Given a large positive integer represented as a string str. The task is to round this number to the nearest multiple of 10.
Examples:
Input: str = “99999999999999993”
Output: 99999999999999990Input: str = “99999999999999996”
Output: 100000000000000000
Approach: A solution to the same problem has been discussed in this article which will not work for large numbers. When the number is large and represented as strings we can process the number digit by digit. The main observation is that if the last digit of the number is ? 5 then only the last digit will get affected i.e. it will be replaced with a 0. If it is something greater than 5 then the number has to be rounded to some next higher multiple of 10 i.e. the last digit will be replaced with a 0 and 1 will have to be added to the rest of the number i.e. the number represented by the sub-string str[0…n-1] which can be done by storing carry generated at every step (digit).
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to round the given number// to the nearest multiple of 10void roundToNearest(string str, int n){ // If string is empty if (str == "") return; // If the last digit is less than or equal to 5 // then it can be rounded to the nearest // (previous) multiple of 10 by just replacing // the last digit with 0 if (str[n - 1] - '0' <= 5) { // Set the last digit to 0 str[n - 1] = '0'; // Print the updated number cout << str.substr(0, n); } // The number hast to be rounded to // the next multiple of 10 else { // To store the carry int carry = 0; // Replace the last digit with 0 str[n - 1] = '0'; // Starting from the second last digit, add 1 // to digits while there is carry int i = n - 2; carry = 1; // While there are digits to consider // and there is carry to add while (i >= 0 && carry == 1) { // Get the current digit int currentDigit = str[i] - '0'; // Add the carry currentDigit += carry; // If the digit exceeds 9 then // the carry will be generated if (currentDigit > 9) { carry = 1; currentDigit = 0; } // Else there will be no carry else carry = 0; // Update the current digit str[i] = (char)(currentDigit + '0'); // Get to the previous digit i--; } // If the carry is still 1 then it must be // inserted at the beginning of the string if (carry == 1) cout << carry; // Print the rest of the number cout << str.substr(0, n); }}// Driver codeint main(){ string str = "99999999999999993"; int n = str.length(); roundToNearest(str, n); return 0;} |
Java
// Java implementation of the approach import java.io.*;class GFG {// Function to round the given number// to the nearest multiple of 10static void roundToNearest(StringBuilder str, int n) { // If string is empty if (str.toString() == "") return; // If the last digit is less than or equal to 5 // then it can be rounded to the nearest // (previous) multiple of 10 by just replacing // the last digit with 0 if (str.charAt(n - 1) - '0' <= 5) { // Set the last digit to 0 str.setCharAt(n - 1, '0'); // Print the updated number System.out.print(str.substring(0, n)); } // The number hast to be rounded to // the next multiple of 10 else { // To store the carry int carry = 0; // Replace the last digit with 0 str.setCharAt(n - 1, '0'); // Starting from the second last digit, // add 1 to digits while there is carry int i = n - 2; carry = 1; // While there are digits to consider // and there is carry to add while (i >= 0 && carry == 1) { // Get the current digit int currentDigit = str.charAt(i) - '0'; // Add the carry currentDigit += carry; // If the digit exceeds 9 then // the carry will be generated if (currentDigit > 9) { carry = 1; currentDigit = 0; } // Else there will be no carry else carry = 0; // Update the current digit str.setCharAt(i, (char)(currentDigit + '0')); // Get to the previous digit i--; } // If the carry is still 1 then it must be // inserted at the beginning of the string if (carry == 1) System.out.print(carry); // Print the rest of the number System.out.print(str.substring(0, n)); }}// Driver codepublic static void main(String[] args) { StringBuilder str = new StringBuilder("99999999999999993"); int n = str.length(); roundToNearest(str, n);}}// This code is contributed by// sanjeev2552 |
Python3
# Python 3 implementation of the approach# Function to round the given number# to the nearest multiple of 10def roundToNearest(str, n): # If string is empty if (str == ""): return # If the last digit is less than or equal to 5 # then it can be rounded to the nearest # (previous) multiple of 10 by just replacing # the last digit with 0 if (ord(str[n - 1]) - ord('0') <= 5): # Set the last digit to 0 str = list(str) str[n - 1] = '0' str = ''.join(str) # Print the updated number print(str[0:n]) # The number hast to be rounded to # the next multiple of 10 else: # To store the carry carry = 0 # Replace the last digit with 0 str = list(str) str[n - 1] = '0' str = ''.join(str) # Starting from the second last digit, # add 1 to digits while there is carry i = n - 2 carry = 1 # While there are digits to consider # and there is carry to add while (i >= 0 and carry == 1): # Get the current digit currentDigit = ord(str[i]) - ord('0') # Add the carry currentDigit += carry # If the digit exceeds 9 then # the carry will be generated if (currentDigit > 9): carry = 1 currentDigit = 0 # Else there will be no carry else: carry = 0 # Update the current digit str[i] = chr(currentDigit + '0') # Get to the previous digit i -= 1 # If the carry is still 1 then it must be # inserted at the beginning of the string if (carry == 1): print(carry) # Print the rest of the number print(str[0:n]) # Driver codeif __name__ == '__main__': str = "99999999999999993" n = len(str) roundToNearest(str, n)# This code is contributed by# Surendra_Gangwar |
C#
// C# implementation of the approach using System;using System.Text;class GFG { // Function to round the given number // to the nearest multiple of 10 static void roundToNearest(StringBuilder str, int n) { // If string is empty if (str.ToString() == "") return; // If the last digit is less than or equal to 5 // then it can be rounded to the nearest // (previous) multiple of 10 by just replacing // the last digit with 0 if (str[n - 1] - '0' <= 5) { // Set the last digit to 0 str[n - 1] = '0'; // Print the updated number Console.Write(str.ToString().Substring(0, n)); } // The number hast to be rounded to // the next multiple of 10 else { // To store the carry int carry = 0; // Replace the last digit with 0 str[n - 1] = '0'; // Starting from the second last digit, // add 1 to digits while there is carry int i = n - 2; carry = 1; // While there are digits to consider // and there is carry to add while (i >= 0 && carry == 1) { // Get the current digit int currentDigit = str[i] - '0'; // Add the carry currentDigit += carry; // If the digit exceeds 9 then // the carry will be generated if (currentDigit > 9) { carry = 1; currentDigit = 0; } // Else there will be no carry else carry = 0; // Update the current digit str[i] = (char)(currentDigit + '0'); // Get to the previous digit i--; } // If the carry is still 1 then it must be // inserted at the beginning of the string if (carry == 1) Console.Write(carry); // Print the rest of the number Console.Write(str.ToString().Substring(0, n)); } } // Driver code public static void Main(String[] args) { StringBuilder str = new StringBuilder("99999999999999993"); int n = str.Length; roundToNearest(str, n); } } // This code is contributed by // Rajnis09 |
Javascript
// JS implementation of the approach // Function to round the given number // to the nearest multiple of 10 function roundToNearest(str, n) { str = Array.from(str) // If string is empty if ((str.join("")).localeCompare("") == 0) return; // If the last digit is less than or equal to 5 // then it can be rounded to the nearest // (previous) multiple of 10 by just replacing // the last digit with 0 if (parseInt(str[n - 1]) <= 5) { // Set the last digit to 0 str[n - 1] = '0'; // Print the updated number process.stdout.write(str.join("")); } // The number hast to be rounded to // the next multiple of 10 else { // To store the carry let carry = 0; // Replace the last digit with 0 str[n - 1] = '0'; // Starting from the second last digit, // add 1 to digits while there is carry let i = n - 2; carry = 1; // While there are digits to consider // and there is carry to add while (i >= 0 && carry == 1) { // Get the current digit let currentDigit = parseInt(str[i]); // Add the carry currentDigit += carry; // If the digit exceeds 9 then // the carry will be generated if (currentDigit > 9) { carry = 1; currentDigit = 0; } // Else there will be no carry else carry = 0; // Update the current digit str[i] = String(currentDigit); // Get to the previous digit i--; } // If the carry is still 1 then it must be // inserted at the beginning of the string if (carry == 1) process.stdout.write(1); // Print the rest of the number process.stdout.write(str.join("")); } } // Driver code let str = "99999999999999993"; let n = str.length; roundToNearest(str, n); // This code is contributed by // phasing17 |
Output:
99999999999999990
Time Complexity: O(n), where n is the length of the given string.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
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