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Given a characters array `letters`

that is sorted in **non-decreasing** order and a character `target`

, return *the smallest character in the array that is larger than *`target`

.

**Note** that the letters wrap around.

- For example, if
`target == 'z'`

and`letters == ['a', 'b']`

, the answer is`'a'`

.

**Example 1:**

```
Input: letters = ["c","f","j"], target = "a"
Output: "c"
```

**Example 2:**

```
Input: letters = ["c","f","j"], target = "c"
Output: "f"
```

**Example 3:**

```
Input: letters = ["c","f","j"], target = "d"
Output: "f"
```

**Constraints:**

`2 <= letters.length <= 10`

^{4}`letters[i]`

is a lowercase English letter.`letters`

is sorted in**non-decreasing**order.`letters`

contains at least two different characters.`target`

is a lowercase English letter.

```
class Solution(object):
def nextGreatestLetter(self, letters, target):
"""
:type letters: List[str]
:type target: str
:rtype: str
"""
# if the number is out of bound
if target >= letters[-1] or target < letters[0]:
return letters[0]
low = 0
high = len(letters)-1
while low <= high:
mid = (high+low)//2
if target >= letters[mid]: # in binary search this would be only greater than
low = mid+1
if target < letters[mid]:
high = mid-1
return letters[low]
```

```
class Solution {
public char nextGreatestLetter(char[] a, char x) {
int n = a.length;
if (x >= a[n - 1]) x = a[0];
else x++;
int lo = 0, hi = n - 1;
while (lo < hi) {
int mid = lo + (hi - lo) / 2;
if (a[mid] == x) return a[mid];
if (a[mid] < x) lo = mid + 1;
else hi = mid;
}
return a[hi];
}
}
```

```
class Solution {
public:
char nextGreatestLetter(vector<char>& letters, char target) {
auto it = upper_bound(letters.begin(), letters.end(), target);
return it == letters.end() ? letters[0] : *it;
}
};
```

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