Print all permutations in sorted (lexicographic) order | GeeksforGeeks

Given a string, print all permutations of it in sorted order.

1. Sort the given string in non-decreasing order and print it. The first permutation is always the string sorted in non-decreasing order.

2. Start generating next higher permutation. Do it until next higher permutation is not possible. If we reach a permutation where all characters are sorted in non-increasing order, then that permutation is the last permutation.

Steps to generate the next higher permutation:
1. Take the previously printed permutation and find the rightmost character in it, which is smaller than its next character. Let us call this character as ‘first character’.

2. Now find the ceiling of the ‘first character’. Ceiling is the smallest character on right of ‘first character’, which is greater than ‘first character’. Let us call the ceil character as ‘second character’.

3. Swap the two characters found in above 2 steps.

4. Sort the substring (in non-decreasing order) after the original index of ‘first character’.

Time Complexity: O(n^2 x n!)

int findCeil (char str[], char first, int l, int h)

{

    // initialize index of ceiling element

    int ceilIndex = l;

    // Now iterate through rest of the elements and find

    // the smallest character greater than 'first'

    for (int i = l+1; i <= h; i++)

      if (str[i] > first && str[i] < str[ceilIndex])

            ceilIndex = i;

    return ceilIndex;

}

// Print all permutations of str in sorted order

void sortedPermutations ( char str[] )

{

    // Get size of string

    int size = strlen(str);

    // Sort the string in increasing order

    qsort( str, size, sizeof( str[0] ), compare );

    // Print permutations one by one

    bool isFinished = false;

    while ( ! isFinished )

    {

        // print this permutation

        printf ("%s \n", str);

        // Find the rightmost character which is smaller than its next

        // character. Let us call it 'first char'

        int i;

        for ( i = size - 2; i >= 0; --i )

           if (str[i] < str[i+1])

              break;

        // If there is no such chracter, all are sorted in decreasing order,

        // means we just printed the last permutation and we are done.

        if ( i == -1 )

            isFinished = true;

        else

        {

            // Find the ceil of 'first char' in right of first character.

            // Ceil of a character is the smallest character greater than it

            int ceilIndex = findCeil( str, str[i], i + 1, size - 1 );

            // Swap first and second characters

            swap( &str[i], &str[ceilIndex] );

            // Sort the string on right of 'first char'

            qsort( str + i + 1, size - i - 1, sizeof(str[0]), compare );

        }

    }

}

We can optimize step 4 of the above algorithm for finding next permutation. Instead of sorting the subarray after the ‘first character’, we can reverse the subarray, because the subarray we get after swapping is always sorted in non-increasing order. This optimization makes the time complexity as O(n x n!).

void sortedPermutations ( char str[] )

{

    // Get size of string

    int size = strlen(str);

    // Sort the string in increasing order

    qsort( str, size, sizeof( str[0] ), compare );

    // Print permutations one by one

    bool isFinished = false;

    while ( ! isFinished )

    {

        // print this permutation

        printf ("%s \n", str);

        // Find the rightmost character which is smaller than its next

        // character. Let us call it 'first char'

        int i;

        for ( i = size - 2; i >= 0; --i )

           if (str[i] < str[i+1])

              break;

        // If there is no such chracter, all are sorted in decreasing order,

        // means we just printed the last permutation and we are done.

        if ( i == -1 )

            isFinished = true;

        else

        {

            // Find the ceil of 'first char' in right of first character.

            // Ceil of a character is the smallest character greater than it

            int ceilIndex = findCeil( str, str[i], i + 1, size - 1 );

            // Swap first and second characters

            swap( &str[i], &str[ceilIndex] );

            // reverse the string on right of 'first char'

            reverse( str, i + 1, size - 1 );

        }

    }

}

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