LeetCode_169. Majority Element

 

169. Majority Element

Easy

Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.

You may assume that the array is non-empty and the majority element always exist in the array.

Example 1:

Input: [3,2,3]
Output: 3

Example 2:

Input: [2,2,1,1,1,2,2]
Output: 2

 

package leetcode.easy;

import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;
import java.util.Random;

public class MajorityElement {
	@org.junit.Test
	public void test() {
		int[] nums1 = { 3, 2, 3 };
		int[] nums2 = { 2, 2, 1, 1, 1, 2, 2 };
		System.out.println(majorityElement1(nums1));
		System.out.println(majorityElement1(nums2));
		System.out.println(majorityElement2(nums1));
		System.out.println(majorityElement2(nums2));
		System.out.println(majorityElement3(nums1));
		System.out.println(majorityElement3(nums2));
		System.out.println(majorityElement4(nums1));
		System.out.println(majorityElement4(nums2));
		System.out.println(majorityElement5(nums1));
		System.out.println(majorityElement5(nums2));
		System.out.println(majorityElement6(nums1));
		System.out.println(majorityElement6(nums2));
	}

	public int majorityElement1(int[] nums) {
		int majorityCount = nums.length / 2;

		for (int num : nums) {
			int count = 0;
			for (int elem : nums) {
				if (elem == num) {
					count += 1;
				}
			}

			if (count > majorityCount) {
				return num;
			}
		}

		return -1;
	}

	private Map countNums(int[] nums) {
		Map counts = new HashMap();
		for (int num : nums) {
			if (!counts.containsKey(num)) {
				counts.put(num, 1);
			} else {
				counts.put(num, counts.get(num) + 1);
			}
		}
		return counts;
	}

	public int majorityElement2(int[] nums) {
		Map counts = countNums(nums);

		Map.Entry majorityEntry = null;
		for (Map.Entry entry : counts.entrySet()) {
			if (majorityEntry == null || entry.getValue() > majorityEntry.getValue()) {
				majorityEntry = entry;
			}
		}

		return majorityEntry.getKey();
	}

	public int majorityElement3(int[] nums) {
		Arrays.sort(nums);
		return nums[nums.length / 2];
	}

	private int randRange(Random rand, int min, int max) {
		return rand.nextInt(max - min) + min;
	}

	private int countOccurences(int[] nums, int num) {
		int count = 0;
		for (int i = 0; i < nums.length; i++) {
			if (nums[i] == num) {
				count++;
			}
		}
		return count;
	}

	public int majorityElement4(int[] nums) {
		Random rand = new Random();

		int majorityCount = nums.length / 2;

		while (true) {
			int candidate = nums[randRange(rand, 0, nums.length)];
			if (countOccurences(nums, candidate) > majorityCount) {
				return candidate;
			}
		}
	}

	private int countInRange(int[] nums, int num, int lo, int hi) {
		int count = 0;
		for (int i = lo; i <= hi; i++) {
			if (nums[i] == num) {
				count++;
			}
		}
		return count;
	}

	private int majorityElementRec(int[] nums, int lo, int hi) {
		// base case; the only element in an array of size 1 is the majority
		// element.
		if (lo == hi) {
			return nums[lo];
		}

		// recurse on left and right halves of this slice.
		int mid = (hi - lo) / 2 + lo;
		int left = majorityElementRec(nums, lo, mid);
		int right = majorityElementRec(nums, mid + 1, hi);

		// if the two halves agree on the majority element, return it.
		if (left == right) {
			return left;
		}

		// otherwise, count each element and return the "winner".
		int leftCount = countInRange(nums, left, lo, hi);
		int rightCount = countInRange(nums, right, lo, hi);

		return leftCount > rightCount ? left : right;
	}

	public int majorityElement5(int[] nums) {
		return majorityElementRec(nums, 0, nums.length - 1);
	}

	public int majorityElement6(int[] nums) {
		int count = 0;
		Integer candidate = null;

		for (int num : nums) {
			if (count == 0) {
				candidate = num;
			}
			count += (num == candidate) ? 1 : -1;
		}

		return candidate;
	}
}