Easy Go Solution | Using Map
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```golang []\nfunc getLargestOutlier(nums []int) int {\n    mp := make(map[int]int) // Frequency map\n\ttotal := 0             // Sum of all numbers\n\tfor _, it := range nums {\n\t\tmp[it]++\n\t\ttotal += it\n\t}\n\n\tans := math.MinInt // Smallest possible integer\n\tfor _, it := range nums {\n\t\t// Temporarily reduce the count of the current number\n\t\tmp[it]--\n\t\tif mp[it] == 0 {\n\t\t\tdelete(mp, it) // Remove the key if count is 0\n\t\t}\n\n\t\tcur := total - it // Current sum excluding `it`\n\n\t\t// Check if `cur` is even and `cur/2` exists in the map\n\t\tif cur%2 == 0 {\n\t\t\tif _, exists := mp[cur/2]; exists {\n\t\t\t\tif it > ans {\n\t\t\t\t\tans = it // Update the answer if `it` is greater\n\t\t\t\t}\n\t\t\t}\n\t\t}\n\n\t\t// Restore the original count of `it`\n\t\tmp[it]++\n\t}\n    return ans\n}\n```
['Go']