Skip to main content

Counting Valleys HackerRank Solution | Java Solution

Counting Valleys HackerRank Solution in Java


An avid hiker keeps meticulous records of their hikes. During the last hike that took exactly steps. , for every step it was noted if it was an Uphil ,U or Downhill, D.
Hikes always start and end at sea level, and each step up or down represents a 1unit 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.

 

In this post, We do discuss two solution of  "Counting Valleys" problem. so lets see first one.

Solution 1

public class CountingValleys{

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
       
        System.out.println("Enter number of steps");
        int number = scanner.nextInt();
        
        System.out.println("Enter valleys string");
        String path = scanner.next();
       
        int steps = 0;
        int valleys = 0;
        
        // For loop until Entered number
        for (int i = 0; i < number; i++) {
           
            if (path.charAt(i) == 'U') {
                // If steps == -1 then it means hiker completed its one valleys
                if (steps == -1) {
                    valleys++;
                }
                steps++;
            }
           
            if (path.charAt(i) == 'D') {
                steps--;
            }
        }
        System.out.println(valleys);
    }
}

Output :-
Enter number of steps
8
Enter valleys string
UDDDUDUU
1

Enter number of steps
12
Enter valleys string
DDUUDDUDUUUD
2

Solution 2

public class SubString {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
       
        System.out.println("Enter valleys string");
        String path = scanner.next();       
        
        int steps = 0;
        int valleys = 0;
        
        for (char chr : path.toCharArray()) {
            if (chr == 'U') steps++;
            if (chr == 'D') steps--;
            
            if(steps == 0  && chr == 'U')
                ++valleys;
        }
        
        System.out.println(valleys);
    }
}

Output :-
Enter valleys string
UDDDUDUU
1

Enter valleys string
DDUUDDUDUUUD
2

 

Comments

Popular posts from this blog

Flipping the Matrix HackerRank Solution in Java with Explanation

Java Solution for Flipping the Matrix | Find Highest Sum of Upper-Left Quadrant of Matrix Problem Description : Sean invented a game involving a 2n * 2n matrix where each cell of the matrix contains an integer. He can reverse any of its rows or columns any number of times. The goal of the game is to maximize the sum of the elements in the n *n submatrix located in the upper-left quadrant of the matrix. Given the initial configurations for q matrices, help Sean reverse the rows and columns of each matrix in the best possible way so that the sum of the elements in the matrix's upper-left quadrant is maximal.  Input : matrix = [[1, 2], [3, 4]] Output : 4 Input : matrix = [[112, 42, 83, 119], [56, 125, 56, 49], [15, 78, 101, 43], [62, 98, 114, 108]] Output : 119 + 114 + 56 + 125 = 414 Full Problem Description : Flipping the Matrix Problem Description   Here we can find solution using following pattern, So simply we have to find Max of same number of box like (1,1,1,1). And last

Plus Minus HackerRank Solution in Java | Programming Blog

Java Solution for HackerRank Plus Minus Problem Given an array of integers, calculate the ratios of its elements that are positive , negative , and zero . Print the decimal value of each fraction on a new line with 6 places after the decimal. Example 1 : array = [1, 1, 0, -1, -1] There are N = 5 elements, two positive, two negative and one zero. Their ratios are 2/5 = 0.400000, 2/5 = 0.400000 and 1/5 = 0.200000. Results are printed as:  0.400000 0.400000 0.200000 proportion of positive values proportion of negative values proportion of zeros Example 2 : array = [-4, 3, -9, 0, 4, 1]  There are 3 positive numbers, 2 negative numbers, and 1 zero in array. Following is answer : 3/6 = 0.500000 2/6 = 0.333333 1/6 = 0.166667 Lets see solution Solution 1 import java.io.*; import java.math.*; import java.security.*; import java.text.*; import java.util.*; import java.util.concurrent.*; import java.util.function.*; import java.util.regex.*; import java.util.stream.*; import static java.util.st