Given an array arr[] of size N representing the size of candles which reduce by 1 unit each day. The room is illuminated using the given N candles. Find the maximum number of days the room is without darkness.
Examples:
Input: N = 3, arr[] = {1, 1, 2}
Output: 2
Explanation: The candles’ length reduces by 1 in 1 day. So, at the end of day 1: Sizes would be 0 0 1, So, at the end of day 2: Sizes would be 0 0 0. This means the room was illuminated for 2 days.Input: N = 5, arr[] = {2, 3, 4, 2, 1}
Output: 4
Approach: An efficient approach to solving the problem is to find the maximum element from the array using a single loop.
Follow the steps to solve the problem:
- Initialize max element as arr[0].
- Iterate over the array and check if arr[i] > max.
- Then update max with arr[i].
- Return max.
Below is the implementation for the above approach:
C++
// C++ code for above approach:#include <bits/stdc++.h>using namespace std;long maxDays(vector<long>& arr, int n){ long max = arr[0]; for (int i = 0; i < n; i++) { if (arr[i] > max) { max = arr[i]; } } return max;}// Driver's Codeint main(){ int n = 3; vector<long> arr = { 1, 1, 2 }; cout << "The maximum number of days the room is " "without darkness are:- " << // function call maxDays(arr, n) << endl; return 0;} |
C
// C code for above idea#include <stdio.h>long maxDays(long arr[], int n) { long max = arr[0]; for (int i = 0; i < n; i++) { if (arr[i] > max) { max = arr[i]; } } return max;}int main() { int n = 3; long arr[] = {1, 1, 2}; printf("The maximum number of days the room is without darkness are:- %ld\n", maxDays(arr, n)); return 0;} |
Java
// Java code for above ideaimport java.io.*;import java.util.*;class GFG { public static void main (String[] args) { int n = 3; long arr[] = {1,1,2}; System.out.println("The maximum number of days the room is without darkness are:- "+maxDays(arr,n)); } static long maxDays(long arr[], int n){ long max = arr[0]; for (int i = 0; i < n; i++) { if(arr[i] > max) { max = arr[i]; } } return max; }} |
Python3
# python code for above ideadef maxDays(arr, n): max = arr[0] for i in range(n): if arr[i] > max: max = arr[i] return maxn = 3arr = [1, 1, 2]print("The maximum number of days the room is without darkness are:- " + str(maxDays(arr, n))) |
C#
using System;using System.Collections.Generic;class GFG{ static long MaxDays(List<long> arr, int n) { long max = arr[0]; for (int i = 0; i < n; i++) { if (arr[i] > max) { max = arr[i]; } } return max; } static void Main(string[] args) { int n = 3; List<long> arr = new List<long>() { 1, 1, 2 }; Console.WriteLine("The maximum number of days the room is without darkness are: " + MaxDays(arr, n)); }} |
Javascript
// JavaScript code for above ideafunction maxDays(arr, n) { let max = arr[0]; for (let i = 0; i < n; i++) { if (arr[i] > max) { max = arr[i]; } } return max;}let n = 3;let arr = [1, 1, 2];console.log("The maximum number of days the room is without darkness are:- " + maxDays(arr, n));// This code is contributed by Tapesh(tapeshdua420) |
The maximum number of days the room is without darkness are:- 2
Time Complexity: O(N), Single loop for traversing the array.
Auxiliary Space: O(1)
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