🌻|| C++|| Easy Approach || PROPER EXPLANATION
ALGORITHM\n\nStep - 1 Initialize  a string s with n number of \'a\' to make it as a smallest possible lexiographic.\nStep - 2 Count of \'a = 1\' , so if i have 
### ALGORITHM\n```\nStep - 1 Initialize  a string s with n number of \'a\' to make it as a smallest possible lexiographic.\nStep - 2 Count of \'a = 1\' , so if i have n number of \'a\' it means it sum is also \'n\'.\nStep - 3 Reduce the value of k by n (i.e k=k-n);\nStep - 4 Start traversing the string from the end because if we append any character from starting then there is chances of not getting smallest possible lexiographic order. \n\t\t Ex if we have to append \'z\' then from starting it would be like zaaa and from end it would be like aaaz which clearly shows that aaaz<zaaa.\nStep - 5 If \'k\' becomes \'0\' stop the traversal because that would be our answer.\nStep - 6 Else if k < 25, it means for getting the sum just replace ith character by \'a\'+k. And update k by k=0.\nStep - 7 Else k > 25 so iterate more and replace ith character by (\'a\'+25) and move ahead. And update k by k-25.\n\nLets take an example.\n\tI/P   n=3 k=27\n\tNow follow step by step.\n\tstep-1  s=aaa  (initialize a string s with n number of a)\n\tstep-2  k=k-n=27-3=24\n\tstep-3 start iterating from right most side.\n\tstep 4 check k, here k  < 25, therefore update ith place of string by \'a\'+k i.e \'y\' because ASCII value of \'a\' is a 97 and 97+24 is 121 i.e \'y\'. Update k by 0. As we get our sum 27.\n\tStep 5 check k which is zero. Therefore stop iteration and return the string as it was our answer.\n```\n### CODE\n\n```\nclass Solution {\npublic:\n    string getSmallestString(int n, int k) {\n        string s="";\n        for(int i=0;i<n;i++)\n            s+=\'a\';\n        k=k-n;\n        for(int i=n-1;i>=0;i--)\n        {\n            if(k==0)\n                break;\n            else if(k<25)\n            {\n                s[i] = (char)(\'a\'+k);          //using (char) to convert ASCII to respective character\n                k=0;\n            }\n            else{\n                s[i] = (char)(\'a\'+25);\n                k = k - 25;\n            }\n        }\n        return s;\n    }\n};\n```
['C']