Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
⌊ n/2 ⌋ times.[3, 30, 34, 5, 9], the largest formed number is 9534330. A -> 1
B -> 2
C -> 3
...
Z -> 26
AA -> 27
AB -> 28
1 -> A
2 -> B
3 -> C
...
26 -> Z
27 -> AA
28 -> AB
gas[i].cost[i] of gas to travel from station i to its next station (i+1). You begin the journey with an empty tank at one of the gas stations.1,2,3 → 1,3,23,2,1 → 1,2,31,1,5 → 1,5,1"PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)P A H N A P L S I I G Y I RAnd then read line by line:
"PAHNAPLSIIGYIR"string convert(string text, int nRows);
convert("PAYPALISHIRING", 3) should return "PAHNAPLSIIGYIR".
gas[i].cost[i] of gas to travel from station i to its next station (i+1). You begin the journey with an empty tank at one of the gas stations.[1,2,0] return 3,[3,4,-1,1] return 2.null.