Given an integer array arr[] of size N, the task is to find the number of times the graph crosses the X-axis, where a positive number means going above its current position by that value and a negative number means going down by that value. Initially, the current position is at the origin.
Examples:
Input: arr[] = {4, -6, 2, 8, -2, 3, -12}
Output: 3
Explanation:
So the graph crosses the X-axis 3 times.
Input: arr[] = {1, 1, -3, 2}
Output: 2
Approach: Iterate over the array and keep the value of the previous level and current level into two variables. Initially, both levels were zero. Increase / Decrease the level by the value given in the array and increase the count in the two cases below.
- If the previous level is less than zero and the current level is greater than or equal to zero.
- If the previous level is greater than zero and the current level is less than or equal to zero.
Below is the implementation of the above approach.
C++
// C++ implementation to count the// number of times the graph// crosses the x-axis.#include <bits/stdc++.h>using namespace std;// Function to count the// number of times the graph// crosses the x-axis.int times(int steps[], int n){ int current_level = 0; int previous_level = 0; int count = 0; // Iterate over the steps array for (int i = 0; i < n; i++) { // Update the previous level and // current level by value given // in the steps array previous_level = current_level; current_level = current_level + steps[i]; // Condition to check that the // graph crosses the origin. if ((previous_level < 0 && current_level >= 0) || (previous_level > 0 && current_level <= 0)) { count++; } } return count;}// Driver Codeint main(){ int steps[12] = { 1, -1, 0, 0, 1, 1, -3, 2 }; int n = sizeof(steps) / sizeof(int); cout << times(steps, n); return 0;} |
Java
// Java implementation to count the // number of times the graph // crosses the x-axis. class GFG { // Function to count the // number of times the graph // crosses the x-axis. static int times(int []steps, int n) { int current_level = 0; int previous_level = 0; int count = 0; // Iterate over the steps array for (int i = 0; i < n; i++) { // Update the previous level and // current level by value given // in the steps array previous_level = current_level; current_level = current_level + steps[i]; // Condition to check that the // graph crosses the origin. if ((previous_level < 0 && current_level >= 0) || (previous_level > 0 && current_level <= 0)) { count++; } } return count; } // Driver Code public static void main (String[] args) { int steps[] = { 1, -1, 0, 0, 1, 1, -3, 2 }; int n = steps.length; System.out.println(times(steps, n)); } }// This code is contributed by AnkitRai01 |
Python3
# Python3 implementation to count the# number of times the graph# crosses the x-axis.# Function to count the# number of times the graph# crosses the x-axis.def times(steps, n): current_level = 0 previous_level = 0 count = 0 # Iterate over the steps array for i in range(n): # Update the previous level and # current level by value given #in the steps array previous_level = current_level current_level = current_level+ steps[i] # Condition to check that the # graph crosses the origin. if ((previous_level < 0 and current_level >= 0) or (previous_level > 0 and current_level <= 0)): count += 1 return count# Driver Codesteps = [1, -1, 0, 0, 1, 1, -3, 2]n = len(steps)print(times(steps, n))# This code is contributed by mohit kumar 29 |
C#
// C# implementation to count the // number of times the graph // crosses the x-axis.using System;class GFG { // Function to count the // number of times the graph // crosses the x-axis. static int times(int []steps, int n) { int current_level = 0; int previous_level = 0; int count = 0; // Iterate over the steps array for (int i = 0; i < n; i++) { // Update the previous level and // current level by value given // in the steps array previous_level = current_level; current_level = current_level + steps[i]; // Condition to check that the // graph crosses the origin. if ((previous_level < 0 && current_level >= 0) || (previous_level > 0 && current_level <= 0)) { count++; } } return count; } // Driver Code public static void Main () { int []steps = { 1, -1, 0, 0, 1, 1, -3, 2 }; int n = steps.Length; Console.WriteLine(times(steps, n)); } }// This code is contributed by AnkitRai01 |
Javascript
<script>// Javascript implementation to count the// number of times the graph// crosses the x-axis.// Function to count the// number of times the graph// crosses the x-axis.function times(steps, n) { let current_level = 0; let previous_level = 0; let count = 0; // Iterate over the steps array for (let i = 0; i < n; i++) { // Update the previous level and // current level by value given // in the steps array previous_level = current_level; current_level = current_level + steps[i]; // Condition to check that the // graph crosses the origin. if ((previous_level < 0 && current_level >= 0) || (previous_level > 0 && current_level <= 0)) { count++; } } return count;}// Driver Codelet steps = [1, -1, 0, 0, 1, 1, -3, 2];let n = steps.length;document.write(times(steps, n));// This code is contributed by _saurabh_jaiswal</script> |
Output
3
Time Complexity: O(n), where n is the size of the given array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
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