Given a string S and an integer K, the task is to find the total number of strings that can be formed by inserting exactly K characters at any position of the string S. Since the answer can be large, print it modulo 109+7.
Examples:
Input: S = “a” K = 1
Output: 51
Explanation:
Since any of the 26 characters can be inserted at before ‘a’ or after ‘a’, a total of 52 possible strings can be formed.
But the string “aa” gets formed twice. Hence count of distinct strings possible is 51.
Input: S = “abc” K = 2
Output: 6376
Approach:
The idea is to find the number of strings that contains the str as a subsequence. Follow the steps below to solve the problem:
- The total number of strings that can be formed by N characters is 26N.
- Calculate 26N using Binary Exponentiation.
- In this problem, only the strings that contain the str as a subsequence needs to be considered.
- Hence, the final count of strings is given by
(total number of strings) – (number of strings that don’t contain the input string as a sub-sequence)
- While calculating such strings that don’t contain the str as a subsequence, observe that the length of the prefix of S is a subsequence of the resulting string can be between 0 to |S|-1.
- For every prefix length from 0 to |S|-1, find the total number of strings that can be formed with such a prefix as a sub-sequence. Then subtract that value from 26N.
- Hence, the final answer is:
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;#define int long long intconst int mod = 1e9 + 7;// Function to calculate and return// x^n in log(n) time using// Binary Exponentiationint binExp(int base, int power){ int x = 1; while (power) { if (power % 2 == 1) x = ((x % mod) * (base % mod)) % mod; base = ((base % mod) * (base % mod)) % mod; power = power / 2; } return x;}// Function to calculate the factorial// of a numberint fact(int num){ int result = 1; for (int i = 1; i <= num; ++i) { result = ((result % mod) * (i % mod)) % mod; } return result;}// Function to calculate combinationint calculate_nCi(int N, int i){ // nCi = ( n! ) / (( n-i )! * i!) int nfact = fact(N); int ifact = fact(i); int dfact = fact(N - i); // Using Euler's theorem of Modular // multiplicative inverse to find // the inverse of a number. // (1/a)%mod=a^(m?2)%mod int inv_ifact = binExp(ifact, mod - 2); int inv_dfact = binExp(dfact, mod - 2); int denm = ((inv_ifact % mod) * (inv_dfact % mod)) % mod; int answer = ((nfact % mod) * (denm % mod)) % mod; return answer;}// Function to find the count of// possible stringsvoid countSubstring(int N, int s, int k){ // Number of ways to form all // possible strings int allWays = binExp(26, N); // Number of ways to form strings // that don't contain the input // string as a subsequence int noWays = 0; // Checking for all prefix length // from 0 to |S|-1. for (int i = 0; i < s; ++i) { // to calculate nCi int nCi = calculate_nCi(N, i); // Select the remaining characters // 25 ^ (N-i) int remaining = binExp(25, N - i); int multiply = ((nCi % mod) * (remaining % mod)) % mod; // Add the answer for this prefix // length to the final answer noWays = ((noWays % mod) + (multiply % mod)) % mod; } // Answer is the difference of // allWays and noWays int answer = ((allWays % mod) - (noWays % mod)) % mod; if (answer < 0) answer += mod; // Print the answer cout << answer;}// Driver Codeint32_t main(){ string str = "abc"; int k = 2; int s = str.length(); int N = s + k; countSubstring(N, s, k);} |
Java
// Java program for the above approachclass GFG{static final long mod = 1000000007;// Function to calculate and return// x^n in log(n) time using// Binary Exponentiationstatic long binExp(long base, long power){ long x = 1; while (power != 0) { if (power % 2 == 1) x = ((x % mod) * (base % mod)) % mod; base = ((base % mod) * (base % mod)) % mod; power = power / 2; } return x;}// Function to calculate the factorial// of a numberstatic long fact(long num){ long result = 1; for(long i = 1; i <= num; ++i) { result = ((result % mod) * (i % mod)) % mod; } return result;}// Function to calculate combinationstatic long calculate_nCi(long N, long i){ // nCi = ( n! ) / (( n-i )! * i!) long nfact = fact(N); long ifact = fact(i); long dfact = fact(N - i); // Using Euler's theorem of Modular // multiplicative inverse to find // the inverse of a number. // (1/a)%mod=a^(m?2)%mod long inv_ifact = binExp(ifact, mod - 2); long inv_dfact = binExp(dfact, mod - 2); long denm = ((inv_ifact % mod) * (inv_dfact % mod)) % mod; long answer = ((nfact % mod) * (denm % mod)) % mod; return answer;}// Function to find the count of// possible stringsstatic void countSubstring(long N, long s, long k){ // Number of ways to form all // possible strings long allWays = binExp(26, N); // Number of ways to form strings // that don't contain the input // string as a subsequence long noWays = 0; // Checking for all prefix length // from 0 to |S|-1. for(long i = 0; i < s; ++i) { // To calculate nCi long nCi = calculate_nCi(N, i); // Select the remaining characters // 25 ^ (N-i) long remaining = binExp(25, N - i); long multiply = ((nCi % mod) * (remaining % mod)) % mod; // Add the answer for this prefix // length to the final answer noWays = ((noWays % mod) + (multiply % mod)) % mod; } // Answer is the difference of // allWays and noWays long answer = ((allWays % mod) - (noWays % mod)) % mod; if (answer < 0) answer += mod; // Print the answer System.out.println(answer);} // Driver code public static void main(String[] args) { String str = "abc"; long k = 2; long s = str.length(); long N = s + k; countSubstring(N, s, k);}}// This code is contributed by rutvik_56 |
Python3
# Python3 program for the above approachmod = 1000000007# Function to calculate and return# x^n in log(n) time using# Binary Exponentiationdef binExp(base, power): x = 1 while (power): if (power % 2 == 1): x = (((x % mod) * (base % mod)) % mod) base = (((base % mod) * (base % mod)) % mod) power = power // 2 return x# Function to calculate the factorial# of a numberdef fact(num): result = 1 for i in range(1, num + 1): result = (((result % mod) * (i % mod)) % mod) return result# Function to calculate combinationdef calculate_nCi(N, i): # nCi = ( n! ) / (( n-i )! * i!) nfact = fact(N) ifact = fact(i) dfact = fact(N - i) # Using Euler's theorem of Modular # multiplicative inverse to find # the inverse of a number. # (1/a)%mod=a^(m?2)%mod inv_ifact = binExp(ifact, mod - 2) inv_dfact = binExp(dfact, mod - 2) denm = (((inv_ifact % mod) * (inv_dfact % mod)) % mod) answer = (((nfact % mod) * (denm % mod)) % mod) return answer# Function to find the count of# possible stringsdef countSubstring(N, s, k): # Number of ways to form all # possible strings allWays = binExp(26, N) # Number of ways to form strings # that don't contain the input # string as a subsequence noWays = 0 # Checking for all prefix length # from 0 to |S|-1. for i in range(s): # To calculate nCi nCi = calculate_nCi(N, i) # Select the remaining characters # 25 ^ (N-i) remaining = binExp(25, N - i) multiply = (((nCi % mod) * (remaining % mod)) % mod) # Add the answer for this prefix # length to the final answer noWays =(((noWays % mod) + (multiply % mod)) % mod) # Answer is the difference of # allWays and noWays answer = (((allWays % mod) - (noWays % mod)) % mod) if (answer < 0): answer += mod # Print the answer print(answer)# Driver Codeif __name__ == "__main__": st = "abc" k = 2 s = len(st) N = s + k countSubstring(N, s, k)# This code is contributed by chitranayal |
C#
// C# program for the above approachusing System;class GFG{static readonly long mod = 1000000007;// Function to calculate and return// x^n in log(n) time using// Binary Exponentiationstatic long binExp(long Base, long power){ long x = 1; while (power != 0) { if (power % 2 == 1) x = ((x % mod) * (Base % mod)) % mod; Base = ((Base % mod) * (Base % mod)) % mod; power = power / 2; } return x;}// Function to calculate the factorial// of a numberstatic long fact(long num){ long result = 1; for(long i = 1; i <= num; ++i) { result = ((result % mod) * (i % mod)) % mod; } return result;}// Function to calculate combinationstatic long calculate_nCi(long N, long i){ // nCi = ( n! ) / (( n-i )! * i!) long nfact = fact(N); long ifact = fact(i); long dfact = fact(N - i); // Using Euler's theorem of Modular // multiplicative inverse to find // the inverse of a number. // (1/a)%mod=a^(m?2)%mod long inv_ifact = binExp(ifact, mod - 2); long inv_dfact = binExp(dfact, mod - 2); long denm = ((inv_ifact % mod) * (inv_dfact % mod)) % mod; long answer = ((nfact % mod) * (denm % mod)) % mod; return answer;}// Function to find the count of// possible stringsstatic void countSubstring(long N, long s, long k){ // Number of ways to form all // possible strings long allWays = binExp(26, N); // Number of ways to form strings // that don't contain the input // string as a subsequence long noWays = 0; // Checking for all prefix length // from 0 to |S|-1. for(long i = 0; i < s; ++i) { // To calculate nCi long nCi = calculate_nCi(N, i); // Select the remaining characters // 25 ^ (N-i) long remaining = binExp(25, N - i); long multiply = ((nCi % mod) * (remaining % mod)) % mod; // Add the answer for this prefix // length to the readonly answer noWays = ((noWays % mod) + (multiply % mod)) % mod; } // Answer is the difference of // allWays and noWays long answer = ((allWays % mod) - (noWays % mod)) % mod; if (answer < 0) answer += mod; // Print the answer Console.WriteLine(answer);} // Driver code public static void Main(String[] args) { String str = "abc"; long k = 2; long s = str.Length; long N = s + k; countSubstring(N, s, k);}}// This code is contributed by Rajput-Ji |
Javascript
<script>// JavaScript program for the above approach var mod = 10007; // Function to calculate and return // x^n in log(n) time using // Binary Exponentiation function binExp(base , power) { var x = 1; while (power != 0) { if (power % 2 == 1) x = (((x % mod) * (base % mod)) % mod); base = (((base % mod) * (base % mod)) % mod); power = parseInt(power / 2); } return x; } // Function to calculate the factorial // of a number function fact(num) { var result = 1; for (var i = 1; i <= num; ++i) { result = (((result % mod) * (i % mod)) % mod); } return result; } // Function to calculate combination function calculate_nCi(N , i) { // nCi = ( n! ) / (( n-i )! * i!) var nfact = fact(N); var ifact = fact(i); var dfact = fact(N - i); // Using Euler's theorem of Modular // multiplicative inverse to find // the inverse of a number. // (1/a)%mod=a^(m?2)%mod var inv_ifact = binExp(ifact, mod - 2); var inv_dfact = binExp(dfact, mod - 2); var denm = (((inv_ifact % mod) * (inv_dfact % mod)) % mod); var answer = (((nfact % mod) * (denm % mod)) % mod); return answer; } // Function to find the count of // possible strings function countSubstring(N , s , k) { // Number of ways to form all // possible strings var allWays = binExp(26, N); // Number of ways to form strings // that don't contain the input // string as a subsequence var noWays = 0; // Checking for all prefix length // from 0 to |S|-1. for (var i = 0; i < s; ++i) { // To calculate nCi var nCi = calculate_nCi(N, i); // Select the remaining characters // 25 ^ (N-i) var remaining = binExp(25, N - i); var multiply = (((nCi % mod) * (remaining % mod)) % mod); // Add the answer for this prefix // length to the final answer noWays = (((noWays % mod) + (multiply % mod)) % mod); } // Answer is the difference of // allWays and noWays var answer = (((allWays % mod) - (noWays % mod)) % mod); if (answer < 0) answer += mod; // Print the answer document.write(answer); } // Driver code var str = "abc"; var k = 2; var s = str.length; var N = s + k; countSubstring(N, s, k);// This code contributed by umadevi9616</script> |
Output:
6376
Time Complexity: O(N), where N is the length of the given string.
Auxiliary Space: O(1)
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