I have tried this problem but I'm getting errors.. Is there anyone who can help me in solving this one ??

Question

Create a function that takes an array of numbers and returns "Hello World!" if the digit 5 appears in the array. Otherwise, return "there is no 5 in the array".

Array of numbers is

12 65 4 67 0987 234 87 89 6745 567 987 476 708 36 5587 798 884 671 7889

We have to check single digit of array element and compare it with 5 if 5 then hello world else no 5

12 so element is 1 and 2
You will get this 2 elements by mod operator 12%10 so you will get 2 and and divide the number by 12/10 you will get 1
And compare 1 and 2 with 5 you get the desired output

a=n%10
n=n/10
If(a==5)
Hello world

void fun(int arr, int n){
bool flag = 0;
for(int i = 0; i < n; i++){
int cnt = 0;
while(arr[i] > 0){
arr[i] = arr[i]/10;
cnt++;
}
if(cnt == 5){
flag = 1;
}
}
if(flag){
cout<<"Hello world";
}
else{
cout<<"there is no five";
}
}

Can anyone help me in solving this problem in c or c++??

Answer 1

Thnks everyone for ur help...

Answer 2

I modified something in that code

void fun(int arr, int n){
bool flag = 0;
int i = 0,a;

while(i<n){
while(arr[i]>0){
a = arr[i]%10;
if(a == 5){
flag = 1;
break;
}
arr[i]=arr[i]/10;
}
if(flag )
break ;
i++;
}
if(flag){
cout<<"Hello world";
}
else{
cout<<"there is no five";
}
}

Hope this works

Answer 3

#include<iostream>
#include<string>
using namespace std;
void fun(int arr[], int n)
{
int flag = 0;
int i = 0,a;
while(i<n)
{
while(arr[i]>0)
{
a = arr[i]%10;
if(a == 5)
{
flag = 1;
break;
}
arr[i]=arr[i]/10;
}
if(flag ) break ;
i++;
}
if(flag)
{
cout<<"Hello World!";
}
else
{
cout<<"there is no five in the array";
}

}
int main(){
int n;
cin>>n;
int arr[n];
for(int i=0;i<n;i++)
cin>>arr[i];
fun(arr,n);
return 0;

}

Answer 4

Thnks

Answer 5

#include<iostream>
#include<string>
std::string fun(int arr, int n)
{
bool flag = 0;
int i = 0,a;
while(i<n)
{
while(arr[i]>0)
{
a = arr[i]%10;
if(a == 5)
{
flag = 1;
break;
}
arr[i]=arr[i]/10;
}
if(flag ) break ;
i++;
}
if(flag)
{
return"Hello world!";
}
else
{
return"there is no five in the array";
}
}

Still it's showing compilation error
Anyone plz give me the full solution so that I can check where I'm going wrong...

Answer 6

char* foo(int* nums, int len) {
for (int i=0,j;i<len;i++) {
for (j=nums[i];j;j/=10) {
if (j % 10 == 5) {
return "Hello World!";
}
}
}
return "there is no five in the array";
}

// Hope this helps

Answer 7

#include<iostream>
#include<string>
using namespace std;
void fun(int arr[], int n)
{
int flag = 0;
int i = 0,a;
while(i<n)
{
while(arr[i]>0)
{
a = arr[i]%10;
if(a == 5)
{
flag = 1;
break;
}
arr[i]=arr[i]/10;
}
if(flag ) break ;
i++;
}
if(flag)
{
cout<<"Hello World!";
}
else
{
cout<<"there is no five in the array";
}

}
int main(){
int n;
cin>>n;
int arr[n];
for(int i=0;i<n;i++)
cin>>arr[i];
fun(arr,n);
return 0;

}

This code executed successfully but it's showing wrong ans....why is it so??

Maybe because in the qs it's asking to return the expression and in this coding we r printing the expression....Maybe this could be the reason... I'm not sure....

Answer 8

Arun Ruban SJ
Create a function that takes an array of numbers and returns "Hello World!" if the digit 5 appears in the array. Otherwise, return "there is no 5 in the array".

Array of numbers is

12 65 4 67 0987 234 87 89 6745 567 987 476 708 36 5587 798 884 671 7889

Answer 9

Sandeep Kushwaha
Ok thnks

Answer 10

https://code.sololearn.com/clbejkMR7T2l/?ref=app

Answer 11

What you did here is just counting digits in a number, for example 15 is made of two digits: 1 and 5, so 2 digits and cnt=2 in this example.
What you have to do is check if one of these digits is equal 5, so in your while loop you should put something like that:
cnt=arr[i]%10;
if(cnt==5){
cout<<"Hello World!";
return 0;
}
arr[i]/=10;

Answer 12

Thnks

Answer 13

Show your exact question then maybe we can find it

Answer 14

In turbo c++ pluss output comes hello why can u explain this

main()
{
for(;0;)
{
printf("hello");
}
}
Output in TC++ ??

Answer 15

Hi