Given a string and a positive integer d, rearrange characters of the given string such that the same characters become at-least 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: "aaaabbbcc", d = 2 Output: "ababacabc" Input: "aacbbc", d = 3 Output: "abcabc" Input: "neveropen", d = 3 Output: egkesfegkeors Input: "aaa", d = 2 Output: Cannot be rearranged
We have already discussed how to put same characters exactly d distance away. This is a extended version where same characters should be moved at-least d distance away.
The idea is to use extra space to store frequencies of all characters and maintain an array for inserting the values at correct distance. Following is the complete algorithm:
- Let the given string be str and size of string be n and alphabet size is be assumed as 256 (a constant).
- We scan input string str and store frequencies of all characters in an array freq.
- We create an array dist[] for inserting the values at correct distance. dist[j] will store the least distance between current position and the next position we can use character ‘j’.
If dist[j] <= 0, character ‘j’ can be inserted in current position.- run a loop n times
- Search for next eligible character with maximum frequency and dist[j] <= 0.
- If we found such character, we put that character at next available position in output array, decrease its frequency and reset its distance as d. If we don’t find any character, string cannot be rearranged and we return false.
- As we move forward in output string, we decrement distance of all characters in dist[] by 1.
Following is the implementation of above algorithm.
C++
// C++ program to rearrange a string so that all same// characters become atleast d distance away#include <bits/stdc++.h>#define MAX_CHAR 256using namespace std;// The function returns next eligible character// with maximum frequency (Greedy!!) and// zero or negative distanceint nextChar(int freq[], int dist[]){ int max = INT_MIN; for (int i = 0; i < MAX_CHAR; i++) if (dist[i] <= 0 && freq[i] > 0 && (max == INT_MIN || freq[i] > freq[max])) max = i; return max;}// The main function that rearranges input string 'str'// such that two same characters become atleast d// distance awayint rearrange(char str[], char out[], int d){ // Find length of input string int n = strlen(str); // Create an array to store all characters and their // frequencies in str[] int freq[MAX_CHAR] = { 0 }; // Traverse the input string and store frequencies // of all characters in freq[] array. for (int i = 0; i < n; i++) freq[str[i]]++; // Create an array for inserting the values at // correct distance dist[j] stores the least distance // between current position and the next position we // can use character 'j' int dist[MAX_CHAR] = { 0 }; for (int i = 0; i < n; i++) { // find next eligible character int j = nextChar(freq, dist); // return 0 if string cannot be rearranged if (j == INT_MIN) return 0; // Put character j at next position out[i] = j; // decrease its frequency freq[j]--; // set distance as d dist[j] = d; // decrease distance of all characters by 1 for (int i = 0; i < MAX_CHAR; i++) dist[i]--; } // null terminate output string out[n] = '\0'; // return success return 1;}// Driver codeint main(){ char str[] = "aaaabbbcc"; int n = strlen(str); // To store output char out[n]; if (rearrange(str, out, 2)) cout << out; else cout << "Cannot be rearranged"; return 0;} |
Java
// Java program to rearrange a string so that all same // characters become atleast d distance away import java.util.*;class GFG {static int MAX_CHAR = 256; // The function returns next eligible character // with maximum frequency (Greedy!!) and // zero or negative distance static int nextChar(int freq[], int dist[]) { int max = Integer.MIN_VALUE; for (int i = 0; i < MAX_CHAR; i++) if (dist[i] <= 0 && freq[i] > 0 && (max == Integer.MIN_VALUE || freq[i] > freq[max])) max = i; return max; } // The main function that rearranges input string 'str' // such that two same characters become atleast d // distance away static int rearrange(char str[], char out[], int d) { // Find length of input string int n = str.length; // Create an array to store all characters and their // frequencies in str[] int []freq = new int[MAX_CHAR]; // Traverse the input string and store frequencies // of all characters in freq[] array. for (int i = 0; i < n; i++) freq[str[i]]++; // Create an array for inserting the values at // correct distance dist[j] stores the least distance // between current position and the next position we // can use character 'j' int []dist = new int[MAX_CHAR]; for (int i = 0; i < n; i++) { // find next eligible character int j = nextChar(freq, dist); // return 0 if string cannot be rearranged if (j == Integer.MIN_VALUE) return 0; // Put character j at next position out[i] = (char) j; // decrease its frequency freq[j]--; // set distance as d dist[j] = d; // decrease distance of all characters by 1 for (int k = 0; k < MAX_CHAR; k++) dist[k]--; } // null terminate output string Arrays.copyOfRange(out, 0, n); // out[n] = '\0'; // return success return 1; } // Driver code public static void main(String[] args){ char str[] = "aaaabbbcc".toCharArray(); int n = str.length; // To store output char []out = new char[n]; if (rearrange(str, out, 2)==1) System.out.println(String.valueOf(out)); else System.out.println("Cannot be rearranged");}}// This code is contributed by Rajput-Ji |
Python3
# Python3 program to rearrange a string so that all # same characters become at least d distance away MAX_CHAR = 256# The function returns next eligible character # with maximum frequency (Greedy!!) # and zero or negative distance def nextChar(freq, dist): Max = float('-inf') for i in range(0, MAX_CHAR): if (dist[i] <= 0 and freq[i] > 0 and (Max == float('-inf') or freq[i] > freq[Max])): Max = i return Max# The main function that rearranges input # string 'str' such that two same characters # become atleast d distance away def rearrange(string, out, d): # Find length of input string n = len(string) # Create an array to store all characters # and their frequencies in str[] freq = [0] * MAX_CHAR # Traverse the input string and store frequencies # of all characters in freq[] array. for i in range(0, n): freq[ord(string[i])] += 1 # Create an array for inserting the values at # correct distance dist[j] stores the least # distance between current position and the # we next position can use character 'j' dist = [0] * MAX_CHAR for i in range(0, n): # find next eligible character j = nextChar(freq, dist) # return 0 if string cannot be rearranged if j == float('-inf'): return 0 # Put character j at next position out[i] = chr(j) # decrease its frequency freq[j] -= 1 # set distance as d dist[j] = d # decrease distance of all characters by 1 for i in range(0, MAX_CHAR): dist[i] -= 1 # return success return 1# Driver code if __name__ == "__main__": string = "aaaabbbcc" n = len(string) # To store output out = [None] * n if rearrange(string, out, 2): print(''.join(out)) else: print("Cannot be rearranged") # This code is contributed by Rituraj Jain |
C#
// C# program to rearrange a string so that all same // characters become atleast d distance away using System; class GFG {static int MAX_CHAR = 256; // The function returns next eligible character // with maximum frequency (Greedy!!) and // zero or negative distance static int nextChar(int []freq, int []dist) { int max = int.MinValue; for (int i = 0; i < MAX_CHAR; i++) if (dist[i] <= 0 && freq[i] > 0 && (max == int.MinValue || freq[i] > freq[max])) max = i; return max; } // The main function that rearranges input string 'str' // such that two same characters become atleast d // distance away static int rearrange(char []str, char []ouT, int d) { // Find length of input string int n = str.Length; // Create an array to store all characters and their // frequencies in str[] int []freq = new int[MAX_CHAR]; // Traverse the input string and store frequencies // of all characters in freq[] array. for (int i = 0; i < n; i++) freq[str[i]]++; // Create an array for inserting the values at // correct distance dist[j] stores the least distance // between current position and the next position we // can use character 'j' int []dist = new int[MAX_CHAR]; for (int i = 0; i < n; i++) { // find next eligible character int j = nextChar(freq, dist); // return 0 if string cannot be rearranged if (j == int.MinValue) return 0; // Put character j at next position ouT[i] = (char) j; // decrease its frequency freq[j]--; // set distance as d dist[j] = d; // decrease distance of all characters by 1 for (int k = 0; k < MAX_CHAR; k++) dist[k]--; } // null terminate output string Array.Copy(ouT,ouT, n); // out[n] = '\0'; // return success return 1; } // Driver code public static void Main(String[] args){ char []str = "aaaabbbcc".ToCharArray(); int n = str.Length; // To store output char []ouT = new char[n]; if (rearrange(str, ouT, 2)==1) Console.WriteLine(String.Join("",ouT)); else Console.WriteLine("Cannot be rearranged");}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript program to rearrange a // string so that all same characters // become atleast d distance away let MAX_CHAR = 256; // The function returns next eligible// character with maximum frequency// (Greedy!!) and zero or negative distance function nextChar(freq, dist){ let max = Number.MIN_VALUE; for(let i = 0; i < MAX_CHAR; i++) if (dist[i] <= 0 && freq[i] > 0 && (max == Number.MIN_VALUE || freq[i] > freq[max])) max = i; return max; }// The main function that rearranges input// string 'str' such that two same characters// become atleast d distance away function rearrange(str, out, d){ // Find length of input string let n = str.length; // Create an array to store all characters // and their frequencies in str[] let freq = new Array(MAX_CHAR); for(let i = 0; i < freq.length; i++) { freq[i] = 0; } // Traverse the input string and store // frequencies of all characters in // freq[] array. for(let i = 0; i < n; i++) freq[str[i].charCodeAt(0)]++; // Create an array for inserting the // values at correct distance dist[j] // stores the least distance between // current position and the next position // we can use character 'j' let dist = new Array(MAX_CHAR); for(let i = 0; i < dist.length; i++) { dist[i] = 0; } for(let i = 0; i < n; i++) { // Find next eligible character let j = nextChar(freq, dist); // return 0 if string cannot // be rearranged if (j == Number.MIN_VALUE) return 0; // Put character j at next position out[i] = String.fromCharCode (j); // Decrease its frequency freq[j]--; // Set distance as d dist[j] = d; // Decrease distance of all // characters by 1 for(let k = 0; k < MAX_CHAR; k++) dist[k]--; } // Null terminate output string // Arrays.copyOfRange(out, 0, n); // out[n] = '\0'; // Return success return 1; }// Driver code let str= "aaaabbbcc".split("");let n = str.length;// To store output let out = new Array(n); if (rearrange(str, out, 2) == 1) document.write(out.join(""));else document.write("Cannot be rearranged");// This code is contributed by rag2127</script> |
Output
ababacabc
Time Complexity : O(N* MAX_CHAR) , here N is length of string
Space Complexity : O(N)
This article is contributed by Aditya Goel. If you like neveropen and would like to contribute, you can also write an article and mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
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