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

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `#include ` `#include ` `#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; ` `    ``if` `(heap_size > 1) ` `    ``{ ` `        ``freq = 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`

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.

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Improved By : PKumar7

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