Code In Ur Hand

  Sum of Digit is Pallindrome or not Given a number N.Find if the digit sum(or sum of digits) of N is a Palindrome number or not. Note:A Palindrome number is a number which stays the same when reversed.Example- 121,131,7 etc. Example 1: Input: N=56 Output: 1 Explanation: The digit sum of 56 is 5+6=11. Since, 11 is a palindrome number.Thus, answer is 1. Example 2: Input: N=98 Output: 0 Explanation: The digit sum of 98 is 9+8=17. Since 17 is not a palindrome,thus, answer is 0. Your Task: You don't need to read input or print aything.Your Task is to complete the function isDigitSumPalindrome() which takes a number N as input parameter and returns 1 if the Digit sum of N is a palindrome.Otherwise it returns 0. Expected Time Complexi

Closest Number Basic Accuracy: 11.21% Submissions: 1615 Points: 1 Given non-zero two integers N and M . The problem is to find the number closest to N and divisible by M . If there are more than one such number, then output the one having maximum absolute value .   Example 1: Input: N = 13 , M = 4 Output: 12 Explanation: 12 is the Closest Number to 13 which is divisible by 4. Example 2: Input: N = -15 , M = 6 Output: -18 Explanation: -12 and -18 are both similarly close to -15 and divisible by 6. but -18 has the maximum absolute value. So, Output is -18   Your Task: You don't need to read input or print anything.

              => Problem Condition =>Solution #include<iostream> using namespace std; int common(int a, int b) {     if (b == 0)         return a;     return common(b, a % b); } int find(int input1[],int input2){ int maxValue;  //by default zero in function for(int i = 0;i <input2; ++i)     {        if(maxValue < input1[i])            maxValue = input1[i];     }     cout << "Largest percentage of Zoombie = " << maxValue<<endl<<endl; int mycount=0; int index[100]; for(int i=0; i<input2; ++i){   if(maxValue==input1[i]){         index[mycount]=i;     cout<<"Find at city no. "<<i<<endl;     mycount++;   } } for(int i=0;i<mycount;++i){     cout<<"index["<<i<<"] : "<<index[i]<<endl; } for(int i=0; i<mycount; ++i){  if(index[i]!=input2-1){      if(common(input1[index[i]],input1[index[i]+1])==1){         cout<<input1[index[i]]<<" co-prime w