Rearrange a string so that all same characters become d distance away

3.7

Given a string and a positive integer d. Some characters may be repeated in the given string. Rearrange characters of the given string such that the same characters become d distance away from each other. Note that there can be many possible rearrangements, the output should be one of the possible rearrangements. If no such arrangement is possible, that should also be reported.
Expected time complexity is O(n) where n is length of input string.

Examples:
Input:  "abb", d = 2
Output: "bab"

Input:  "aacbbc", d = 3
Output: "abcabc"

Input: "geeksforgeeks", d = 3
Output: egkegkesfesor

Input:  "aaa",  d = 2
Output: Cannot be rearranged

Hint: Alphabet size may be assumed as constant (256) and extra space may be used.

Solution: The idea is to count frequencies of all characters and consider the most frequent character first and place all occurrences of it as close as possible. After the most frequent character is placed, repeat the same process for remaining characters.

1) Let the given string be str and size of string be n

2) Traverse str, store all characters and their frequencies in a Max Heap MH. The value of frequency decides the order in MH, i.e., the most frequent character is at the root of MH.

3) Make all characters of str as ‘\0’.

4) Do following while MH is not empty.
a) Extract the Most frequent character. Let the extracted character be x and its frequency be f.
b) Find the first available position in str, i.e., find the first ‘\0’ in str.
c) Let the first position be p. Fill x at p, p+d,.. p+(f-1)d

Following are C++ and Python implementations of above algorithm.

C/C++

// C++ program to rearrange a string so that all same 
// characters become at least d distance away
#include <iostream>
#include <cstring>
#include <cstdlib>
#define MAX 256
using namespace std;

// A structure to store a character 'c' and its frequency 'f'
// in input string
struct charFreq {
    char c;
    int f;
};

// A utility function to swap two charFreq items.
void swap(charFreq *x, charFreq *y) {
    charFreq z = *x;
    *x = *y;
    *y = z;
}

// A utility function to maxheapify the node freq[i] of a heap 
// stored in freq[]
void maxHeapify(charFreq freq[], int i, int heap_size)
{
    int l = i*2 + 1;
    int r = i*2 + 2;
    int largest = i;
    if (l < heap_size && freq[l].f > freq[i].f)
        largest = l;
    if (r < heap_size && freq[r].f > freq[largest].f)
        largest = r;
    if (largest != i)
    {
        swap(&freq[i], &freq[largest]);
        maxHeapify(freq, largest, heap_size);
    }
}

// A utility function to convert the array freq[] to a max heap
void buildHeap(charFreq freq[], int n)
{
    int i = (n - 1)/2;
    while (i >= 0)
    {
        maxHeapify(freq, i, n);
        i--;
    }
}

// A utility function to remove the max item or root from max heap
charFreq extractMax(charFreq freq[], int heap_size)
{
    charFreq root = freq[0];
    if (heap_size > 1)
    {
        freq[0] = freq[heap_size-1];
        maxHeapify(freq, 0, heap_size-1);
    }
    return root;
}

// The main function that rearranges input string 'str' such that
// two same characters become d distance away 
void rearrange(char str[], int d)
{
    // Find length of input string
    int n = strlen(str);

    // Create an array to store all characters and their
    // frequencies in str[]
    charFreq freq[MAX] = {{0, 0}};

    int m = 0; // To store count of distinct characters in str[]

    // Traverse the input string and store frequencies of all
    // characters in freq[] array.
    for (int i = 0; i < n; i++)
    {        
        char x = str[i];

        // If this character has occurred first time, increment m
        if (freq[x].c == 0)
            freq[x].c = x, m++;

        (freq[x].f)++;
        str[i] = '\0';  // This change is used later 
    }

    // Build a max heap of all characters
    buildHeap(freq, MAX);

    // Now one by one extract all distinct characters from max heap
    // and put them back in str[] with the d distance constraint
    for (int i = 0; i < m; i++)
    {
        charFreq x = extractMax(freq, MAX-i);

        // Find the first available position in str[]
        int p = i;
        while (str[p] != '\0')
            p++;

        // Fill x.c at p, p+d, p+2d, .. p+(f-1)d
        for (int k = 0; k < x.f; k++)
        {
            // If the index goes beyond size, then string cannot
            // be rearranged.
            if (p + d*k >= n)
            {
                cout << "Cannot be rearranged";
                exit(0);
            }
            str[p + d*k] = x.c;
        }
    }
}

// Driver program to test above functions
int main()
{
    char str[] = "aabbcc";
    rearrange(str, 3);
    cout << str;
}

Python

// Python program to rearrange a string so that all same 
// characters become at least d distance away
MAX = 256

# A structure to store a character 'c' and its frequency 'f'
# in input string
class charFreq(object):
    def __init__(self,c,f):
        self.c = c
        self.f = f

# A utility function to swap two charFreq items.
def swap(x, y):
    return y, x

# A utility function
def toList(string):
    t = []
    for x in string:
        t.append(x)

    return t

# A utility function
def toString(l):
    return ''.join(l)

# A utility function to maxheapify the node freq[i] of a heap
# stored in freq[]
def maxHeapify(freq, i, heap_size):
    l = i*2 + 1
    r = i*2 + 2
    largest = i
    if l < heap_size and freq[l].f > freq[i].f:
        largest = l
    if r < heap_size and freq[r].f > freq[largest].f:
        largest = r
    if largest != i:
        freq[i], freq[largest] = swap(freq[i], freq[largest])
        maxHeapify(freq, largest, heap_size)

# A utility function to convert the array freq[] to a max heap
def buildHeap(freq, n):
    i = (n - 1)/2
    while i >= 0:
        maxHeapify(freq, i, n)
        i-=1

# A utility function to remove the max item or root from max heap
def extractMax(freq, heap_size):
    root = freq[0]
    if heap_size > 1:
        freq[0] = freq[heap_size-1]
        maxHeapify(freq, 0, heap_size-1)

    return root

# The main function that rearranges input string 'str' such that
# two same characters become d distance away
def rearrange(string, d):
    # Find length of input string
    n = len(string)

    # Create an array to store all characters and their
    # frequencies in str[]
    freq = []
    for x in xrange(MAX):
        freq.append(charFreq(0,0))

    m = 0

    # Traverse the input string and store frequencies of all
    # characters in freq[] array.
    for i in xrange(n):
        x = ord(string[i])

        # If this character has occurred first time, increment m
        if freq[x].c == 0:
            freq[x].c = chr(x)
            m+=1

        freq[x].f+=1
        string[i] = '\0'

    # Build a max heap of all characters
    buildHeap(freq, MAX)

    # Now one by one extract all distinct characters from max heap
    # and put them back in str[] with the d distance constraint
    for i in xrange(m):
        x = extractMax(freq, MAX-i)

        # Find the first available position in str[]
        p = i
        while string[p] != '\0':
            p+=1

        # Fill x.c at p, p+d, p+2d, .. p+(f-1)d
        for k in xrange(x.f):

            # If the index goes beyond size, then string cannot
            # be rearranged.
            if p + d*k >= n:
                print "Cannot be rearranged"
                return

            string[p + d*k] = x.c

    return toString(string)

# Driver program
string = "aabbcc"
print rearrange(toList(string), 3)

# This code is contributed by BHAVYA JAIN


Output:
abcabc

Algorithmic Paradigm: Greedy Algorithm

Time Complexity: Time complexity of above implementation is O(n + mLog(MAX)). Here n is the length of str, m is count of distinct characters in str[] and MAX is maximum possible different characters. MAX is typically 256 (a constant) and m is smaller than MAX. So the time complexity can be considered as O(n).

More Analysis:
The above code can be optimized to store only m characters in heap, we have kept it this way to keep the code simple. So the time complexity can be improved to O(n + mLogm). It doesn’t much matter through as MAX is a constant.

Also, the above algorithm can be implemented using a O(mLogm) sorting algorithm. The first steps of above algorithm remain same. Instead of building a heap, we can sort the freq[] array in non-increasing order of frequencies and then consider all characters one by one from sorted array.

We will soon be covering an extended version where same characters should be moved at least d distance away.

This article is contributed by Himanshu Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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