Merge pull request neetcode-gh#3042 from benmak11/229

sujal-goswami · web-flow · commit 0ac8f44f58c5 · 2023-10-05T23:28:06.000+05:30
Create 0229-majority-element-ii.java
diff --git a/java/0229-majority-element-ii.java b/java/0229-majority-element-ii.java
@@ -0,0 +1,65 @@
+class Solution {
+    /**
+     * First solution utilizes a hashmap and then does the due diligience of adding the 
+     * appropriate values that appear more than n/3 times
+     * Runtime O(n) : Space O(n)
+     */
+    public List<Integer> majorityElement(int[] nums) {
+        List<Integer> res = new ArrayList<>();
+        Map<Integer, Integer> map = new HashMap<>();
+        for (int i = 0; i < nums.length; i++) {
+            if (map.containsKey(nums[i])) 
+                map.put(nums[i], map.get(nums[i]) + 1);
+            else 
+                map.put(nums[i], 1);
+        }
+
+        for (Map.Entry<Integer, Integer> entry: map.entrySet()) {
+            int potentialCandidate = entry.getValue();
+            if (potentialCandidate > nums.length / 3)
+                res.add(entry.getKey());
+        }
+
+        return res;
+    }
+
+
+    /**
+     * This is called Boyer-Moore Vote algorithm and the idea here is having candidates 
+     * with diff values and two counters. 
+     * For each number in the array we see if it equals the candidate and increment the count. 
+     * The two numbers left after this process are the majority candidates. 
+     * Loop through the array again then make sure that each candidate does indeed have more than n/3 occurrences
+     * 
+     * Runtime O(n) : Space O(1)
+     */
+    public List<Integer> majorityElement_2(int[] nums) {
+        int candidate1 = 0, candidate2 = 0, count1 = 0, count2 = 0;
+
+        for (int num : nums) {
+            if (num == candidate1) count1++;
+            else if (num == candidate2) count2++;
+            else if (count1 == 0) {
+                candidate1 = num;
+                count1++;
+            } else if (count2 == 0) {
+                candidate2 = num;
+                count2++;
+            } else {
+                count1--;
+                count2--;
+            }
+        }
+
+        count1 = count2 = 0;
+        for (int num : nums) {
+            if (num == candidate1) count1++;
+            else if (num == candidate2) count2++;
+        }
+
+        List<Integer> res = new ArrayList<>();
+        if (count1 > nums.length / 3) res.add(candidate1);
+        if (count2 > nums.length / 3) res.add(candidate2);
+        return res;
+    }
+}
