JS easy solution with O(n)
\nvar maxPower = function(s) {\n    let output = 1;\n    let currentPower = 1;\n    \n    for (let i = 0; i < s.length; i++) {\n        if (s[i] === s[i+1]) {\n
```\nvar maxPower = function(s) {\n    let output = 1;\n    let currentPower = 1;\n    \n    for (let i = 0; i < s.length; i++) {\n        if (s[i] === s[i+1]) {\n            currentPower++;\n        } else {\n            output = Math.max(output, currentPower);\n            currentPower = 1;\n        }\n    }\n    \n    return output;\n};\n```
['JavaScript']