chore(java): add majority element (#354)
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2. [Kadanes Algorithm](arrays/kadanes-algorithm.java)
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3. [Left Rotation](arrays/left-rotation.java)
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4. [Unique Digits of Large Number](arrays/unique-digits-of-large-number.java)
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5. [Majority Element](arrays/majority-element.java)
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## Graphs
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@ -0,0 +1,70 @@
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//Given an array A of N elements. Find the majority element in the array.
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//A majority element in an array A of size N is an element that appears more than N/2 times in the array
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// Moore’s Voting Algorithm => Time Complexity: O(n), Space Complexity: O(1)
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import java.util.*;
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import java.io.*;
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import java.lang.*;
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class Main {
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public static void main(String args[]) {
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Scanner sc = new Scanner(System.in);
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int t = sc.nextInt();
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while(t-- > 0) {
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int n =sc.nextInt();
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int arr[] = new int[n];
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for(int i = 0; i < n; i++)
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arr[i] = sc.nextInt();
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System.out.println(new Solution().majorityElement(arr, n));
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}
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}
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}
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class Solution
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{
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static int majorityElement(int a[], int size){
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int cand = findCand(a, size);
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if(isMajority(a, size, cand))
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return cand;
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return -1;
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}
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static int findCand(int arr[], int n){
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int max_index=0; int count=1;
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for(int i=1; i<n; i++){
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if(arr[max_index]==arr[i])
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count++;
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else count--;
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if(count==0){
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max_index=i;
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count=1;
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}
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}
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return arr[max_index];
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}
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static boolean isMajority(int arr[], int n, int cand){
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int count=0;
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for(int i=0; i<n; i++){
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if(cand==arr[i])
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count++;
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}
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if(count > n/2) return true;
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return false;
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}
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}
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/* Test Case:
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Input:
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N = 5 A[] = {3,1,3,3,2}
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Output: 3
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Explanation:
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Since, 3 is present more
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than N/2 times, so it is
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the majority element. */
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