Difference between the summation of numbers whose frequency of all digits are same and different

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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; ` `} `

Output:

```2790
```

My Personal Notes arrow_drop_up

Computer Science student during day Full stack developer at night

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.