TCS NQT Set 2 | Day 1 2020 | odd and even difference

TCS NQT Set 2 | Day 1 2020 | odd and even difference

TCS NQT 2020 | Day 1 | Set 2

Given a maximum of 100 digit numbers as input, find the difference between the sum of odd
and even position digits.

Logic:

  • Get the input as a string because we need up to 100 digits for eg 4567.
  • Store added value of the odd position in variable a and even position in variable b
  • Then convert the char value to integer value using ASCII value.
  • Check the odd position the value is 4 added into variable a=4
  • Check the next position which is the even position the value is 5 added into variable b=5
  • Check the next position which is the odd position the value is 6 added into variable a=10
  • Check the next position which is the even position the value is 7 added into variable b=12
  • By using the abs() function we can find the difference of a and b to eliminate the negative value
  • And then print the difference value a-b is 2

TCS NQT Set 2:

Program:

[code lang=”c”]

#include<stdio.h>
void main()
{
int a=0,b=0,i=0,n;
char num[100];
printf("Enter the number:");
scanf("%s",&num); //get the input up to 100 digit
n=strlen(num);
while(n>0)
{
if(i==0) //add even digits when no of digit is even and vise versa
{
a+=num[n-1]-48;
n–;
i=1;
}
else //add odd digits when no of digit is even and vise versa
{
b+=num[n-1]-48;
n–;
i=0;
}
}
printf("%d",abs(a-b)); //print the difference of odd and even
}

[/code]

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Vignesh

A Computer Science graduate who likes to make things simpler. When he’s not working, you can find him surfing the web, learning facts, tricks and life hacks. He also enjoys movies in his leisure time.

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