Consecutive steps to roof top
Given heights of consecutive buildings, find the maximum number of consecutive steps one can put forward such that he gain a increase in altitude while going from roof of one building to next adjacent one.
Examples :
Input : arr[] = {1, 2, 2, 3, 2}
Output : 1
Explanation :
Maximum consecutive steps from 1 to 2 OR 2 to 3.
Input : arr[] = {1, 2, 3, 4}
Output : 3
This problem is basically a variation of Longest increasing subarray
Approach:-
initialize count = 0
initialize maximum = 0
if arr[i]>a[i-1]
then count increment
else
maximum = max(maximum, count)
at the end maximum=max(maximum, count)
C++
// CPP code to find maximum // number of consecutive steps. #include <bits/stdc++.h> using namespace std; // Function to count consecutive steps int find_consecutive_steps(int arr[], int len) { int count = 0; int maximum = 0; for (int index = 1; index < len; index++) { // count the number of consecutive // increasing height building if (arr[index] > arr[index - 1]) count++; else { maximum = max(maximum, count); count = 0; } } return max(maximum, count); } // Driver code int main() { int arr[] = { 1, 2, 3, 4 }; int len = sizeof(arr) / sizeof(arr[0]); cout << find_consecutive_steps(arr, len); } |
Java
// Java code to find maximum // number of consecutive steps. import java.io.*; class GFG { // Function to count consecutive steps static int find_consecutive_steps(int arr[], int len) { int count = 0; int maximum = 0; for (int index = 1; index < len; index++) { // count the number of consecutive // increasing height building if (arr[index] > arr[index - 1]) count++; else { maximum = Math.max(maximum, count); count = 0; } } return Math.max(maximum, count); } // Driver code public static void main (String[] args) { int arr[] = { 1, 2, 3, 4 }; int len = arr.length; System.out.println(find_consecutive_steps(arr, len)); } } // This code is contributed by Gitanjali. |
Python3
# Python3 code to find maximum # number of consecutive steps import math # Function to count consecutive steps def find_consecutive_steps(arr, len): count = 0; maximum = 0 for index in range(1, len): # count the number of consecutive # increasing height building if (arr[index] > arr[index - 1]): count += 1 else: maximum = max(maximum, count) count = 0 return max(maximum, count) # Driver code arr = [ 1, 2, 3, 4 ] len = len(arr) print(find_consecutive_steps(arr, len)) # This code is contributed by Gitanjali. |
C#
// C# code to find maximum // number of consecutive steps. using System; class GFG { // Function to count consecutive steps static int find_consecutive_steps(int []arr, int len) { int count = 0; int maximum = 0; for (int index = 1; index < len; index++) { // count the number of consecutive // increasing height building if (arr[index] > arr[index - 1]) count++; else { maximum = Math.Max(maximum, count); count = 0; } } return Math.Max(maximum, count); } // Driver code public static void Main () { int []arr = { 1, 2, 3, 4 }; int len = arr.Length; Console.WriteLine(find_consecutive_steps(arr, len)); } } // This code is contributed by vt_m. |
PHP
<?php // PHP code to find maximum // number of consecutive steps. // Function to count // consecutive steps function find_consecutive_steps($arr, $len) { $count = 0; $maximum = 0; for ($index = 1; $index < $len; $index++) { // count the number of consecutive // increasing height building if ($arr[$index] > $arr[$index - 1]) $count++; else { $maximum = max($maximum, $count); $count = 0; } } return max($maximum, $count); } // Driver code $arr = array( 1, 2, 3, 4 ); $len = count($arr); echo find_consecutive_steps($arr, $len); // This code is contributed by vt_m. ?> |
Output :
3
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