Consider below list where each digit from 1 to 9 maps to few characters.
1 -> ['A', 'B', 'C'] 2 -> ['D', 'E', 'F'] 3 -> ['G', 'H', 'I'] 4 -> ['J', 'K', 'L'] 5 -> ['M', 'N', 'O'] 6 -> ['P', 'Q', 'R'] 7 -> ['S', 'T', 'U'] 8 -> ['V', 'W', 'X'] 9 -> ['Y', 'Z']
Given a number, replace its digits with corresponding characters in given list and print all strings possible. Same character should be considered for every occurrence of a digit in the number. Input number is positive and doesn’t contain 0.
Examples :
Input : 121 Output : ADA BDB CDC AEA BEB CEC AFA BFB CFC Input : 22 Output : DD EE FF
The idea is for each digit in the input number, we consider strings formed by previous digit and append characters mapped to current digit to them. If this is not the first occurrence of the digit, we append same character as used in its first occurrence.
C++
// C++ program to find all strings formed from a given // number where each digit maps to given characters. #include <bits/stdc++.h> using namespace std; // Function to find all strings formed from a given // number where each digit maps to given characters. vector<string> findCombinations(vector<int> input, vector<char> table[]) { // vector of strings to store output vector<string> out, temp; // stores index of first occurrence // of the digits in input unordered_map<int, int> mp; // maintains index of current digit considered int index = 0; // for each digit for (int d: input) { // store index of first occurrence // of the digit in the map if (mp.find(d) == mp.end()) mp[d] = index; // clear vector contents for future use temp.clear(); // do for each character thats maps to the digit for (int i = 0; i < table[d - 1].size(); i++) { // for first digit, simply push all its // mapped characters in the output list if (index == 0) { string s(1, table[d - 1].at(i)); out.push_back(s); } // from second digit onwards if (index > 0) { // for each string in output list // append current character to it. for(string str: out) { // convert current character to string string s(1, table[d - 1].at(i)); // Imp - If this is not the first occurrence // of the digit, use same character as used // in its first occurrence if(mp[d] != index) s = str[mp[d]]; str = str + s; // store strings formed by current digit temp.push_back(str); } // nothing more needed to be done if this // is not the first occurrence of the digit if(mp[d] != index) break; } } // replace contents of output list with temp list if(index > 0) out = temp; index++; } return out; } // Driver program int main() { // vector to store the mappings vector<char> table[] = { { 'A', 'B', 'C' }, { 'D', 'E', 'F' }, { 'G', 'H', 'I' }, { 'J', 'K', 'L' }, { 'M', 'N', 'O' }, { 'P', 'Q', 'R' }, { 'S', 'T', 'U' }, { 'V', 'W', 'X' }, { 'Y', 'Z' } }; // vector to store input number vector<int> input = { 1, 2, 1}; vector<string> out = findCombinations(input, table); // print all possible strings for (string it: out) cout << it << " "; return 0; } |
Python3
# Python program to find all strings formed from a given # number where each digit maps to given characters. # Function to find all strings formed from a given # number where each digit maps to given characters. def findCombinations(input: list, table: list) -> list: # vector of strings to store output out, temp = [], [] # stores index of first occurrence # of the digits in input mp = dict() # maintains index of current digit considered index = 0 # for each digit for d in input: # store index of first occurrence # of the digit in the map if d not in mp: mp[d] = index # clear vector contents for future use temp.clear() # do for each character thats maps to the digit for i in range(len(table[d - 1])): # for first digit, simply push all its # mapped characters in the output list if index == 0: s = table[d - 1][i] out.append(s) # from second digit onwards if index > 0: # for each string in output list # append current character to it. for string in out: # convert current character to string s = table[d - 1][i] # Imp - If this is not the first occurrence # of the digit, use same character as used # in its first occurrence if mp[d] != index: s = string[mp[d]] string = string + s # store strings formed by current digit temp.append(string) # nothing more needed to be done if this # is not the first occurrence of the digit if mp[d] != index: break # replace contents of output list with temp list if index > 0: out = temp.copy() index += 1 return out # Driver Code if __name__ == "__main__": # vector to store the mappings table = [['A', 'B', 'C'], ['D', 'E', 'F'], ['G', 'H', 'I'], ['J', 'K', 'L'], ['M', 'N', 'O'], ['P', 'Q', 'R'], ['S', 'T', 'U'], ['V', 'W', 'X'], ['Y', 'Z']] # vector to store input number input = [1, 2, 1] out = findCombinations(input, table) # print all possible strings for it in out: print(it, end=" ") # This code is contributed by # sanjeev2552 |
ADA BDB CDC AEA BEB CEC AFA BFB CFC
This article is contributed by Aditya Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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