Golang Simple Solution BEAT 100% Runtime
Approach\n Describe your approach to solving the problem. \nBrutforce all variation, choose the min value if count is equal\n# Complexity\n- Time complexity:\n 
# Approach\n<!-- Describe your approach to solving the problem. -->\nBrutforce all variation, choose the min value if count is equal\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n$$O(N\u2217M)$$\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n$$O(1)$$\n# Code\n```golang []\nfunc maxDivScore(nums []int, divisors []int) int {\n    var minInt int\n\tmaxCount := -1\n\tfor _, divisor := range divisors {\n\n\t\tcounter := 0\n\t\tfor _, num := range nums {\n\t\t\tif num >= divisor && num%divisor == 0 {\n\t\t\t\tcounter++\n\t\t\t}\n\t\t}\n\t\tif counter > maxCount {\n\t\t\tmaxCount = counter\n\t\t\tminInt = divisor\n\t\t} else if counter == maxCount {\n\t\t\tminInt = min(divisor, minInt)\n\t\t}\n\t}\n\n\treturn minInt\n}\n```
['Go']