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left to 1 and right to the maximum value in quantities.left is less than or equal to right:
a. Calculate mid as the average of left and right.
b. Initialize a variable requiredStores to 0.
c. Iterate over each quantity in quantities:
i. Increment requiredStores by the ceiling of the current quantity divided by mid.
d. If requiredStores is less than or equal to n, update right to mid - 1 (try a smaller x).
e. Otherwise, update left to mid + 1 (try a larger x).left as the minimum possible x.