# Counting Valleys

### Problem Statement :

```An avid hiker keeps meticulous records of their hikes. During the last hike that took exactly steps steps, for every step it was noted if it was an uphill, U , or a downhill, D step. Hikes always start and end at sea level, and each step up or down represents a 1 unit change in altitude. We define the following terms:

A mountain is a sequence of consecutive steps above sea level, starting with a step up from sea level and ending with a step down to sea level.
A valley is a sequence of consecutive steps below sea level, starting with a step down from sea level and ending with a step up to sea level.
Given the sequence of up and down steps during a hike, find and print the number of valleys walked through.

Function Description

Complete the countingValleys function in the editor below.

countingValleys has the following parameter(s):

int steps: the number of steps on the hike
string path: a string describing the path

Returns

int: the number of valleys traversed
Input Format

The first line contains an integer steps, the number of steps in the hike.
The second line contains a single string path, of steps  characters that describe the path.

Constraints

2  <=   steps  <=  10^6```

### Solution :

```                            ```Solution in C :

In   C  :

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {
int n;
scanf("%i", &n);
char str[n];
scanf("%s", str);
int level = 0, result = 0, valley = 0;
for (int i = 0; i < n; i++) {
if(str[i] == 'U') {
level++;
if(level == 0 && valley) {
valley = 0;
result++;
}
}
else if(str[i] == 'D') {
if(level == 0)
valley = 1;
level--;
}
}
printf("%i", result);
return 0;
}```
```

```                        ```Solution in C++ :

In   C  ++ :

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

int main() {
int l;
string str; cin>>l>>str;
int height = 0;
int count = 0;
for(int i=0;i<l;i++){
if (str[i]=='U') height++;
else {
if (height==0) count++;
height--;
}
}
if (height<0) count--;
cout<<count<<endl;

return 0;
}```
```

```                        ```Solution in Java :

In   Java  :

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
String s = in.next();
int valeys = 0;
int level = 0;
for(int i=0; i<s.length(); i++){
if(s.charAt(i)=='D'){
if(level==0){
valeys++;
}
level--;
}
else{
level++;
}

}
System.out.println(valeys);
}
}```
```

```                        ```Solution in Python :

In  Python3  :

#!/bin/python3

import math
import os
import random
import re
import sys

# Complete the countingValleys function below.
def countingValleys(n, s):
lv = 0
v = 0
for i in s:
if i == 'U':
lv += 1
if i == 'D':
lv -= 1
if lv == 0 and i == 'U':
v += 1

return v

if __name__ == '__main__':
fptr = open(os.environ['OUTPUT_PATH'], 'w')

n = int(input())

s = input()

result = countingValleys(n, s)

fptr.write(str(result) + '\n')

fptr.close()```
```

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