Check whether occurrence of individual digit in a number is same or not

Question

We need to check whether the occurrence of an individual digit in a number is same or not.For e.g. for 2244 (2 occur 2 times and 4 occur 2 times).Therefore, occurrence of both digits are same.

//This will return true if occurrence of individual digit in //a number is same else false bool stable(int no) { vector<int> v1; int k , count = 1; int arr[10]; //Initializing all arr[] -> 0 for(int k = 0 ; k < 10 ;k++) { arr[k] = 0; } while(no != 0) { k=no%10; arr[k]++; no=no/10; } for(int i = 0 ; i < 10 ; i++) { if(arr[i] != 0) { v1.push_back(arr[i]); //storing count of individual digits } } vector<int>::iterator it , it2; for(it = v1.begin()+1 ,it2 = v1.begin(); it != v1.end() ; it++,it2++) { if(*it == *it2) //if all the values are same return true else false { count++; } } if(count == v1.size()) return true; return false; }

But this code doesn't work for no like 2222,1111,444. Also, would you suggest some good way to optimize the code?

Your code returns true for 2222, as 2222 is stable (2 is the only digit with occurrence 4), your code seems right. — Rajeev Singh
– Rajeev Singh, Commented May 9, 2017 at 7:06
Shubh, can you explain how the code does not work for those numbers? Have you done any debugging? — halfer
– halfer, Commented May 9, 2017 at 10:10
Thanx @RajeevSingh , Actually doing this on my machine I had set count = 0 that's y I was getting an error. — Shubh Tripathi
– Shubh Tripathi, Commented May 9, 2017 at 14:56

WhozCraig · Accepted Answer · 2017-05-09 20:08:58Z

I think you're making this harder than it needs to be (or I'm severely misunderstanding the question, which happens often).

The assumption is the requirements are as specified: Given a non-zero positive value, a number is qualified if all digits appear with equal frequency, including 1 (ex: 1122, 2222, 1234 all qualify, as no digit has higher frequency than any other).

The algorithm is simple:

Quick return for any value less than 10; a single digit number is immediately qualified.
Build a counter array of modulus residuals.
Find the first non-zero value in the array
From that point on, find the first non-zero value that does NOT match the value from (4). If you reach the end of the sequence without finding such a difference, all digits in the original number must have the same count.

In all the complexity is logarithmic base-10 to the input number plus a single-pass scan of the constant-sized array (10 elements).

Example Code

#include <algorithm> #include <iterator> bool stable(unsigned value) { if (value < 10) // single digit only, including zero return true; unsigned ar[10]={0}; do { ++ar[value%10]; } while (value /= 10); auto it = std::find_if(std::begin(ar), std::end(ar), [](auto n) { return n != 0; }); return std::find_if(std::next(it), std::end(ar), [it](auto n){ return n && (n != *it);}) == std::end(ar); }

You could always further this by retaining the maximum digit count and leaving without the find operations if it ultimately was 1 (ex: 1234, 102938 are such examples). I'll leave that as an exercise for you to benchmark to determine if there is any performance benefit. I honestly doubt there will be.

Kostas · Accepted Answer · 2017-05-09 06:24:41Z

Try this:

bool stable(int no) { std::vector<int> v1; while (no != 0){ v1.push_back(no%10); no /= 10; } std::sort(v1.begin(), v1.end()); //SORTING DIGITS IN ASCENDING ORDER std::vector<int> index = {0}; //BUILDING VEC WITH INDEXES WHERE CHANGES HAPPEN for (unsigned int i = 0; i < v1.size()-1; ++i){ if (v1[i] != v1[i+1]) index.push_back(i+1); } //EDGE CASE WHEN ONLY 1 DIGIT APPEARS (e.g. 555) if (index.size() == 1) return true; //CHECKING THAT ALL INDEXES ARE EQUALLY SEPARATED int diff = index[1] - index[0]; for (unsigned int i = 1; i < index.size()-1; ++i){ if (index[i+1] - index[i] != diff) return false; } return true; }

Rajeev Singh · Accepted Answer · 2017-05-09 06:32:42Z

While checking if all the repeated counts are same, you can directly return false if counts do not match, no need to check further. If the vector contains only a single count for numbers like 2222, 1111 it will return true.

vector<int>::iterator it , it2; for(it = v1.begin()+1 ,it2 = v1.begin(); it != v1.end() ; it++,it2++) { if(*it != *it2) //if two values are not same return false { return false; } } return true;

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