pre>
code>
};
}
return ans;
}
ans += ret;
long long ret = getCnt(cnt, n);
}
cnt[c-'0']++;
for(char c : vals){
int cnt[10] = {0};
for(auto vals : s){
long long ans = 0;
makePalindrome(n, k, 0, 0);
long long countGoodIntegers(int n, int k) {
}
return ret;
}
n -= cnt[i];
}
ret *= combi(n, cnt[i]);
} else{
ret *= combi(n-1, cnt[i]);
if(i == 0){
if(cnt[i] == 0) continue;
for(int i = 0 ; i < 10; i++){
long long ret = 1;
long long getCnt(int cnt[10], int n){
}
return ret;
}<= i;
ret code>
for(int i = 1; i <= r; i++){
}
n--;
rr--;
ret *= n;
while(rr){
int rr = r;
long long ret = 1;
long long combi(int n, int r){
}
}
makePalindrome(n, k, curNum*10 + i, depth + 1);
if(depth == 0 && i == 0) continue;
for(int i = 0; i <= 9; i++){
}
return;
}
push(curNum);
if((curNum%k) == 0){
}<= 10;
tmp code>
curNum += tmp %10;
curNum *= 10;
while(tmp){
}<= 10;
tmp code>
if((n%2) == 1){
long long tmp = curNum;<2){
if(depth == (n+1)code>
void makePalindrome(int n, int k, long long curNum, int depth){
}
s.insert(str);
}
str += to_string(tmp[i]);
for(int i = 0 ; i < tmp.size(); i++){
string str = "";
sort(tmp.begin(), tmp.end(), [](long long a, long long b) -> long long {return a > b; });
}<= 10;
nn code>
tmp.push_back(nn%10);
while(nn){
long long nn = n;
vector tmp;
void push(long long n){
set s;
public:
class Solution {
Solution Code:
The time complexity is O(10^n * n^2 * log(n)) due to the sorting and combination operations.
Then it counts the number of good integers that can be formed by rearranging their digits using the combination formula and a helper function getCnt()
The solution first generates all possible k-palindromic integers with n digits by recursively generating all possible combinations of digits and then sorting them in descending order
Explanation
O(10^n * n^2 * log(n))
Complexity
The approach used in the solution is to generate all possible k-palindromic integers with n digits, and then count the number of good integers that can be formed by rearranging their digits.
Approach
This problem is about counting the number of good integers that can be rearranged to form a k-palindromic integer.
Summary<" target="blank">Find the Count of Good Integers
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