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Oct 5, 2023
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Create 0229-majority-element-ii.java #3042

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65 changes: 65 additions & 0 deletions java/0229-majority-element-ii.java
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Original file line number Diff line number Diff line change
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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;
}
}