how to find nth permutation of a string in c

how to find nth permutation of a string in c.write a c program to find nth permutation of a string. For example if given string is “1256” then all permutation of string “1256” would be

1 2 5 6 
1 2 6 5 
1 5 2 6 
1 5 6 2 
1 6 2 5 
1 6 5 2 
2 1 5 6 
2 1 6 5 
2 5 1 6 
2 5 6 1 
2 6 1 5 
2 6 5 1 
5 1 2 6 
5 1 6 2 
5 2 1 6 
5 2 6 1 
5 6 1 2 
5 6 2 1 
6 1 2 5 
6 1 5 2 
6 2 1 5 
6 2 5 1 
6 5 1 2 
6 5 2 1

So if nth is 8 then nth permutation of string “1256” will be 2 1 6 5 ( i.e 7th for 0 base index ).

Let us write c program to find all permutation of given string and later on we would modify this program to find nth permutation of string.

#include<stdio.h>
#include<string.h>

int arr[4] = { 1, 2, 5, 6 };
// helper array
int parr[4];

int visited[4];

void perm(int k, int size, int nthp) {
	//base case
	if (k == size) {
		for (int i = 0; i < size ; i++)
			printf("%d", parr[i]);
		printf("\n");
		return;
	}
	for (int i = 0; i < size; i++) {
		if (!visited[i]) {
			visited[i] = 1;
			parr[k] = arr[i];
			perm(k + 1, size, nthp);
			visited[i] = 0; // backtrack
		}
	}
}

int main() {
	//reset visited
	memset(visited, 0, 4);
	int nthp = 8;
	perm(0, 4, nthp);
	return 0;
}

The above program will print all permutation of “1256”.

how to find nth permutation of a string in c

From above output followings are the observation

• The length of given string “125”  is 3 and its total permutation count is 6 i.e factorial 3.
• In this way we can modify the base case of above program to find nth permutation. See below code.

#include<stdio.h>
#include<string.h>

int arr[4] = { 1, 2, 5, 6 };
// helper array
int parr[4];

int visited[4];
int kth = 0;
void perm(int k, int size, int nthp) {
	//base case
	if (k == size) {
		kth++;
		for (int i = 0; i < size && nthp == kth; i++)
			printf("%d", parr[i]);
		return;
	}
	for (int i = 0; i < size; i++) {
		if (!visited[i]) {
			visited[i] = 1;
			parr[k] = arr[i];
			perm(k + 1, size, nthp);
			visited[i] = 0; // backtrack
		}
	}
}

int main() {
	//reset visited
	memset(visited, 0, 4);
	int nthp = 8;
	perm(0, 4, nthp);
	return 0;
}

Output : 2165

optimization opportunity

The above program to find nth permutation is computing all permutation of a given string. can you think what are some optimizations require for faster execution or about reducing complexity.

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