Merge pull request #202 from shoebxsiddiqui/patch-8

Create BoyerMooreVotingAlgo.java

Author: kodiakhq[bot]

JAVA/BoyerMooreVotingAlgo.java

@@ -0,0 +1,85 @@
+//Ques. Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times.
+
+
+// Basic idea is to use Boyer Moore Voting Algo
+// there can be atmost 2 elements in the array which can become maximum element because
+// 3*(n/3) = n it is possible that only 3 elements can occir exactly n/3 times in the array
+// if any element is appearing more than n/3 times atmost 2 elements can be the maximun element...
+
+// now we will maintain two variables num1 , num 2 to store the value of frequently appearing elements using extended boyer moore algo
+
+				class BoyerMooreVotingAlgo {
+          
+                                        public static void main(String[] args) {
+                                           int[] arr = {1,2,3,3,3,4,4,5,4,3};
+                                           System.out.println(majorityElement(arr).toString());
+                                        }
+          
+					public static List<Integer> majorityElement(int[] nums) {
+					   List<Integer> list = new ArrayList<>();
+						 if(nums == null) return list;
+						int num1=0;
+						int num2=0;
+						int count1=0;
+						int count2=0;
+						for(int i=0;i<nums.length;i++){
+							int num3 = nums[i];
+							if(count1>0 && count2>0){
+
+								if(num3 == num1){
+									count1++;
+								}else if(num3 ==num2){
+									count2++;
+								}else{
+									count1--;
+									count2--;
+								}
+
+							}else if(count1 >0){
+
+								if(num3 == num1){
+									count1++;
+								}else{
+									num2 = num3;
+									count2++;
+								}
+
+							}else if(count2>0){
+
+								if(num3 == num2){
+									count2++;
+								}else{
+									num1 = num3;
+									count1++;
+								}
+
+							}else{
+								num1 = num3;
+								count1++;
+							}
+						}
+
+				//         now we have the values that appears frequently in the array but it doesnot means that they both are maximum elements .........
+				//         so we have to iterate again over the array to cehck for maximum element out of both .......
+
+						count1=0;
+						count2=0;
+
+						for(int i=0;i<nums.length;i++){
+							if(nums[i] == num1){
+								count1++;
+							}
+							else if(nums[i] ==num2){
+								count2++;
+							}
+						}
+						if(count1>nums.length/3){
+							list.add(num1);
+						}
+						if(count2>nums.length/3){
+							list.add(num2);
+						}
+
+						return list;
+					}
+				}