Add majority element problem

Author: sudheerj

README.md

@@ -58,6 +58,7 @@ List of Programs related to data structures and algorithms
 | 17  | Pascal's Triangle                  |        [Source](https://github.com/sudheerj/datastructures-algorithms/blob/master/src/javascript/algorithms/array/17.pascalsTriangle/pascalsTriangle.js)        | [JavaScript](https://livecodes.io/?console=open&x=https://github.com/sudheerj/datastructures-algorithms/blob/master/src/javascript/algorithms/array/17.pascalsTriangle/pascalsTriangle.js)               |        [Documentation](https://github.com/sudheerj/datastructures-algorithms/blob/master/src/javascript/algorithms/array/17.pascalsTriangle/pascalsTriangle.md)        | Easy | Array traversal and sequential sum |
 | 18  | Remove Element                   |        [Source](https://github.com/sudheerj/datastructures-algorithms/blob/master/src/javascript/algorithms/array/18.removeElement/removeElement.js)        | [JavaScript](https://livecodes.io/?console=open&x=https://github.com/sudheerj/datastructures-algorithms/blob/master/src/javascript/algorithms/array/18.removeElement/removeElement.js)               |        [Documentation](https://github.com/sudheerj/datastructures-algorithms/blob/master/src/javascript/algorithms/array/18.removeElement/removeElement.md)        | Easy | Array traversal |
 | 19  | Can place flowers                  |        [Source](https://github.com/sudheerj/datastructures-algorithms/blob/master/src/javascript/algorithms/array/19.canPlaceFlowers/canPlaceFlowers.js)        | [JavaScript](https://livecodes.io/?console=open&x=https://github.com/sudheerj/datastructures-algorithms/blob/master/src/javascript/algorithms/array/19.canPlaceFlowers/canPlaceFlowers.js)               |        [Documentation](https://github.com/sudheerj/datastructures-algorithms/blob/master/src/javascript/algorithms/array/19.canPlaceFlowers/canPlaceFlowers.md)        | Easy | Array traversal and comparison |
+| 20  | Majority Element                 |        [Source](https://github.com/sudheerj/datastructures-algorithms/blob/master/src/javascript/algorithms/array/20.majorityElement/majorityElement.js)        | [JavaScript](https://livecodes.io/?console=open&x=https://github.com/sudheerj/datastructures-algorithms/blob/master/src/javascript/algorithms/array/20.majorityElement/majorityElement.js)               |        [Documentation](https://github.com/sudheerj/datastructures-algorithms/blob/master/src/javascript/algorithms/array/20.majorityElement/majorityElement.md)        | Easy | Boyer Moore Voting algorithm |
 
 ### String
 

src/java1/algorithms/array/majorityElement/MajorityElement.java

@@ -0,0 +1,24 @@
+package java1.algorithms.array.majorityElement;
+
+public class MajorityElement {
+    private static int majorityElement(int[] nums){
+        int count = 0;
+        int candidate = 0;
+
+        for (int num : nums) {
+            if(count == 0) {
+                candidate = num;
+            }
+
+            count += num == count ? 1 : -1;
+        }
+
+        return candidate;
+    }
+    public static void main(String[] args) {
+        int[] nums1 = new int[]{5,4,5,5};
+        int[] nums2 = new int[]{3,4,3,5,3,3,1,3};
+        System.out.println(majorityElement(nums1));
+        System.out.println(majorityElement(nums2));
+    }
+}

src/java1/algorithms/array/majorityElement/MajorityElement.md

@@ -0,0 +1,36 @@
+**Description:**
+Given an integer array `nums` of size `n`, find the majority element from an array. The majority element is defined as the element that appears more than ⌊n / 2⌋ times.
+
+**Note:** The majority element always exists in an array.
+
+### Examples
+Example 1:
+
+Input: nums = [5,4,5,5]
+Output: 5
+
+Example 2:
+
+Input: nums = [3,4,3,5,3,3,1,3]
+Output: 3
+
+**Algorithmic Steps**
+This problem is solved with the help of Boyer Moore Voting algorithm. The algorithmic approach can be summarized as follows:
+
+1. Create a function(`majorityElement`) by accepting integer array(`nums`) to determine the majority element.
+
+2. Initialize the counting variable(`count`) to determine number of occurances of an element. The default of majority element(`candidate`) is initialized to `0`.
+
+3. Iterate over the array, and for each iteration compare the counting variable equals to 0 or not. 
+    1. If count value is equals to zero, update the candidate value with current element. 
+
+    2. The counter value will be incremented by 1 if both candidate and current element are same, otherwise the counter is decremented by 1.
+   
+    3. Since there is a gurantee of majority element, you don't need to verify whether the calculated element is majority element or not. i.e, majority element occurs more than half of array(`n/2`) times, where `n` is the length of given array.
+   
+ 4. After completion of iteration, return candidate value to represent majority element in the given number.
+
+**Time and Space complexity:**
+This algorithm has a time complexity of `O(n)`, where `n` is the number of elements in an array. This is because we need to iterate over all the elements at most once.
+ 
+It takes constant time complexity of `O(1)` due to no additional datastructure been used other than counting variable.

src/javascript/algorithms/array/20.majorityElement/majorityElement.js

@@ -0,0 +1,19 @@
+function majorityElement(nums){
+    let count = 0;
+    let candidate = null;
+
+    for (const num of nums) {
+        if(count === 0) {
+            candidate = num;
+        }
+
+        count += num === candidate ? 1 : -1;
+    }
+
+    return candidate;
+}
+
+const nums1 = [5,4,5,5];
+const nums2 = [3,4,3,5,3,3,1,3];
+console.log(majorityElement(nums1));
+console.log(majorityElement(nums2));

src/javascript/algorithms/array/20.majorityElement/majorityElement.md

@@ -0,0 +1,36 @@
+**Description:**
+Given an integer array `nums` of size `n`, find the majority element from an array. The majority element is defined as the element that appears more than ⌊n / 2⌋ times.
+
+**Note:** The majority element always exists in an array.
+
+### Examples
+Example 1:
+
+Input: nums = [5,4,5,5]
+Output: 5
+
+Example 2:
+
+Input: nums = [3,4,3,5,3,3,1,3]
+Output: 3
+
+**Algorithmic Steps**
+This problem is solved with the help of Boyer Moore Voting algorithm. The algorithmic approach can be summarized as follows:
+
+1. Create a function(`majorityElement`) by accepting integer array(`nums`) to determine the majority element.
+
+2. Initialize the counting variable(`count`) to determine number of occurances of an element. The default of majority element(`candidate`) is initialized to `null`.
+
+3. Iterate over the array, and for each iteration compare the counting variable equals to 0 or not. 
+    1. If count value is equals to zero, update the candidate value with current element. 
+
+    2. The counter value will be incremented by 1 if both candidate and current element are same, otherwise the counter is decremented by 1.
+   
+    3. Since there is a gurantee of majority element, you don't need to verify whether the calculated element is majority element or not. i.e, majority element occurs more than half of array(`n/2`) times, where `n` is the length of given array.
+   
+ 4. After completion of iteration, return candidate value to represent majority element in the given number.
+
+**Time and Space complexity:**
+This algorithm has a time complexity of `O(n)`, where `n` is the number of elements in an array. This is because we need to iterate over all the elements at most once.
+ 
+It takes constant time complexity of `O(1)` due to no additional datastructure been used other than counting variable.