Given a number N in the form of a string, the task is to check whether the given number is balanced or not.
Balanced Number: A number is said to be balanced if the sum of digits in the first half of it is equal to the sum of the digits in the second half.
Note: All Palindromic numbers are balanced numbers.
Examples:
Input: N = 19091
Output: Balanced
Explanation:
middle element is 0
Sum of left half = 1 + 9 = 10
Sum of right half = 9 + 1 = 10
Hence, the given number is a Balanced number.Input: N = 133423
Output: Not Balanced
Explanation:
Sum of left half = 1 + 3 + 3 (7)
Sum of right half = 4 + 2 + 3 (9)
Hence, the given number is not Balanced
Approach:
Iterate over half the length of the number from the beginning. Calculate the sum of digits of the first half and the second half simultaneously by adding s[i] and s[number of digits – 1 – i] to leftSum and rightSum respectively. Finally, check if the leftSum and rightSum are equal or not.
Below is the implementation of the above approach.
C++
// C++ program to check// if a number is// Balanced or not#include <bits/stdc++.h>using namespace std;// Function to check whether N is// Balanced Number or notvoid BalancedNumber(string s){ int Leftsum = 0; int Rightsum = 0; // Calculating the Leftsum // and rightSum simultaneously for (int i = 0; i < s.size() / 2; i++) { // Typecasting each character // to integer and adding the // digit to respective sums Leftsum += int(s[i] - '0'); Rightsum += int(s[s.size() - 1 - i] - '0'); } if (Leftsum == Rightsum) cout << "Balanced" << endl; else cout << "Not Balanced" << endl;}// Driver Codeint main(){ string s = "12321"; // Function call BalancedNumber(s); return 0;} |
Java
// Java program to check if a number // is Balanced or notimport java.io.*;class GFG{ // Function to check whether N is// Balanced Number or notprivate static void BalancedNumber(String s){ int Leftsum = 0; int Rightsum = 0; // Calculating the Leftsum // and rightSum simultaneously for(int i = 0; i < s.length() / 2; i++) { // Typecasting each character // to integer and adding the // digit to respective sums Leftsum += (int)(s.charAt(i) - '0'); Rightsum += (int)(s.charAt( s.length() - 1 - i) - '0'); } if (Leftsum == Rightsum) System.out.println("Balanced"); else System.out.println("Not Balanced");}// Driver Codepublic static void main (String[] args) { String s = "12321"; // Function call BalancedNumber(s);}}// This code is contributed by jithin |
Python3
# Python3 program to check # if a number is # Balanced or not # Function to check whether N is # Balanced Number or not def BalancedNumber(s): Leftsum = 0 Rightsum = 0 # Calculating the Leftsum # and rightSum simultaneously for i in range(0, int(len(s) / 2)): # Typecasting each character # to integer and adding the # digit to respective sums Leftsum = Leftsum + int(s[i]) Rightsum = (Rightsum + int(s[len(s) - 1 - i])) if (Leftsum == Rightsum): print("Balanced", end = '\n') else: print("Not Balanced", end = '\n')# Driver Code s = "12321"# Function call BalancedNumber(s)# This code is contributed by PratikBasu |
C#
// C# program to check // if a number is // Balanced or not using System;class GFG{ // Function to check whether N is // Balanced Number or not static void BalancedNumber(string s) { int Leftsum = 0; int Rightsum = 0; // Calculating the Leftsum // and rightSum simultaneously for (int i = 0; i < s.Length / 2; i++) { // Typecasting each character // to integer and adding the // digit to respective sums Leftsum += (int)(Char.GetNumericValue(s[i]) - Char.GetNumericValue('0')); Rightsum += (int)(Char.GetNumericValue(s[s.Length - 1 - i]) - Char.GetNumericValue('0')); } if (Leftsum == Rightsum) Console.WriteLine("Balanced"); else Console.WriteLine("Not Balanced"); } // Driver codestatic void Main() { string s = "12321"; // Function call BalancedNumber(s); }}// This code is contributed by divyeshrabadiya07 |
Javascript
<script>// JavaScript program to check if a number// is Balanced or not// Function to check whether N is// Balanced Number or notfunction BalancedNumber(s){ let Leftsum = 0; let Rightsum = 0; // Calculating the Leftsum // and rightSum simultaneously for(let i = 0; i < s.length / 2; i++) { // Typecasting each character // to integer and adding the // digit to respective sums Leftsum += (s[i] - '0'); Rightsum += (s[ s.length - 1 - i] - '0'); } if (Leftsum == Rightsum) document.write("Balanced"); else document.write("Not Balanced");} // Driver Code let s = "12321"; // Function call BalancedNumber(s); </script> |
Output
Balanced
Time Complexity: O(len), where len is the length of the string.
Auxiliary Space: O(1)
Approach 2 (Two Pointers) : The idea is to use 2 pointers from the starting index and the ending index and will check whether the number is palindromic in nature or not. To make it simple will convert the number to a string first and apply the described algorithm.
Below is the code for the implementation :
C++
// C++ program to check// if a number is// Balanced or not#include <bits/stdc++.h>using namespace std;// Function to check whether N is// Balanced Number or notvoid BalancedNumber(int n){ // Convert the number to string string s = to_string(n); // Set both the start and end pointers int si = 0, ei = s.size() - 1; // Keep a flag variable bool flag = true; while (si < ei) { // if both side characters are same // update the pointers if (s[si] == s[ei]) { si++; ei--; } // if violates update flag and break else { flag = false; break; } } // If no violation occurred if (flag) cout << "Balanced"; // In case of violation else cout << "Not Balanced";}// Driver Codeint main(){ int n = 1235321; // Function call BalancedNumber(n); return 0;}// This code is contributed by Rajdeep Mallick(rajdeep999) |
Java
// Java program for the above approach public class BalancedNumber { public static void isBalancedNumber(int n) { // Convert the number to string String s = Integer.toString(n); // Set both the start and end pointers int si = 0; int ei = s.length() - 1; // Keep a flag variable boolean flag = true; while (si < ei) { // if both side characters are same // update the pointers if (s.charAt(si) == s.charAt(ei)) { si++; ei--; } // if violates update flag and break else { flag = false; break; } } // If no violation occurred if (flag) { System.out.println("Balanced"); } // In case of violation else { System.out.println("Not Balanced"); } } public static void main(String[] args) { int n = 1235321; isBalancedNumber(n); }}// This code is contributed adityashatmfh |
Python3
# Python program to check # if a number is balanced # or notdef is_balanced_number(n): # Convert the number to string s = str(n) # Set both the start and end pointers si = 0 ei = len(s) - 1 # Keep a flag variable flag = True while si < ei: # if both side characters are same # update the pointers if s[si] == s[ei]: si += 1 ei -= 1 # if violates update flag and break else: flag = False break # If no violation occurred if flag: print("Balanced") # In case of violation else: print("Not Balanced")# Driver coden = 1235321# Function callis_balanced_number(n)# This code is contributed by princecp152 |
C#
// C# program for the above approach using System;public class Program{ // Function to check whether N is // Balanced Number or not static void BalancedNumber(int n) { // Convert the number to string string s = n.ToString(); // Set both the start and end pointers int si = 0, ei = s.Length - 1; // Keep a flag variable bool flag = true; while (si < ei) { // if both side characters are same // update the pointers if (s[si] == s[ei]) { si++; ei--; } // if violates update flag and break else { flag = false; break; } } // If no violation occurred if (flag) Console.WriteLine("Balanced"); // In case of violation else Console.WriteLine("Not Balanced"); } // Driver Code public static void Main() { int n = 1235321; // Function call BalancedNumber(n); }} |
Javascript
// Javascript program to check if a number is balanced or notfunction is_balanced_number(n) { // Convert the number to string let s = n.toString(); // Set both the start and end pointers let si = 0; let ei = s.length - 1; // Keep a flag variable let flag = true; while (si < ei) { // if both side characters are same // update the pointers if (s[si] == s[ei]) { si++; ei--; } // if violates update flag and break else { flag = false; break; } } // If no violation occurred if (flag) { console.log("Balanced"); } // In case of violation else { console.log("Not Balanced"); }}// Driver codelet n = 1235321;// Function callis_balanced_number(n); |
Output
Balanced
Time Complexity: O(len), where len is the length of the string.
Auxiliary Space: O(1)
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