << endl;
return 0;
}
import java.util.HashMap;
import java.util.Map;
class GfG {
static int majorityElement(int[] arr) {
int n = arr.length;
Map<Integer, Integer> countMap = new HashMap<>();
// Traverse the array and count occurrences using the hash map
for (int num : arr) {
countMap.put(num, countMap.getOrDefault(num, 0) + 1);
// Check if current element count exceeds n / 2
if (countMap.get(num) > n / 2) {
return num;
}
}
// If no majority element is found, return -1
return -1;
}
public static void main(String[] args) {
int[] arr = {1, 1, 2, 1, 3, 5, 1};
System.out.println(majorityElement(arr));
}
}
from collections import defaultdict
def majorityElement(arr):
n = len(arr)
countMap = defaultdict(int)
# Traverse the list and count occurrences using the hash map
for num in arr:
countMap[num] += 1
# Check if current element count exceeds n / 2
if countMap[num] > n / 2:
return num
# If no majority element is found, return -1
return -1
if __name__ == "__main__":
arr = [1, 1, 2, 1, 3, 5, 1]
print(majorityElement(arr))
using System;
using System.Collections.Generic;
class GfG {
static int majorityElement(int[] arr) {
int n = arr.Length;
Dictionary<int, int> countMap = new Dictionary<int, int>();
// Traverse the array and count occurrences using the hash map
foreach (int num in arr) {
if (countMap.ContainsKey(num))
countMap[num]++;
else
countMap[num] = 1;
// Check if current element count exceeds n / 2
if (countMap[num] > n / 2) {
return num;
}
}
// If no majority element is found, return -1
return -1;
}
public static void Main() {
int[] arr = {1, 1, 2, 1, 3, 5, 1};
Console.WriteLine(majorityElement(arr));
}
}
function majorityElement(arr) {
const n = arr.length;
const countMap = new Map();
// Traverse the array and count occurrences using the hash map
for (const num of arr) {
countMap.set(num, (countMap.get(num) || 0) + 1);
// Check if current element count exceeds n / 2
if (countMap.get(num) > n / 2) {
return num;
}
}
// If no majority element is found, return -1
return -1;
}
// Driver Code
const arr = [1, 1, 2, 1, 3, 5, 1];
console.log(majorityElement(arr));
[Expected Approach] Using Moore's Voting Algorithm - O(n) Time and O(1) Space
The idea is to use the Boyer-Moore Voting Algorithm to efficiently find a potential majority element by canceling out different elements. If a majority element exists, it will remain as the candidate. Then verify it.
This is a two-step process:
- The first step gives the element that may be the majority element in the array. If there is a majority element in an array, then this step will definitely return majority element, otherwise, it will return candidate for majority element.
- Check if the element obtained from the above step is the majority element. This step is necessary as there might be no majority element.
Step By Step Approach:
- Initialize a candidate variable and a count variable.
- Traverse the array once:
-> If count is zero, set the candidate to the current element and set count to one.
-> If the current element equals the candidate, increment count.
-> If the current element differs from the candidate, decrement count. - Traverse the array again to count the occurrences of the candidate.
- If the candidate's count is greater than n / 2, return the candidate as the majority element.
C++
#include <iostream>
#include <vector>
using namespace std;
int majorityElement(vector<int>& arr) {
int n = arr.size();
int candidate = -1;
int count = 0;
// Find a candidate
for (int num : arr) {
if (count == 0) {
candidate = num;
count = 1;
}
else if (num == candidate) {
count++;
}
else {
count--;
}
}
// Validate the candidate
count = 0;
for (int num : arr) {
if (num == candidate) {
count++;
}
}
// If count is greater than n / 2, return the
// candidate; otherwise, return -1
if (count > n / 2) {
return candidate;
} else {
return -1;
}
}
int main() {
vector<int> arr = {1, 1, 2, 1, 3, 5, 1};
cout << majorityElement(arr) << endl;
return 0;
}
#include <stdio.h>
int majorityElement(int arr[], int n) {
int candidate = -1;
int count = 0;
// Find a candidate
for (int i = 0; i < n; i++) {
if (count == 0) {
candidate = arr[i];
count = 1;
}
else if (arr[i] == candidate) {
count++;
}
else {
count--;
}
}
// Validate the candidate
count = 0;
for (int i = 0; i < n; i++) {
if (arr[i] == candidate) {
count++;
}
}
// If count is greater than n / 2, return
// the candidate; otherwise, return -1
if (count > n / 2) {
return candidate;
} else {
return -1;
}
}
int main() {
int arr[] = {1, 1, 2, 1, 3, 5, 1};
int n = sizeof(arr) / sizeof(arr[0]);
printf("%d\n", majorityElement(arr, n));
return 0;
}
class GfG {
static int majorityElement(int[] arr) {
int n = arr.length;
int candidate = -1;
int count = 0;
// Find a candidate
for (int num : arr) {
if (count == 0) {
candidate = num;
count = 1;
}
else if (num == candidate) {
count++;
}
else {
count--;
}
}
// Validate the candidate
count = 0;
for (int num : arr) {
if (num == candidate) {
count++;
}
}
// If count is greater than n / 2, return
// the candidate; otherwise, return -1
if (count > n / 2) {
return candidate;
} else {
return -1;
}
}
public static void main(String[] args) {
int[] arr = {1, 1, 2, 1, 3, 5, 1};
System.out.println(majorityElement(arr));
}
}
def majorityElement(arr):
n = len(arr)
candidate = -1
count = 0
# Find a candidate
for num in arr:
if count == 0:
candidate = num
count = 1
elif num == candidate:
count += 1
else:
count -= 1
# Validate the candidate
count = 0
for num in arr:
if num == candidate:
count += 1
# If count is greater than n / 2, return
# the candidate; otherwise, return -1
if count > n // 2:
return candidate
else:
return -1
if __name__ == "__main__":
arr = [1, 1, 2, 1, 3, 5, 1]
print(majorityElement(arr))
using System;
class GfG {
static int majorityElement(int[] arr) {
int n = arr.Length;
int candidate = -1;
int count = 0;
// Find a candidate
foreach (int num in arr) {
if (count == 0) {
candidate = num;
count = 1;
} else if (num == candidate) {
count++;
} else {
count--;
}
}
// Validate the candidate
count = 0;
foreach (int num in arr) {
if (num == candidate) {
count++;
}
}
// If count is greater than n / 2, return
// the candidate; otherwise, return -1
if (count > n / 2) {
return candidate;
} else {
return -1;
}
}
static void Main() {
int[] arr = {1, 1, 2, 1, 3, 5, 1};
Console.WriteLine(majorityElement(arr));
}
}
function majorityElement(arr) {
const n = arr.length;
let candidate = -1;
let count = 0;
// Find a candidate
for (const num of arr) {
if (count === 0) {
candidate = num;
count = 1;
}
else if (num === candidate) {
count++;
}
else {
count--;
}
}
// Validate the candidate
count = 0;
for (const num of arr) {
if (num === candidate) {
count++;
}
}
// If count is greater than n / 2, return
// the candidate; otherwise, return -1
if (count > n / 2) {
return candidate;
} else {
return -1;
}
}
// Driver Code
const arr = [1, 1, 2, 1, 3, 5, 1];
console.log(majorityElement(arr));
Solving the Majority Element Problem
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