Given two numbers N and M. Construct maximal number by permuting (changing order) the digits of N, not exceeding M.
Note : It is allowed to leave N as it is.
Examples:
Input : N = 123, M = 222
Output : 213
There are total 3! permutations possible for N = 123, But the only permutation that satisfies the given condition is 213. Similarly, In example 2, there are total 4! permutations possible for N = 3921, But the only permutation that satisfies the given condition is 9321.Input : N = 3921, M = 10000
Output : 9321
Approach : Let’s construct the answer digit by digit starting from the leftmost. We are asked to build lexicographically maximal answer. So in this order, we should choose the greatest digit on each step. The approach is to iterate over all possible digits starting from the greatest. For each digit check if it’s possible to put it in this position and compare the resulting number with number M. If it comes less than or equal to the value of M, then proceed to the next digit.
Below is the CPP implementation:
// CPP program to Maximize the given number. #include <bits/stdc++.h> using namespace std; // Function to maximize the number N with // limit as M. string maximizeNumber(string N, int M) { // Sorting the digits of the // number in increasing order. sort(N.begin(), N.end()); for (int i = 0; i < N.size(); i++) for (int j = i + 1; j < N.size(); j++) { // Copying the string into another // temp string. string t = N; // Swaping the j-th char(digit) // with i-th char(digit) swap(t[j], t[i]); // Sorting the temp string // from i-th pos to end. sort(t.begin() + i + 1, t.end()); // Checking if the string t is // greater than string N and less // than or equal to the number M. if (stoll(t) > stoll(N) && stoll(t) <= M) // If yes then, we will permanently // swap the i-th char(or digit) // with j-th char(digit). swap(N[i], N[j]); } // Returns the maximized number. return N; } // Driver function int main() { string N = "123"; int M = 222; cout << maximizeNumber(N, M) << endl; return 0; } |
Output:
213
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