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 equal to zero.
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.
Approach:
One by one traverse digits and find the two sums. If difference between two sums is 0, print yes, else no.
Below is the implementation of the above approach:
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 codepublic 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 approachdef 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 coden = 1243if(isDiff0(n)): print("Yes")else: print("No") |
C#
// C# Program for the above approachusing 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 approachfunction 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: O(log n),
Auxiliary space: O(1)
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