**Monotonous String Groups π - Google Top Interview Questions**

### Problem Statement :

You are given a lowercase alphabet string s. Return the minimum numbers of contiguous substrings in which s must be broken into such that each substring is either non-increasing or non-decreasing. For example, "acccf" is a non-decreasing string, and "bbba" is a non-increasing string. Constraints n β€ 100,000 where n is the length of s Example 1 Input s = "abcdcba" Output 2 Explanation We can break s into "abcd" + "cba" Example 2 Input s = "zzz" Output 1 Explanation We can break s into just "zzz"

### Solution :

` ````
Solution in C++ :
int solve(string s) {
int ret = 0;
for (int i = 0; i < s.size();) {
ret++;
int j = i + 1;
bool noninc = true;
bool nondec = true;
while (j < s.size()) {
if (s[j] > s[j - 1]) noninc = false;
if (s[j] < s[j - 1]) nondec = false;
if (!noninc && !nondec) break;
j++;
}
i = j;
}
return ret;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(String s) {
// If empty string, there are no substrings
if (s.length() < 1) {
return 0;
}
// Counter to keep track of number of substrings
int count = 1;
// Number to indicate whether current substring is
// non-decreasing, non-increasing, or neither decreasing nor
// increasing (staying the same).
int current = 0;
for (int i = 1; i < s.length(); i++) {
// Find out out whether behavior is decreasing, staying the same
// or increasing between current character and previous character
int diff = s.charAt(i) - s.charAt(i - 1);
// If the current substring's behavior has not been determined yet
// (can still either be non-decreasing or non-increasing)
if (current == 0) {
// Set the current string's behavior to non-increasing
if (diff < 0) {
current = -1;
// Set the current string's behavior to non-decreasing
} else if (diff > 0) {
current = 1;
}
// Note that if diff is 0, the current string's behavior can be
// determined at a later time
// If the current substring's behavior has already been determined
// (either non-decreasing or non-increasing)
} else {
// If the behavior between this index and the previous index does
// not match the behavior of the current substring, it is time
// to start a new substring
if (diff < 0 && current == 1 || diff > 0 && current == -1) {
count++;
current = 0;
}
}
}
return count;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, s):
if not s:
return 0
count = 1
increasing = None
for i in range(len(s) - 1):
if increasing is None and s[i] != s[i + 1]:
increasing = s[i] < s[i + 1]
elif (increasing and s[i] > s[i + 1]) or (increasing is False and s[i] < s[i + 1]):
count += 1
increasing = None
return count
```

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