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 :



title-img 


                            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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