A simple O(n)-Python
\nif k>len(arr)//2:\n\treturn(max(arr))\nelse:\n\tcount=0\n\tcurr_max=arr[0]\n\tfor i in range(1,len(arr)):\n\t  if count==k:\n\t\treturn(curr_max)\n\t\tbreak\n
```\nif k>len(arr)//2:\n\treturn(max(arr))\nelse:\n\tcount=0\n\tcurr_max=arr[0]\n\tfor i in range(1,len(arr)):\n\t  if count==k:\n\t\treturn(curr_max)\n\t\tbreak\n\t  else:\n\t\tif arr[i]>curr_max:\n\t\t  curr_max=arr[i]\n\t\t  count=1\n\t\telse:\n\t\t  count+=1\nreturn(curr_max)\n```
['Python']