we can use binary search for the second array. lets assume that n > m then meaning, we would need to sort 2 arrays (O(nlogn)+O(mlogm) => 2O(nlogn))

then on each iteration of the keyboards we will do another log(n) operation

===> 2O(nlogn) + nlogn time, and O(1) space

def findMaxValue(arr, value) -> int: low,high= 0,(len(arr) - 1) while low <= high: mid = (low + high) // 2 if arr[mid] == value: return value elif arr[mid] > value: high = mid - 1 else: low = mid + 1 return None if arr[high] > value else arr[high]

maxValue = -1
keyboards.sort()
drives.sort()
for keyboard in keyboards:
    if keyboard >= b:   
        continue
    maxDrive = findMaxValue(drives, b - keyboard)
    if maxDrive:
        maxValue = max(maxValue, keyboard + maxDrive)    

return maxValue