---
description: only for case where majority has more than n/2  occurrences
---

# Finding the majority element

The Boyer-Moore voting method determines the majority element i.e with more than $$N/2$$ occurrences  among other elements.

{% hint style="info" %}
If there is no majority, the algorithm will not recognise this and will continue to output one of the items. To avoid this we can add a check
{% endhint %}



## Algorithm

First pass identifies an element as a majority

```java
// we count the votes 
// number of votes will be > 0 only for the majority element
int candidate = -1, votes = 0; 
for (int i = 0; i < nums.length; ++i) { 
	// if votes is 0 then replace the candiate
	if(votes == 0) { 
		candidate = nums[i];
	} 
	
	// if the element is different consider it as vote againts 	
	votes+= (candidate == nums[i]) ? +1 : -1;
} 
```

Second pass confirms that the element identified in the first pass is indeed a majority

```java
int count = 0;
for (int i = 0; i < nums.length; ++i) { 
	if(candidate == nums[i]) count++;
}

if(count < (nums.length >> 1) ) return -1;

return candidate; 
```



## Complexity

* Time Complexity : $$\text{O}(n)$$
  * Only 2 iteration are made over the array
* Space Complexity : $$\text{O}(1)$$
  * No extra space is needed

## Reference

* [Boyer-Moore majority vote algorithm](https://www.topcoder.com/thrive/articles/boyer-moore-majority-vote-algorithm)
