Print all the combinations of a string in lexicographical order
Given a string str, print of all the combinations of a string in lexicographical order.
Examples:
Input: str = "ABC" Output: A AB ABC AC ACB B BA BAC BC BCA C CA CAB CB CBA Input: ED Output: D DE E ED
Approach: Count the occurrences of all the characters in the string using a map, then using recursion all the possible combinations can be printed. Store the elements and their counts in two different arrays. Three arrays are used, input[] array which has the characters, count[] array has the count of characters and result[] is a temporary array which is used in recursion to generate all the combinations. Using recursion and backtracking all the combinations can be printed.
Below is the implementation of the above approach.
C++
// C++ program to find all combinations// of a string in lexicographical order#include <bits/stdc++.h>using namespace std;// function to print stringvoid printResult(char* result, int len){ for (int i = 0; i <= len; i++) cout << result[i]; cout << endl;}// Method to found all combination// of string it is based in treevoid stringCombination(char result[], char str[], int count[], int level, int size, int length){ // return if level is equal size of string if (level == size) return; for (int i = 0; i < length; i++) { // if occurrence of char is 0 then // skip the iteration of loop if (count[i] == 0) continue; // decrease the char occurrence by 1 count[i]--; // store the char in result result[level] = str[i]; // print the string till level printResult(result, level); // call the function from level +1 stringCombination(result, str, count, level + 1, size, length); // backtracking count[i]++; }}void combination(string str){ // declare the map for store // each char with occurrence map<char, int> mp; for (int i = 0; i < str.size(); i++) { if (mp.find(str[i]) != mp.end()) mp[str[i]] = mp[str[i]] + 1; else mp[str[i]] = 1; } // initialize the input array // with all unique char char* input = new char[mp.size()]; // initialize the count array with // occurrence the unique char int* count = new int[mp.size()]; // temporary char array for store the result char* result = new char[str.size()]; map<char, int>::iterator it = mp.begin(); int i = 0; for (it; it != mp.end(); it++) { // store the element of input array input[i] = it->first; // store the element of count array count[i] = it->second; i++; } // size of map(no of unique char) int length = mp.size(); // size of original string int size = str.size(); // call function for print string combination stringCombination(result, input, count, 0, size, length);}// Driver codeint main(){ string str = "ABC"; combination(str); return 0;} |
Java
// Java program to find all combinations// of a string in lexicographical orderimport java.util.HashMap;class GFG{ // function to print string static void printResult(char[] result, int len) { for (int i = 0; i <= len; i++) System.out.print(result[i]); System.out.println(); } // Method to found all combination // of string it is based in tree static void stringCombination(char[] result, char[] str, int[] count, int level, int size, int length) { // return if level is equal size of string if (level == size) return; for (int i = 0; i < length; i++) { // if occurrence of char is 0 then // skip the iteration of loop if (count[i] == 0) continue; // decrease the char occurrence by 1 count[i]--; // store the char in result result[level] = str[i]; // print the string till level printResult(result, level); // call the function from level +1 stringCombination(result, str, count, level + 1, size, length); // backtracking count[i]++; } } static void combination(String str) { // declare the map for store // each char with occurrence HashMap<Character, Integer> mp = new HashMap<>(); for (int i = 0; i < str.length(); i++) mp.put(str.charAt(i), mp.get(str.charAt(i)) == null ? 1 : mp.get(str.charAt(i)) + 1); // initialize the input array // with all unique char char[] input = new char[mp.size()]; // initialize the count array with // occurrence the unique char int[] count = new int[mp.size()]; // temporary char array for store the result char[] result = new char[str.length()]; int i = 0; for (HashMap.Entry<Character, Integer> entry : mp.entrySet()) { // store the element of input array input[i] = entry.getKey(); // store the element of count array count[i] = entry.getValue(); i++; } // size of map(no of unique char) int length = mp.size(); // size of original string int size = str.length(); // call function for print string combination stringCombination(result, input, count, 0, size, length); } // Driver code public static void main (String[] args) { String str = "ABC"; combination(str); }}// This code is contributed by// sanjeev2552 |
C#
// C# program to find all combinations// of a string in lexicographical orderusing System;using System.Collections.Generic;class GFG{ // function to print string static void printResult(char[] result, int len) { for (int i = 0; i <= len; i++) Console.Write(result[i]); Console.WriteLine(); } // Method to found all combination // of string it is based in tree static void stringCombination(char[] result, char[] str, int[] count, int level, int size, int length) { // return if level is equal size of string if (level == size) return; for (int i = 0; i < length; i++) { // if occurrence of char is 0 then // skip the iteration of loop if (count[i] == 0) continue; // decrease the char occurrence by 1 count[i]--; // store the char in result result[level] = str[i]; // print the string till level printResult(result, level); // call the function from level +1 stringCombination(result, str, count, level + 1, size, length); // backtracking count[i]++; } } static void combination(String str) { int i; // declare the map for store // each char with occurrence Dictionary<char,int> mp = new Dictionary<char,int>(); for (i= 0; i < str.Length; i++) if(mp.ContainsKey(str[i])) mp[str[i]] = mp[str[i]] + 1; else mp.Add(str[i], 1); // initialize the input array // with all unique char char[] input = new char[mp.Count]; // initialize the count array with // occurrence the unique char int[] count = new int[mp.Count]; // temporary char array for store the result char[] result = new char[str.Length]; i = 0; foreach(KeyValuePair<char, int> entry in mp) { // store the element of input array input[i] = entry.Key; // store the element of count array count[i] = entry.Value; i++; } // size of map(no of unique char) int length = mp.Count; // size of original string int size = str.Length; // call function for print string combination stringCombination(result, input, count, 0, size, length); } // Driver code public static void Main(String[] args) { String str = "ABC"; combination(str); }}// This code is contributed by Rajput-Ji |
Javascript
<script> // JavaScript program to find all combinations // of a string in lexicographical order // function to print string function printResult(result, len) { for (var i = 0; i <= len; i++) document.write(result[i]); document.write("<br>"); } // Method to found all combination // of string it is based in tree function stringCombination(result, str, count, level, size, len) { // return if level is equal size of string if (level === size) return; for (var i = 0; i < len; i++) { // if occurrence of char is 0 then // skip the iteration of loop if (count[i] === 0) continue; // decrease the char occurrence by 1 count[i]--; // store the char in result result[level] = str[i]; // print the string till level printResult(result, level); // call the function from level +1 stringCombination(result, str, count, level + 1, size, len); // backtracking count[i]++; } } function combination(str) { var i; // declare the map for store // each char with occurrence var mp = {}; for (i = 0; i < str.length; i++) if (mp.hasOwnProperty(str[i])) mp[str[i]] = mp[str[i]] + 1; else mp[str[i]] = 1; // initialize the input array // with all unique char var input = new Array(Object.keys(mp).length).fill(0); // initialize the count array with // occurrence the unique char var count = new Array(Object.keys(mp).length).fill(0); // temporary char array for store the result var result = new Array(str.length).fill(0); i = 0; for (const [key, value] of Object.entries(mp)) { // store the element of input array input[i] = key; // store the element of count array count[i] = value; i++; } // size of map(no of unique char) var len = Object.keys(mp).length; // size of original string var size = str.length; // call function for print string combination stringCombination(result, input, count, 0, size, len); } // Driver code var str = "ABC"; combination(str); </script> |
Output:
A AB ABC AC ACB B BA BAC BC BCA C CA CAB CB CBA
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