Difference between sums of odd and even digits
Given a long integer, we need to find if the difference between sum of odd digits and sum of even digits is 0 or not. The indexes start from zero (0 index is for leftmost digit).
Examples:
Input : 1212112 Output : Yes Explanation:- the odd position element is 2+2+1=5 the even position element is 1+1+1+2=5 the difference is 5-5=0.so print yes. Input :12345 Output : No Explanation:- the odd position element is 1+3+5=9 the even position element is 2+4=6 the difference is 9-6=3 not equal to zero. So print no.
Method 1: One by one traverse digits and find the two sums. If difference between two sums is 0, print yes, else no.
C++
// C++ program for above approach #include<bits/stdc++.h> using namespace std; bool isDiff0( int n){ int first = 0; int second = 0; bool flag = true ; while (n > 0){ int digit = n % 10; if (flag) first += digit; else second += digit; flag = !flag; n = n/10; } if (first - second == 0) return true ; return false ; } int main(){ int n = 1243; if (isDiff0(n)) cout<< "Yes" ; else cout<< "No" ; return 0; } // This code is contributed by Kirti Agarwal(kirtiagarwal23121999) |
Java
// Java equivalent of above code public class Main { public static boolean isDiff0( int n) { int first = 0 ; int second = 0 ; boolean flag = true ; while (n > 0 ) { int digit = n % 10 ; if (flag) first += digit; else second += digit; flag = !flag; n = n / 10 ; } if (first - second == 0 ) return true ; return false ; } public static void main(String[] args) { int n = 1243 ; if (isDiff0(n)) System.out.println( "Yes" ); else System.out.println( "No" ); } } |
Python
# Python program for the above approach def isDiff0(n): first = 0 second = 0 flag = True while (n > 0 ): digit = n % 10 if (flag): first + = digit else : second + = digit if (flag): flag = False else : flag = True n = int (n / 10 ) if (first - second = = 0 ): return True return False # driver code n = 1243 if (isDiff0(n)): print ( "Yes" ) else : print ( "No" ) |
C#
// C# Program for the above approach using System; public class BinaryTree{ static bool isDiff0( int n){ int first = 0; int second = 0; bool flag = true ; while (n > 0){ int digit = n % 10; if (flag) first += digit; else second += digit; flag = !flag; n = n/10; } if (first - second == 0) return true ; return false ; } public static void Main(){ int n = 1243; if (isDiff0(n)) Console.Write( "Yes" ); else Console.Write( "No" ); } } |
Javascript
// JavaScript prgraom for the above approach function isDiff0(n){ let first = 0; let second = 0; let flag = true ; while (n > 0){ let digit = n % 10; if (flag) first += digit; else second += digit; flag = !flag; n = parseInt(n/10); } if (first - second == 0) return true ; return false ; } let n = 1243; if (isDiff0(n)) console.log( "Yes" ); else console.log( "No" ); // THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGAWRAL2852002) |
Yes
Time Complexity:
The given approach traverses the digits of the input integer once. Hence the time complexity of this approach is O(log n), where n is the value of the input integer.
Space Complexity: O(1)
Method 2 : This can be easily solved using divisibility of 11. This condition is only satisfied if the number is divisible by 11. So check the number is divisible by 11 or not.
CPP
// C++ program to check if difference between sum of // odd digits and sum of even digits is 0 or not #include <bits/stdc++.h> using namespace std; bool isDiff0( long long int n) { return (n % 11 == 0); } int main() { long int n = 1243 if (isDiff0(n)) cout << "Yes" ; else cout << "No" ; return 0; } |
Java
// Java program to check if difference between sum of // odd digits and sum of even digits is 0 or not import java.io.*; import java.util.*; class GFG { public static boolean isDiff( int n) { return (n % 11 == 0 ); } public static void main (String[] args) { int n = 1243 ; if (isDiff(n)) System.out.print( "Yes" ); else System.out.print( "No" ); } } |
Python
# Python program to check if difference between sum of # odd digits and sum of even digits is 0 or not def isDiff(n): return (n % 11 = = 0 ) # Driver code n = 1243 ; if (isDiff(n)): print ( "Yes" ) else : print ( "No" ) # Mohit Gupta_OMG <0_o> |
C#
// C# program to check if difference // between sum of odd digits and sum // of even digits is 0 or not using System; class GFG { public static bool isDiff( int n) { return (n % 11 == 0); } public static void Main () { int n = 1243; if (isDiff(n)) Console.WriteLine( "Yes" ); else Console.WriteLine( "No" ); } } // This code is contributed by vt_m. |
PHP
<?php // PHP program to check if // difference between sum of // odd digits and sum of // even digits is 0 or not function isDiff0( $n ) { return ( $n % 11 == 0); } // Driver Code $n = 1243; if (isDiff0( $n )) echo "Yes" ; else echo "No" ; // This code is contributed by nitin mittal ?> |
Javascript
<script> // Javascript program to check if difference between sum of // odd digits and sum of even digits is 0 or not function isDiff0(n) { return (n % 11 == 0); } let n = 1243 if (isDiff0(n)) document.write( "Yes" ); else document.write( "No" ); // This code is contributed by Mayank Tyagi </script> |
Output:
Yes
Time Complexity: O(1)
Auxiliary Space: O(1)
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