[tabby title=”Task”]

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:

 [3,2,3]

 3

 [2,2,1,1,1,2,2]

 2

This problem was taken from Leetcode

[tabby title=”Solution”]

The solution:

Create an object (or associative array, depends of the language), iterate through all elements in the array adding them to the newly created object using the element value as a key. If element exists increase the value +1.
Keep track of mostly repeated element and return it at the end.

var arr = [3, 3, 4, 2, 4, 2, 4, 4];


/**
 * @param {number[]} nums
 * @return {number}
 */
var majorityElement = function(nums) {
  var elements = {};
  var majorityElement = nums[0];
  var majorityElementCount = 1;
  for(var i in nums) {
  	var element = nums[i];
  	elements[element] = typeof elements[element] == 'undefined' ? 1 : elements[element] + 1;
  	if(elements[element] > majorityElementCount) {
  		majorityElement = element;
  		majorityElementCount = elements[element];
  	}
  }    
  return majorityElement;
};

console.log(">>", majorityElement(arr));

– Create an object (line 9)
– Iterate through each element in the array feeling up the object with the elements from the array where the element value would be the key of the element (line 14)
– Each time when the element exists, add + 1
– Keep track of mostly repeated element (Lines 16-17)

Could we optimize the code? Slightly. When we add + 1 we could check if the value is already greater than the length of the array / 2 and if se we know that this is the majority element because no other element could have greater value.

...
    if(elements[element] > nums.length / 2) {
        return element;
    }
...

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