Given an array arr[] consisting of N characters, the task is to generate all possible combinations of at most X elements ( 1 ? X ? N).
Examples:
Input: N = 3, X = 2, arr[] = {‘a’, ‘b’, ‘a’}
Output: a b c bc ca ab cb ac ba
Explanation: All possible combinations using 1 character is 3 {‘a’, ‘b’, ‘c’}. All possible combinations using 2 characters are {“bc” “ca” “ab” “cb” “ac” “ba”}.Input: N = 3, X = 3, arr[] = {‘d’, ‘a’, ‘b’}
Output: d a b da ab bd ad ba db dab dba abd adb bda bad
Approach: The given problem can be solved using the Dynamic Programming approach. Follow the below steps to solve the problem:
- Generate all possible permutations that can be created with 1 character, which is the given array arr[].
- Store all permutations.
- Once stored, generate all possible permutations of 2 characters and store them.
- Once the last step is completed, discard all permutations of a single character.
- Iteratively, in the same way, calculate the permutations until X is reached.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h>using namespace std;// Function to generate permutations of // at most X elements from array arr[] void differentFlagPermutations(int X, vector<string> arr){ vector<string> ml; ml = arr; for(int i = 0; i < ml.size(); i++) { cout << ml[i] << " "; } int count = ml.size(); // Traverse all possible lengths for(int z = 0; z < X - 1; z++) { // Stores all combinations // of length z vector<string> tmp; // Traverse the array for(int i = 0; i < arr.size(); i++) { for(int k = 0; k < ml.size(); k++) { if (arr[i] != ml[k]) { // Generate all // combinations of length z tmp.push_back(ml[k] + arr[i]); count += 1; } } } // Print all combinations of length z for(int i = 0; i < tmp.size(); i++) { cout << tmp[i] << " "; } // Replace all combinations of length z - 1 // with all combinations of length z ml = tmp; }}// Driver Codeint main(){ // Given array vector<string> arr{ "c", "a", "b" }; // Given X int X = 2; differentFlagPermutations(X, arr); return 0;}// This code is contributed by divyeshrabadiya07 |
Java
// Java program for the above approach import java.util.*;class GFG{// Function to generate permutations of // at most X elements from array arr[] static void differentFlagPermutations(int X, String[] arr){ String[] ml = arr; for(int i = 0; i < ml.length; i++) { System.out.print(ml[i] + " "); } int count = ml.length; // Traverse all possible lengths for(int z = 0; z < X - 1; z++) { // Stores all combinations // of length z Vector<String> tmp = new Vector<String>(); // Traverse the array for(int i = 0; i < arr.length; i++) { for(int k = 0; k < ml.length; k++) { if (arr[i] != ml[k]) { // Generate all // combinations of length z tmp.add(ml[k] + arr[i]); count += 1; } } } // Print all combinations of length z for(int i = 0; i < tmp.size(); i++) { System.out.print(tmp.get(i) + " "); } // Replace all combinations of length z - 1 // with all combinations of length z ml = tmp.toArray(new String[tmp.size()]);; }}// Driver Codepublic static void main(String[] args){ // Given array String []arr = { "c", "a", "b" }; // Given X int X = 2; differentFlagPermutations(X, arr); }}// This code is contributed by shikhasingrajput |
Python3
# Python3 program for the above approach# Function to generate permutations of# at most X elements from array arr[]def differentFlagPermutations(X, arr): ml = arr.copy() print(" ".join(ml), end =" ") count = len(ml) # Traverse all possible lengths for z in range(X-1): # Stores all combinations # of length z tmp = [] # Traverse the array for i in arr: for k in ml: if i not in k: # Generate all # combinations of length z tmp.append(k + i) count += 1 # Print all combinations of length z print(" ".join(tmp), end =" ") # Replace all combinations of length z - 1 # with all combinations of length z ml = tmp# Given arrayarr = ['c', 'a', 'b']# Given XX = 2differentFlagPermutations(X, arr) |
C#
// C# program for the above approach using System;using System.Collections.Generic;class GFG{ // Function to generate permutations of // at most X elements from array arr[] static void differentFlagPermutations(int X, List<string> arr) { List<string> ml = new List<string>(); ml = arr; for(int i = 0; i < ml.Count; i++) { Console.Write(ml[i] + " "); } int count = ml.Count; // Traverse all possible lengths for(int z = 0; z < X - 1; z++) { // Stores all combinations // of length z List<string> tmp = new List<string>(); // Traverse the array for(int i = 0; i < arr.Count; i++) { for(int k = 0; k < ml.Count; k++) { if (arr[i] != ml[k]) { // Generate all // combinations of length z tmp.Add(ml[k] + arr[i]); count += 1; } } } // Print all combinations of length z for(int i = 0; i < tmp.Count; i++) { Console.Write(tmp[i] + " "); } // Replace all combinations of length z - 1 // with all combinations of length z ml = tmp; } } // Driver code static void Main() { // Given array List<string> arr = new List<string>(new string[] { "c", "a", "b" }); // Given X int X = 2; differentFlagPermutations(X, arr); }}// This code is contributed by divyesh072019 |
Javascript
<script> // Javascript program for the above approach // Function to generate permutations of // at most X elements from array arr[] function differentFlagPermutations(X, arr) { let ml = []; ml = arr; for(let i = 0; i < ml.length; i++) { document.write(ml[i] + " "); } let count = ml.length; // Traverse all possible lengths for(let z = 0; z < X - 1; z++) { // Stores all combinations // of length z let tmp = []; // Traverse the array for(let i = 0; i < arr.length; i++) { for(let k = 0; k < ml.length; k++) { if (arr[i] != ml[k]) { // Generate all // combinations of length z tmp.push(ml[k] + arr[i]); count += 1; } } } // Print all combinations of length z for(let i = 0; i < tmp.length; i++) { document.write(tmp[i] + " "); } // Replace all combinations of length z - 1 // with all combinations of length z ml = tmp; } } // Given array let arr = [ "c", "a", "b" ]; // Given X let X = 2; differentFlagPermutations(X, arr);</script> |
Output:
c a b ac bc ca ba cb ab
Time Complexity: O(X*N2)
Auxiliary Space: O(N2)
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