## Global Scholar Interview Question for Developer Program Engineers

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1
of 1 vote

Hope this code works.

int main()
{
int n,a[4]={0}, i=3, b[4]={0}, flag=1;
scanf("%d",&n);
while(n>0)
{
a[i--]=n%10;
n=n/10;

}
for(i=0;i<4;i++)
b[a[i]]++;
for(i=0;i<4;i++)
if(a[i]!=b[i])
{
flag=0;
break;
}
if(flag)
printf("given number has same property");
else printf("given number doesn't possess the property");

}
}}}

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0

vamathi is right i hope

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0
of 0 vote

it seems a little confusing, can u plz give another example

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0
of 0 vote

I think counting the frequency of each 'number' in given number .Representing all missing numbers by '0' at their respective index and others with their frequencies(number of times they have occurred in given number)..

Please correct me if I am missing somethingg

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0
of 0 vote

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0

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0

interviewer told me to use only number

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0
of 0 vote

public static boolean numberCheck(int num){
String numStr = ""+num; //先把这个数字包装到字符串中，以方便用每个字符查找
if(numStr.length()>10){
return false; //显然如果这个数是十一位数或者更大，根本不能满足
}
//char[] numChars = numStr.toCharArray();
for(int i =0;i<numStr.length();i++){
int times = Integer.parseInt(numStr.substring(i,i+1)); //i在num出现的次数
int count =0;
for(int j = 0;j<numStr.length();j++){
if((String.valueOf(i).equals(String.valueOf(numStr.charAt(j))))){
++count;
}
}
if(times!=count){
return false;
}
}
return true;
}

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0
of 0 vote

int indexproperty_check(int data)
{
int data1=data;
int a[9]={0};
int digit_size=0;
while(data1)
{
a[data1%10]++;
data1=data1/10;
digit_size++;
}
for(int i=0;i<digit_size;i++)
{
data1 = a[i]+data1*10;
}
if(data1==data)
{
cout<<"Index property is satisfied";
return 1;
}
else
{
cout<<"Index property is not satisfied";
return 0;
}
}

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0
of 0 vote

``````public static boolean isStructure(int input){
int divider=1;
int digits=0;
int max=Integer.MIN_VALUE;
while(input/divider>0){

if((input/divider)%10>max)
max=(input/divider)%10;
divider*=10;
digits++;
}
if(max>=digits)
return false;
int a[]= new int[digits];
int inputcopy=input;
while(inputcopy>0){
a[inputcopy%10]++;
inputcopy/=10;
}

int result=0;
int i=0;
while(--digits>=0){
result+=a[i++]*Math.pow(10,digits);
}
return result==input;
}

public static void main(String args[]) {
for(int i=0; i<10000;i++){
if(Numbers.isStructure(i))
System.out.println(i);
}
}``````

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0
of 0 vote

``````#include <iostream>
#include <conio.h>
using namespace std;
#include <math.h>
#include <vector>

int _tmain(int argc, _TCHAR* argv[])
{
int n;
cin >> n;

/* Find Number of Digits in the Input Number */
int NumOfDig = ceil(log10(n));

/* Vector to maintain the Value of the Input Digit in the Index */
vector<int> arr(NumOfDig);

/* Vector to maintain Number of Occurences of a value (0-9) in the Input Number */
vector<int> NumOfOccurences(10);

int counter = NumOfDig - 1;
while(counter >= 0)
{
int rem = n % 10;
arr[counter] = rem;
NumOfOccurences[rem] += 1;
n /= 10;
counter--;
}

vector<int>::iterator it;
int index;
for(index = 0, it = arr.begin(); it != arr.end(); it++, index++)
{
if((*it) != NumOfOccurences[index])
{
cout << endl << "NO";
_getch();
return 0;
}
}
cout << endl << "YES";
_getch();
return 0;
}``````

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