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

Difficulty Level : Hard
Last Updated : 19 Nov, 2020

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 the length of the 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

The approach to solve this problem 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 the remaining characters.

Let the given string be str and size of string be n
Traverse str, store all characters and their frequencies in a Max Heap MH(implemented using priority queue). The value of frequency decides the order in MH, i.e., the most frequent character is at the root of MH.
Make all characters of str as ‘\0’.
Do the following while MH is not empty.
- Extract the Most frequent character. Let the extracted character be x and its frequency be f.
- Find the first available position in str, i.e., find the first ‘\0’ in str.
- Let the first position be p. Fill x at p, p+d,.. p+(f-1)d

Below is the implementation of the above algorithm.

C

// C program to rearrange a string so that all same
// characters become at least d distance away
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define MAX 256
  
// A structure to store a character 'c' and its frequency
// 'f' in input string
typedef struct charFreq {
    char c;
    int f;
} charFreq ;
  
// 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) {
                printf("Cannot be rearranged");
                exit(0);
            }
            str[p + d * k] = x.c;
        }
    }
}
  
// Driver Code
int main()
{
    char str[] = "aabbcc";
    
      // Function Call
    rearrange(str, 3);
    printf("%s",str);
}

C++

// C++ program to rearrange a string so that all same
// characters become at least d distance away using STL
#include <bits/stdc++.h>
#include <iostream>
using namespace std;
typedef pair<char, int> PAIR;
  
// Comparator of priority_queue
struct cmp {
    bool operator()(const PAIR& a, const PAIR& b)
    {
        if(a.second < b.second) return true;
          else if(a.second > b.second) return false;
          else return a.first > b.first;
    }
};
  
void rearrange(char* str, int d)
{
    // Length of the string
    int n = strlen(str);
  
    // A structure to store a character and its frequency
    unordered_map<char, int> m;
  
    // Traverse the input string and store frequencies of
    // all characters.
    for (int i = 0; i < n; i++) {
        m[str[i]]++;
        str[i] = '\0';
    }
  
    // max-heap
    priority_queue<PAIR, vector<PAIR>, cmp> pq(m.begin(),
                                               m.end());
  
    // Now one by one extract all distinct characters from
    // heap and put them back in str[] with the d
    // distance constraint
    while (pq.empty() == false) {
        char x = pq.top().first;
          
          // Find the first available position in str[]
          int p = 0;
        while (str[p] != '\0')
            p++;
            
          // Fill x at p, p+d, p+2d, .. p+(frequency-1)d
        for (int k = 0; k < pq.top().second; 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;
        }
        pq.pop();
    }
}
  
// Driver Code
int main()
{
    char str[] = "aabbcc";
    
      // Function call
    rearrange(str, 3);
    cout << str;
}

Python 3

# 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 range(MAX):
        freq.append(charFreq(0, 0))
  
    m = 0
  
    # Traverse the input string and store frequencies of all
    # characters in freq[] array.
    for i in range(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 range(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 range(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.
This article is contributed by RAVI TEJA NAMA. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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