Merge pull request #279 from nik132-eng/patch-4

Create Majority_Element.java

Author: SarthakKeshari

LeetCode/Majority_Element.java

@@ -0,0 +1,68 @@
+//169. Majority Element
+//
+//Given an array nums of size n, return the majority element.
+//
+//The majority element is the element that appears more than ⌊n / 2⌋ times. 
+//You may assume that the majority element always exists in the array.
+//
+//Example 1:
+//
+//Input: nums = [3,2,3]
+//Output: 3
+//
+//Example 2:
+//
+//Input: nums = [2,2,1,1,1,2,2]
+//Output: 2
+// 
+//
+//
+//Constraints:
+//
+//n == nums.length
+//1 <= n <= 5 * 104
+//-231 <= nums[i] <= 231 - 1
+
+import java.util.Scanner;
+
+public class Majority_Element {
+
+	public static void main(String[] args) {
+		Scanner sc = new Scanner(System.in);
+
+		System.out.println("Enter Number of element : ");
+		int n = sc.nextInt();
+
+		int nums[] = new int[n];
+		System.out.println("Enter " + n + " values :");
+		for (int i = 0; i < n; i++) {
+			nums[i] = sc.nextInt();
+		}
+		int ans = majorityElement(nums);
+
+		System.out.println("Majority Element is : " + ans);
+
+	}
+
+	private static int majorityElement(int[] nums) {
+		
+//		take two variables for as key value pair 
+		int count = 0;
+		int candidate = 0;
+//		loop
+		for (int num : nums) {
+//			if the number apears for the first-time after count value is zero than make it candidate
+			if (count == 0) {
+				candidate = num;
+			}
+			
+//			if the key(candidate) and the number from the array are same than add one in count else reduce one
+			if (candidate == num)
+				count++;
+			else
+				count--;
+		}
+		return candidate;
+	}
+
+}