Java: Rearrange a string so that all same characters become d distance away
Write a Java program to rearrange a string so that the same characters are positioned a distance away.
Visual Presentation:
Sample Solution:
Java Code:
// Importing necessary Java utilities.
import java.util.*;
// Define a class named Main.
public class Main {
// Define a constant value for character count.
public static int MX = 26;
// Define a class to store character frequency.
static class freqOfChar {
char c;
int f;
// Constructor for freqOfChar class.
public freqOfChar(char c, int f) {
this.c = c;
this.f = f;
}
}
// Main method to execute the program.
public static void main(String[] args) {
String strr = "accessories";
System.out.println("The given string is: " + strr);
System.out.println("The string after arranging newly is: ");
String strg = charRearrange(strr, 4);
if (strg != null)
System.out.println(strg);
}
// Method to rearrange characters in a string based on frequency and distance k.
public static String charRearrange(String strg, int k) {
if (strg.length() <= 1) return strg;
// Create an array of frequency of characters.
freqOfChar[] chr_fre = new freqOfChar[MX];
int ctr = 0;
// Initialize the frequency array.
for (int i = 0; i < MX; i++) {
chr_fre[i] = new freqOfChar((char)('a' + i), 0);
}
// Calculate frequency of each character in the string.
for (int i = 0; i < strg.length(); i++) {
char ch = strg.charAt(i);
chr_fre[ch - 'a'].f++;
if (chr_fre[ch - 'a'].f == 1) ctr++;
}
// Build a max heap based on character frequencies.
bldOfHeap(chr_fre, MX);
// Create arrays to store rearranged characters and their occurrences.
char[] str1 = new char[strg.length()];
boolean[] occu = new boolean[strg.length()];
// Rearrange characters according to frequency and distance k.
for (int i = 0; i < ctr; i++) {
freqOfChar chfreq = extractMax(chr_fre, MX - i);
int ptr = i;
while (occu[ptr]) ptr++;
for (int j = 0; j < chfreq.f; j++) {
if (ptr >= str1.length)
return null;
str1[ptr] = chfreq.c;
occu[ptr] = true;
ptr += k;
}
}
return new String(str1);
}
// Method to build a max heap.
private static void bldOfHeap(freqOfChar[] chr_fre, int size) {
int i = (size - 1) / 2;
while (i >= 0) {
maxHeapify(chr_fre, i, size);
i--;
}
}
// Method to swap two elements in the frequency array.
private static void swap(freqOfChar cf1, freqOfChar cf2) {
char c = cf1.c;
int f = cf1.f;
cf1.c = cf2.c;
cf1.f = cf2.f;
cf2.c = c;
cf2.f = f;
}
// Method to maintain the max heap property.
private static void maxHeapify(freqOfChar[] chr_fre, int node, int size) {
int l = node * 2 + 1;
int r = node * 2 + 2;
int largest = node;
if (l < size && chr_fre[l].f > chr_fre[node].f) {
largest = l;
}
if (r < size && chr_fre[r].f > chr_fre[largest].f) {
largest = r;
}
if (largest != node) {
swap(chr_fre[node], chr_fre[largest]);
maxHeapify(chr_fre, largest, size);
}
}
// Method to extract the maximum frequency element from the heap.
private static freqOfChar extractMax(freqOfChar[] chr_fre, int size) {
freqOfChar max = chr_fre[0];
chr_fre[0] = chr_fre[size - 1];
chr_fre[size - 1] = null;
maxHeapify(chr_fre, 0, size - 1);
return max;
}
}
Sample Output:
The given string is: accessories The string after arrange newly is: secrsecisao
Flowchart: 1
Flowchart: 2
Java Code Editor:
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