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
1 public class Solution{
2
3 public static int MAX = 26;
4 static class CharFreq {
5 char c;
6 int f;
7 public CharFreq(char c, int f) {
8 this.c = c;
9 this.f = f;
10 }
11 }
12
13 public static void main(String []args){
14 String str = rearrange("geeksforgeeks",3);
15 if (str != null)
16 System.out.println(str);
17 }
18
19 public static String rearrange(String str, int k) {
20 if (str.length() <= 1) return str;
21
22 CharFreq[] cf = new CharFreq[MAX];
23 int count = 0; // number of different characters
24
25 for (int i=0; i<MAX; i++) {
26 cf[i] = new CharFreq((char) (‘a‘+i), 0);
27 }
28
29 for (int i=0; i<str.length(); i++) {
30 char ch = str.charAt(i);
31 cf[ch-‘a‘].f++;
32 if (cf[ch-‘a‘].f == 1) count++;
33 }
34
35 buildHeap(cf, MAX);
36
37 char[] str1 = new char[str.length()];
38 boolean[] occu = new boolean[str.length()]; // judge if the seat is occupied
39 for (int i = 0; i<count; i++) {
40 CharFreq chfreq = extractMax(cf, MAX-i);
41 int pt = i;
42 while (occu[pt]) pt++; // find the first place that is not occupied
43
44 for (int j=0; j<chfreq.f; j++) {
45 if (pt >= str1.length)
46 return null;
47 str1[pt] = chfreq.c;
48 occu[pt] = true;
49 pt += k;
50 }
51 }
52 return new String(str1);
53 }
54
55 private static void buildHeap(CharFreq[] cf, int size) {
56 int i = (size-1) / 2;
57 while (i>=0) {
58 maxHeapify(cf, i, size);
59 i--;
60 }
61 }
62
63 private static void swap(CharFreq cf1, CharFreq cf2) {
64 char c = cf1.c;
65 int f = cf1.f;
66 cf1.c = cf2.c;
67 cf1.f = cf2.f;
68 cf2.c = c;
69 cf2.f = f;
70 }
71
72 private static void maxHeapify(CharFreq[] cf, int node, int size) {
73 int l = node * 2 + 1;
74 int r = node * 2 + 2;
75 int largest = node;
76 if (l < size && cf[l].f > cf[node].f) {
77 largest = l;
78 }
79 if (r < size && cf[r].f > cf[largest].f) {
80 largest = r;
81 }
82 if (largest != node) {
83 swap(cf[node], cf[largest]);
84 maxHeapify(cf, largest, size);
85 }
86 }
87
88 private static CharFreq extractMax(CharFreq[] cf, int size) {
89 CharFreq max = cf[0];
90 cf[0] = cf[size-1];
91 cf[size-1] = null;
92 maxHeapify(cf, 0, size-1);
93 return max;
94 }
95 }
Analysis:
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.