Given an array of N integers, find the difference between the summation of numbers whose frequency of all digits are same and different. For e.g. 8844, 1001, 56, 77, 34764673 are the examples of numbers where all digits have the same frequency. Similarly, 545, 44199, 76672, 202 are the examples of numbers whose frequency of digits are not the same.
Examples:
Input: a[] = {24, 787, 2442, 101, 1212}
Output: 2790
(2442 + 24 + 1212) – (787 + 101) = 2790
Input: a[]= {12321, 786786, 110022, 47, 22895}
Output: 861639
Approach: Traverse for every element in the array. Keep a count of all digits in a map. Once all digits are traversed in the element, check if the map contains the same frequency for all digits. If it contains the same frequency, then add it to same else add it to diff. After the array has been traversed completely, return the difference of both the elements.
Below is the implementation of the above approach:
// C++ Difference between the // summation of numbers // in which the frequency of // all digits are same and different #include <bits/stdc++.h> using namespace std;
// Function that returns the difference int difference( int a[], int n)
{ // Stores the sum of same
// and different frequency digits
int same = 0;
int diff = 0;
// traverse in the array
for ( int i = 0; i < n; i++) {
// duplicate of array element
int num = a[i];
unordered_map< int , int > mp;
// traverse for every digit
while (num) {
mp[num % 10]++;
num = num / 10;
}
// iterator pointing to the
// first element in the array
auto it = mp.begin();
// count of the smallest digit
int freqdigit = (*it).second;
int flag = 0;
// check if all digits have same frequency or not
for ( auto it = mp.begin(); it != mp.end(); it++) {
if ((*it).second != freqdigit) {
flag = 1;
break ;
}
}
// add to diff if not same
if (flag)
diff += a[i];
else
same += a[i];
}
return same - diff;
} // Driver Code int main()
{ int a[] = { 24, 787, 2442, 101, 1212 };
int n = sizeof (a) / sizeof (a[0]);
cout << difference(a, n);
return 0;
} |
// Java Difference between the // summation of numbers // in which the frequency of // all digits are same and different import java.util.*;
class GFG
{ // Function that returns the difference static int difference( int a[], int n)
{ // Stores the sum of same
// and different frequency digits
int same = 0 ;
int diff = 0 ;
// traverse in the array
for ( int i = 0 ; i < n; i++)
{
// duplicate of array element
int num = a[i];
HashMap<Integer,Integer> mp = new HashMap<Integer,Integer>();
// traverse for every digit
while (num > 0 )
{
if (mp.containsKey(num % 10 ))
mp.put(num % 10 , (mp.get(num % 10 ) + 1 ));
else
mp.put(num % 10 , 1 );
num = num / 10 ;
}
// iterator pointing to the
// first element in the array
Iterator<Map.Entry<Integer, Integer>> it = mp.entrySet().iterator();
// count of the smallest digit
int freqdigit = it.next().getValue();
int flag = 0 ;
// check if all digits have same frequency or not
for (Map.Entry<Integer,Integer> its : mp.entrySet())
{
if (its.getValue() != freqdigit)
{
flag = 1 ;
break ;
}
}
// add to diff if not same
if (flag == 1 )
diff += a[i];
else
same += a[i];
}
return same - diff;
} // Driver Code public static void main(String[] args)
{ int a[] = { 24 , 787 , 2442 , 101 , 1212 };
int n = a.length;
System.out.print(difference(a, n));
} } // This code is contributed by PrinciRaj1992 |
# Python3 Difference between the # summation of numbers # in which the frequency of # all digits are same and different # Function that returns the difference def difference(a, n):
# Stores the sum of same
# and different frequency digits
same = 0
diff = 0
# traverse in the array
for i in range (n):
# duplicate of array element
num = a[i]
mp = {}
# traverse for every digit
while (num):
if num % 10 not in mp:
mp[num % 10 ] = 0
mp[num % 10 ] + = 1
num = num / / 10
# iterator pointing to the
# first element in the array
it = list (mp.keys())
# count of the smallest digit
freqdigit = mp[it[ 0 ]]
flag = 0
# check if all digits have same frequency or not
for it in mp:
if mp[it] ! = freqdigit:
flag = 1
break
# add to diff if not same
if (flag):
diff + = a[i]
else :
same + = a[i]
return same - diff
# Driver Code a = [ 24 , 787 , 2442 , 101 , 1212 ]
n = len (a)
print (difference(a, n))
# This code is contributed by SHUBHAMSINGH10 |
// C# implementation of the above Java code using System;
using System.Collections.Generic;
class GFG {
// Function that returns the difference
static int Difference( int [] a, int n)
{
// Stores the sum of same
// and different frequency digits
int same = 0;
int diff = 0;
// traverse in the array
for ( int i = 0; i < n; i++) {
// duplicate of array element
int num = a[i];
Dictionary< int , int > mp
= new Dictionary< int , int >();
// traverse for every digit
while (num > 0) {
if (mp.ContainsKey(num % 10))
mp[num % 10] = mp[num % 10] + 1;
else
mp.Add(num % 10, 1);
num = num / 10;
}
// iterator pointing to the
// first element in the array
IEnumerator<KeyValuePair< int , int > > it
= mp.GetEnumerator();
it.MoveNext();
// count of the smallest digit
int freqdigit = it.Current.Value;
int flag = 0;
// check if all digits have same frequency or
// not
while (it.MoveNext()) {
if (it.Current.Value != freqdigit) {
flag = 1;
break ;
}
}
// add to diff if not same
if (flag == 1)
diff += a[i];
else
same += a[i];
}
return same - diff;
}
// Driver Code
public static void Main( string [] args)
{
int [] a = { 24, 787, 2442, 101, 1212 };
int n = a.Length;
Console.WriteLine(Difference(a, n));
}
} // This code is contributed by phasing17 |
// Javascript Difference between the // summation of numbers // in which the frequency of // all digits are same and different // Function that returns the difference function difference(a, n)
{ // Stores the sum of same
// and different frequency digits
let same = 0;
let diff = 0;
// traverse in the array
for (let i = 0; i < n; i++) {
// duplicate of array element
let num = a[i];
let mp = {};
// traverse for every digit
while (num) {
if (mp[num%10]){
mp[num%10]++;
}
else {
mp[num%10] = 1;
}
num = Math.floor(num / 10);
}
// count of the smallest digit
let freqdigit = Object.values(mp)[0];;
let flag = 0;
// check if all digits have same frequency or not
for (let it in mp) {
if (mp[it] != freqdigit){
flag = 1;
break ;
}
}
// add to diff if not same
if (flag)
diff += a[i];
else
same += a[i];
}
return same - diff;
} // Driver Code let a = [ 24, 787, 2442, 101, 1212 ]; let n = a.length; console.log(difference(a, n)); // The code is contributed by Nidhi goel. |
2790