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This solution uses the Boyer-Moore Voting Algorithm to find the majority element (an element that appears more than n/2 times) in linear time with constant space. The algorithm maintains a count (i) and a candidate (j). It iterates through the array, updating the candidate whenever the count reaches zero. If the current element matches the candidate, the count increases; otherwise, it decreases. At the end of the iteration, the candidate is guaranteed to be the majority element.

Time Complexity: O(n)
Space Complexity: O(1)

Author: coderpheonix

169.Majority Element.cpp

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+class Solution {
+public:
+    int majorityElement(vector<int>& nums) {
+        // i is the count for tracking the majority candidate
+        int i = 0;
+
+        // j is the variable to store the current majority candidate
+        int j = 0;
+
+        // Boyer-Moore Voting Algorithm: This loop determines the majority element
+        for (int num : nums) {
+            // If count 'i' becomes 0, assign the current element 'num' as the new candidate
+            if (i == 0) {
+                j = num;
+            }
+            // If 'num' is the current candidate 'j', increment the count
+            // Otherwise, decrement the count
+            i += (num == j) ? 1 : -1;
+        }
+
+        // Return the candidate which is the majority element
+        return j;
+    }
+};
+