Given an integer U denoting the amount of KWh units of electricity consumed, the task is to calculate the electricity bill with the help of the below charges:
- 1 to 100 units –
- 100 to 200 units –
- 200 to 300 units –
- above 300 units –
Examples:
Input: U = 250
Output: 3500
Explanation:
Charge for the first 100 units – 10*100 = 1000
Charge for the 100 to 200 units – 15*100 = 1500
Charge for the 200 to 250 units – 20*50 = 1000
Total Electricity Bill = 1000 + 1500 + 1000 = 3500
Input: U = 95
Output: 950
Explanation:
Charge for the first 100 units – 10*95 = 950
Total Electricity Bill = 950
Approach 1: The idea is to identify the charge bar in which it falls and then calculate the bill according to the charges mentioned above. Below is the illustration of the steps:
- Check units consumed is less than equal to the 100, If yes then the total electricity bill will be:
- Else if, check that units consumed is less than equal to the 200, if yes then total electricity bill will be:
- Else if, check that units consumed is less than equal to the 300, if yes then total electricity bill will be:
- Else if, check that units consumed greater than 300, if yes then total electricity bill will be:
Below is the implementation of the above approach:
C++
// C++ implementation to calculate the// electricity bill#include<bits/stdc++.h>using namespace std;// Function to calculate the// electricity billint calculateBill(int units){ // Condition to find the charges // bar in which the units consumed // is fall if (units <= 100) { return units * 10; } else if (units <= 200) { return (100 * 10) + (units - 100) * 15; } else if (units <= 300) { return (100 * 10) + (100 * 15) + (units - 200) * 20; } else if (units > 300) { return (100 * 10) + (100 * 15) + (100 * 20) + (units - 300) * 25; } return 0;}// Driver Codeint main(){ int units = 250; cout << calculateBill(units);}// This code is contributed by spp____ |
Java
// Java implementation to calculate the// electricity billimport java.util.*;class ComputeElectricityBill { // Function to calculate the // electricity bill public static int calculateBill(int units) { // Condition to find the charges // bar in which the units consumed // is fall if (units <= 100) { return units * 10; } else if (units <= 200) { return (100 * 10) + (units - 100) * 15; } else if (units <= 300) { return (100 * 10) + (100 * 15) + (units - 200) * 20; } else if (units > 300) { return (100 * 10) + (100 * 15) + (100 * 20) + (units - 300) * 25; } return 0; } // Driver Code public static void main(String args[]) { int units = 250; System.out.println( calculateBill(units)); }} |
Python3
# Python3 implementation to calculate the# electricity bill# Function to calculate the# electricity billdef calculateBill(units): # Condition to find the charges # bar in which the units consumed # is fall if (units <= 100): return units * 10; elif (units <= 200): return ((100 * 10) + (units - 100) * 15); elif (units <= 300): return ((100 * 10) + (100 * 15) + (units - 200) * 20); elif (units > 300): return ((100 * 10) + (100 * 15) + (100 * 20) + (units - 300) * 25); return 0;# Driver Codeunits = 250;print(calculateBill(units));# This code is contributed by Code_Mech |
C#
// C# implementation to calculate the// electricity billusing System;class ComputeElectricityBill{// Function to calculate the// electricity billpublic static int calculateBill(int units){ // Condition to find the charges // bar in which the units consumed // is fall if (units <= 100) { return units * 10; } else if (units <= 200) { return (100 * 10) + (units - 100) * 15; } else if (units <= 300) { return (100 * 10) + (100 * 15) + (units - 200) * 20; } else if (units > 300) { return (100 * 10) + (100 * 15) + (100 * 20) + (units - 300) * 25; } return 0;}// Driver Codepublic static void Main(String []args){ int units = 250; Console.WriteLine(calculateBill(units));}}// This code is contributed by spp____ |
Javascript
<script>// Javascript implementation to calculate the// electricity bill// Function to calculate the// electricity billfunction calculateBill(units){ // Condition to find the charges // bar in which the units consumed // is fall if (units <= 100) { return units * 10; } else if (units <= 200) { return (100 * 10) + (units - 100) * 15; } else if (units <= 300) { return (100 * 10) + (100 * 15) + (units - 200) * 20; } else if (units > 300) { return (100 * 10) + (100 * 15) + (100 * 20) + (units - 300) * 25; } return 0;} // Driver Codevar units = 250;document.write(calculateBill(units));// This code is contributed by Khushboogoyal499</script> |
3500
Time Complexity: O(1)
Auxiliary Space: O(1)
Approach 2 : In this approach, we can use an array to store the different rate of charges and their respective range of units. This approach can make the code more readable and easier to maintain. Here’s how the code would look like:
C++
#include <bits/stdc++.h>using namespace std;const int n = 4;// Function to calculate the electricity billint calculateBill(int units){ int charges[n] = { 10, 15, 20, 25 }; int range[n] = { 100, 100, 100, INT_MAX }; int bill = 0; for (int i = 0; i < n; i++) { if (units <= range[i]) { bill += charges[i] * units; break; } else { bill += charges[i] * range[i]; units -= range[i]; } } return bill;}// Driver codeint main(){ int units = 250; cout << calculateBill(units); return 0;} |
Java
import java.util.*;public class Main { public static final int n = 4; // Function to calculate the electricity bill public static int calculateBill(int units) { int[] charges = {10, 15, 20, 25}; int[] range = {100, 100, 100, Integer.MAX_VALUE}; int bill = 0; for (int i = 0; i < n; i++) { if (units <= range[i]) { bill += charges[i] * units; break; } else { bill += charges[i] * range[i]; units -= range[i]; } } return bill; } // Driver code public static void main(String[] args) { int units = 250; System.out.println(calculateBill(units)); }}// This code is contributed by divyansh2212 |
Python3
import sysMAX_INT = sys.maxsizen = 4# Function to calculate the electricity billdef calculateBill(units): charges = [10, 15, 20, 25] # changed variable name from 'range' to 'range_' range_ = [100, 100, 100, MAX_INT] bill = 0 for i in range(n): if units <= range_[i]: bill += charges[i] * units break else: bill += charges[i] * range_[i] units -= range_[i] return bill# Driver codeunits = 250print(calculateBill(units))# This code is contributed by shivhack999 |
C#
using System;public class Program { const int n = 4; // Function to calculate the electricity bill static int CalculateBill(int units) { int[] charges = { 10, 15, 20, 25 }; int[] range = { 100, 100, 100, int.MaxValue }; int bill = 0; for (int i = 0; i < n; i++) { if (units <= range[i]) { bill += charges[i] * units; break; } else { bill += charges[i] * range[i]; units -= range[i]; } } return bill; } // Driver code public static void Main() { int units = 250; Console.WriteLine(CalculateBill(units)); }}// This code is contributed by sarojmcy2e |
Javascript
const n = 4;// Function to calculate the electricity billfunction calculateBill(units) { const charges = [10, 15, 20, 25]; const range = [100, 100, 100, Number.MAX_VALUE]; let bill = 0; for (let i = 0; i < n; i++) { if (units <= range[i]) { bill += charges[i] * units; break; } else { bill += charges[i] * range[i]; units -= range[i]; } } return bill;}// Driver codeconst units = 250;console.log(calculateBill(units));// This code is contributed by shiv1o43g |
3500
Time Complexity : The time complexity of the calculateBill function is O(n), where n is the number of ranges of units and their respective charges. This is because the function uses a for loop to iterate through the range and charges arrays, and for each iteration, it performs a constant amount of work (calculating the bill based on the units consumed).
Since n is a constant value, the time complexity can be considered as O(1) in the best-case scenario. The function takes a constant amount of time to run, regardless of the number of units consumed.
Auxiliary Space : The space complexity of this code is O(n), where n is the number of rate of charges. This is because the program uses two arrays, charges and range, both of which have a size of n elements. The arrays take up 2 * n * sizeof(int) bytes of memory. In this case, n = 4, so the total memory occupied by the arrays is 2 * 4 * sizeof(int).
