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.


Example
steps = 8 path = [DDUUUUDD]
 
The hiker first enters a valley 2 units deep. Then they climb out and up onto a mountain 2 units high. Finally, the hiker returns to sea level and ends the hike.


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
path[i] belongs to {U, D}


Solution :



title-img 


                            Solution in C :

python 3  :

N=int(input())
S=input()
L=0
V=0
for s in S:
    if s == 'U':
        L += 1
        if L == 0:
            V += 1
    else:
        L -= 1
        
print(V)









Java  :

import java.io.*;
import java.util.*;

public class Solution {

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









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;
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    return 0;
}









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;
}








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