Given two positive integers N and K, the task is to find the number of K length strings that can be generated from the first N alphabets such that the characters in the string are sorted lexicographically.
Examples:
Input: N = 5, K = 2
Output: 15
Explanation: All possible strings are {“AA”, “AB”, “AC”, “AD”, “AE”, “BB”, “BC”, “BD”, “BE”, “CC”, “CD”, “CE”, “DD”, “DE”, “EE”}.Input: N = 7, K = 10
Output: 8008
Naive Approach: The simplest approach to solve the problem is to use Recursion and Backtracking to generate all possible arrangements of characters and for each string, check if the characters follow a lexicographically increasing order or not. Print the count of all such strings.
C++
#include <iostream>#include <vector>#include <algorithm>using namespace std;vector<string> generateStrings(string curr_str, int curr_char, int N, int K) { if (curr_str.length() == K) { vector<string> res; res.push_back(curr_str); return res; } vector<string> res; for (int i = curr_char; i < N; i++) { vector<string> suffixes = generateStrings(curr_str + char(65 + i), i, N, K); res.insert(res.end(), suffixes.begin(), suffixes.end()); } return res;}int countStrings(int N, int K) { vector<string> all_strings = generateStrings("", 0, N, K); vector<string> sorted_strings; for (int i = 0; i < all_strings.size(); i++) { string s = all_strings[i]; if (is_sorted(s.begin(), s.end())) { sorted_strings.push_back(s); } } return sorted_strings.size();}int main() { cout << countStrings(5, 2) << endl; // expected output: 15 cout << countStrings(7, 10) << endl; // expected output: 8008 return 0;} |
Java
import java.util.ArrayList;import java.util.List;public class Main { // Function to generate strings of length K using characters from 'A' to 'A + N - 1' static List<String> generateStrings(String currStr, int currChar, int N, int K) { if (currStr.length() == K) { List<String> res = new ArrayList<>(); res.add(currStr); return res; } List<String> res = new ArrayList<>(); for (int i = currChar; i < N; i++) { List<String> suffixes = generateStrings(currStr + (char) ('A' + i), i, N, K); res.addAll(suffixes); } return res; } // Function to count strings that are sorted lexicographically static int countStrings(int N, int K) { List<String> allStrings = generateStrings("", 0, N, K); List<String> sortedStrings = new ArrayList<>(); for (String s : allStrings) { if (isSorted(s)) { sortedStrings.add(s); } } return sortedStrings.size(); } // Helper function to check if a string is sorted static boolean isSorted(String s) { for (int i = 1; i < s.length(); i++) { if (s.charAt(i - 1) > s.charAt(i)) { return false; } } return true; } public static void main(String[] args) { System.out.println(countStrings(5, 2)); // expected output: 15 System.out.println(countStrings(7, 10)); // expected output: 8008 }} |
Python
def count_strings(N, K): def generate_strings(curr_str, curr_char, K): if len(curr_str) == K: return [curr_str] res = [] for i in range(curr_char, N): res.extend(generate_strings(curr_str + chr(65 + i), i, K)) return res all_strings = generate_strings("", 0, K) sorted_strings = [s for s in all_strings if list(s) == sorted(list(s))] return len(sorted_strings)# example usageprint(count_strings(5, 2)) # expected output: 15print(count_strings(7, 10)) # expected output: 8008 |
Javascript
function countStrings(N, K) { function generateStrings(currStr, currChar, K) { if (currStr.length === K) { return [currStr]; } let res = []; for (let i = currChar; i < N; i++) { res.push(...generateStrings(currStr + String.fromCharCode(65 + i), i, K)); } return res; } let allStrings = generateStrings("", 0, K); let sortedStrings = allStrings.filter(s => [...s].sort().join("") === s); return sortedStrings.length;}// example usageconsole.log(countStrings(5, 2)); // expected output: 15console.log(countStrings(7, 10)); // expected output: 8008 |
Output
15 8008
Time Complexity: O(KN)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is to use Dynamic Programming as there are overlapping subproblems that can be memoized or tabulated in the recursive calls by using an auxiliary 2D array dp[][] and calculate the value of each state in the bottom-up approach.
dp[i][j] represents the number of ways to arrange “i” length strings with the “j” distinct letters.
dp[i][j] = dp[i][j – 1] (Choose not to start with first letter)
+ dp[i – 1][j – 1] (Choose first 1 letter in string as first letter)
+ dp[i – 2][j – 1] (Choose first 2 letters in string as first letter)
+ ….
+ ….
+ dp[0][j – 1] (Choose first i letters in string as first letter)
dp[i][j] = Sum of all values of (j-1)th column for “i” rows
Follow the steps below to solve this problem:
- Initialize an array columnSum[] of size (N+1), where columnSum[i] is sum of all values in column “j” of the array dp[][].
- Initialize a dp[][] table of size (K + 1)*(N + 1).
- Initialize dp[0][i] as 1 and subsequently update array columnSum[].
- Iterate two nested loops over K and using the variable i and j respectively:
- Store dp[i][j] as columnSum[j – 1].
- Update columnSum[j] as columnSum[j] + dp[i][j].
- After the above steps, print the value of dp[K][N] as the resultant number of ways.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to count K-length// strings from first N alphabetsvoid waysToArrangeKLengthStrings( int N, int K){ // To keep track of column sum in dp int column_sum[N + 1] = { 0 }, i, j; // Auxiliary 2d dp array int dp[K + 1][N + 1] = { 0 }; // Initialize dp[0][i] = 1 and // update the column_sum for (i = 0; i <= N; i++) { dp[0][i] = 1; column_sum[i] = 1; } // Iterate for K times for (i = 1; i <= K; i++) { // Iterate for N times for (j = 1; j <= N; j++) { // dp[i][j]: Stores the number // of ways to form i-length // strings consisting of j letters dp[i][j] += column_sum[j - 1]; // Update the column_sum column_sum[j] += dp[i][j]; } } // Print number of ways to arrange // K-length strings with N alphabets cout << dp[K][N];}// Driver Codeint main(){ // Given N and K int N = 5, K = 2; // Function Call waysToArrangeKLengthStrings(N, K); return 0;} |
Java
// Java program for the above approachimport java.io.*;import java.util.*; class GFG{ // Function to count K-length// strings from first N alphabetsstatic void waysToArrangeKLengthStrings( int N, int K){ // To keep track of column sum in dp int[] column_sum = new int[N + 1]; int i, j; for(i = 1; i < N + 1; i++) { column_sum[i] = 0; } // Auxiliary 2d dp array int dp[][] = new int[K + 1][N + 1]; for(i = 1; i < K + 1; i++) { for(j = 1; j < N + 1; j++) { dp[i][j] = 0; } } // Initialize dp[0][i] = 1 and // update the column_sum for(i = 0; i <= N; i++) { dp[0][i] = 1; column_sum[i] = 1; } // Iterate for K times for(i = 1; i <= K; i++) { // Iterate for N times for(j = 1; j <= N; j++) { // dp[i][j]: Stores the number // of ways to form i-length // strings consisting of j letters dp[i][j] += column_sum[j - 1]; // Update the column_sum column_sum[j] += dp[i][j]; } } // Print number of ways to arrange // K-length strings with N alphabets System.out.print(dp[K][N]);} // Driver Codepublic static void main(String[] args){ // Given N and K int N = 5, K = 2; // Function Call waysToArrangeKLengthStrings(N, K);}} // This code is contributed by susmitakundugoaldanga |
Python3
# Python3 program for the above approach# Function to count K-length# strings from first N alphabetsdef waysToArrangeKLengthStrings(N, K): # To keep track of column sum in dp column_sum = [0 for i in range(N + 1)] i = 0 j = 0 # Auxiliary 2d dp array dp = [[0 for i in range(N + 1)] for j in range(K + 1)] # Initialize dp[0][i] = 1 and # update the column_sum for i in range(N + 1): dp[0][i] = 1 column_sum[i] = 1 # Iterate for K times for i in range(1, K + 1): # Iterate for N times for j in range(1, N + 1): # dp[i][j]: Stores the number # of ways to form i-length # strings consisting of j letters dp[i][j] += column_sum[j - 1] # Update the column_sum column_sum[j] += dp[i][j] # Print number of ways to arrange # K-length strings with N alphabets print(dp[K][N])# Driver Codeif __name__ == '__main__': # Given N and K N = 5 K = 2 # Function Call waysToArrangeKLengthStrings(N, K)# This code is contributed by SURENDRA_GANGWAR |
C#
// C# program for the above approachusing System; class GFG{ // Function to count K-length// strings from first N alphabetsstatic void waysToArrangeKLengthStrings(int N, int K){ // To keep track of column sum in dp int[] column_sum = new int[N + 1]; int i, j; for(i = 1; i < N + 1; i++) { column_sum[i] = 0; } // Auxiliary 2d dp array int[,] dp = new int[K + 1, N + 1]; for(i = 1; i < K + 1; i++) { for(j = 1; j < N + 1; j++) { dp[i, j] = 0; } } // Initialize dp[0][i] = 1 and // update the column_sum for(i = 0; i <= N; i++) { dp[0, i] = 1; column_sum[i] = 1; } // Iterate for K times for(i = 1; i <= K; i++) { // Iterate for N times for(j = 1; j <= N; j++) { // dp[i][j]: Stores the number // of ways to form i-length // strings consisting of j letters dp[i, j] += column_sum[j - 1]; // Update the column_sum column_sum[j] += dp[i, j]; } } // Print number of ways to arrange // K-length strings with N alphabets Console.Write(dp[K, N]);} // Driver Codepublic static void Main(){ // Given N and K int N = 5, K = 2; // Function Call waysToArrangeKLengthStrings(N, K);}}// This code is contributed by code_hunt |
Javascript
<script>// Javascript program to implement// the above approach// Function to count K-length// strings from first N alphabetsfunction waysToArrangeKLengthStrings(N, K){ // To keep track of column sum in dp let column_sum = []; let i, j; for(i = 1; i < N + 1; i++) { column_sum[i] = 0; } // Auxiliary 2d dp array let dp = new Array(K + 1); // Loop to create 2D array using 1D array for (i = 0; i < dp.length; i++) { dp[i] = new Array(2); } for(i = 1; i < K + 1; i++) { for(j = 1; j < N + 1; j++) { dp[i][j] = 0; } } // Initialize dp[0][i] = 1 and // update the column_sum for(i = 0; i <= N; i++) { dp[0][i] = 1; column_sum[i] = 1; } // Iterate for K times for(i = 1; i <= K; i++) { // Iterate for N times for(j = 1; j <= N; j++) { // dp[i][j]: Stores the number // of ways to form i-length // strings consisting of j letters dp[i][j] += column_sum[j - 1]; // Update the column_sum column_sum[j] += dp[i][j]; } } // Print number of ways to arrange // K-length strings with N alphabets document.write(dp[K][N]);} // Driver Code // Given N and K let N = 5, K = 2; // Function Call waysToArrangeKLengthStrings(N, K);// This code is contributed by splevel62.</script> |
Output
15
Time Complexity: O(N*K)
Auxiliary Space: O(N*K)
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