# Number of distinct words of size N with at most K contiguous vowels

• Difficulty Level : Hard
• Last Updated : 06 Aug, 2021

Given two integers N and K, the task is to find the number of distinct strings consisting of lowercase alphabets of length N that can be formed with at-most K contiguous vowels. As the answer may be too large, print answer%1000000007.

Input: N = 1, K = 0
Output: 21
Explanation: All the 21 consonants are there which has 0 contiguous vowels and are of length 1.

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Input: N = 1, K = 1
Output: 26

Approach: The idea to solve this problem is based on dynamic programming. Follow the steps below to solve the problem:

• Let dp[i][j] be the number of ways to make distinct strings of length i where the last j characters of the string are vowels.
• So the states of dynamic programming are:
• If j = 0, then dp[i][j] = (dp[i-1] + dp[i-1] +……+ dp[i-1][K])*21(represented by the integer variable sum) because the last added character should be a consonant than only the value of j will become 0 irrespective of its value on previous states.
• If i<j then dp[i][j] = 0. Since it is not possible to create a string containing j vowels and has a length less than j.
• If i == j, then dp[i][j] = 5i because the number of characters in the string is equal to the number of vowels, therefore all the characters should be vowels.
• If j<i then dp[i][j] = dp[i-1][j-1]*5 because a string of length i with last j characters vowel can be made only if the last character is the vowel and the string of length i-1 has last j – 1 character as vowels.
• Print the sum of dp[n] + dp[n] + …… + dp[n][K] as the answer.

Below is the implementation of the above Approach

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Power function to calculate``// long powers with mod``long` `long` `int` `power(``long` `long` `int` `x,``                    ``long` `long` `int` `y,``                    ``long` `long` `int` `p)``{``    ``long` `long` `int` `res = 1ll;` `    ``x = x % p;` `    ``if` `(x == 0)``        ``return` `0;` `    ``while` `(y > 0) {` `        ``if` `(y & 1)``            ``res = (res * x) % p;``        ``y = y >> 1;``        ``x = (x * x) % p;``    ``}``    ``return` `res;``}` `// Function for finding number of ways to``// create string with length N and atmost``// K contiguous vowels``int` `kvowelwords(``int` `N, ``int` `K)``{` `    ``long` `long` `int` `i, j;``    ``long` `long` `int` `MOD = 1000000007;` `    ``// Array dp to store number of ways``    ``long` `long` `int` `dp[N + 1][K + 1] = { 0 };` `    ``long` `long` `int` `sum = 1;``    ``for` `(i = 1; i <= N; i++) {` `        ``// dp[i] = (dp[i-1]+dp[i-1]..dp[i-1][k])*21``        ``dp[i] = sum * 21;``        ``dp[i] %= MOD;` `        ``// Now setting sum to be dp[i]``        ``sum = dp[i];` `        ``for` `(j = 1; j <= K; j++) {``            ``// If j>i, no ways are possible to create``            ``// a string with length i and vowel j``            ``if` `(j > i)``                ``dp[i][j] = 0;``            ``else` `if` `(j == i) {``                ``// If j = i all the character should``                ``// be vowel``                ``dp[i][j] = power(5ll, i, MOD);``            ``}``            ``else` `{``                ``// dp[i][j] relation with dp[i-1][j-1]``                ``dp[i][j] = dp[i - 1][j - 1] * 5;``            ``}` `            ``dp[i][j] %= MOD;` `            ``// Adding dp[i][j] in the sum``            ``sum += dp[i][j];``            ``sum %= MOD;``        ``}``    ``}` `    ``return` `sum;``}``// Driver Program``int` `main()``{``    ``// Input``    ``int` `N = 3;``    ``int` `K = 3;` `    ``// Function Call``    ``cout << kvowelwords(N, K) << endl;``    ``return` `0;``}`

## Java

 `// Java program for the above approach``class` `GFG{` `// Power function to calculate``// long powers with mod``static` `int` `power(``int` `x, ``int` `y, ``int` `p)``{``    ``int` `res = ``1``;``    ``x = x % p;`` ` `    ``if` `(x == ``0``)``        ``return` `0``;`` ` `    ``while` `(y > ``0``)``    ``{``        ``if` `((y & ``1``) != ``0``)``            ``res = (res * x) % p;``            ` `        ``y = y >> ``1``;``        ``x = (x * x) % p;``    ``}``    ``return` `res;``}`` ` `// Function for finding number of ways to``// create string with length N and atmost``// K contiguous vowels``static` `int` `kvowelwords(``int` `N, ``int` `K)``{``    ``int` `i, j;``    ``int` `MOD = ``1000000007``;`` ` `    ``// Array dp to store number of ways``    ``int``[][] dp = ``new` `int``[N + ``1``][K + ``1``] ;`` ` `    ``int` `sum = ``1``;``    ``for``(i = ``1``; i <= N; i++)``    ``{``        ` `        ``// dp[i] = (dp[i-1]+dp[i-1]..dp[i-1][k])*21``        ``dp[i][``0``] = sum * ``21``;``        ``dp[i][``0``] %= MOD;`` ` `        ``// Now setting sum to be dp[i]``        ``sum = dp[i][``0``];`` ` `        ``for``(j = ``1``; j <= K; j++)``        ``{``            ` `            ``// If j>i, no ways are possible to create``            ``// a string with length i and vowel j``            ``if` `(j > i)``                ``dp[i][j] = ``0``;``                ` `            ``else` `if` `(j == i)``            ``{``                ` `                ``// If j = i all the character should``                ``// be vowel``                ``dp[i][j] = power(``5``, i, MOD);``            ``}``            ``else``            ``{``                ` `                ``// dp[i][j] relation with dp[i-1][j-1]``                ``dp[i][j] = dp[i - ``1``][j - ``1``] * ``5``;``            ``}`` ` `            ``dp[i][j] %= MOD;`` ` `            ``// Adding dp[i][j] in the sum``            ``sum += dp[i][j];``            ``sum %= MOD;``        ``}``    ``}``    ``return` `sum;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ` `    ``// Input``    ``int` `N = ``3``;``    ``int` `K = ``3``;`` ` `    ``// Function Call``    ``System.out.println( kvowelwords(N, K));``}``}` `// This code is contributed by target_2`

## Python3

 `# Python3 program for the above approach` `# Power function to calculate``# long powers with mod``def` `power(x, y, p):``    ` `    ``res ``=` `1``    ``x ``=` `x ``%` `p` `    ``if` `(x ``=``=` `0``):``        ``return` `0` `    ``while` `(y > ``0``):``        ``if` `(y & ``1``):``            ``res ``=` `(res ``*` `x) ``%` `p``            ` `        ``y ``=` `y >> ``1``        ``x ``=` `(x ``*` `x) ``%` `p``        ` `    ``return` `res` `# Function for finding number of ways to``# create string with length N and atmost``# K contiguous vowels``def` `kvowelwords(N, K):` `    ``i, j ``=` `0``, ``0``    ``MOD ``=` `1000000007` `    ``# Array dp to store number of ways``    ``dp ``=` `[[``0` `for` `i ``in` `range``(K ``+` `1``)]``             ``for` `i ``in` `range``(N ``+` `1``)]` `    ``sum` `=` `1``    ``for` `i ``in` `range``(``1``, N ``+` `1``):``        ` `        ``#dp[i] = (dp[i-1]+dp[i-1]..dp[i-1][k])*21``        ``dp[i][``0``] ``=` `sum` `*` `21``        ``dp[i][``0``] ``%``=` `MOD` `        ``# Now setting sum to be dp[i]``        ``sum` `=` `dp[i][``0``]` `        ``for` `j ``in` `range``(``1``, K ``+` `1``):``            ` `            ``# If j>i, no ways are possible to create``            ``# a string with length i and vowel j``            ``if` `(j > i):``                ``dp[i][j] ``=` `0``            ``elif` `(j ``=``=` `i):``                ` `                ``# If j = i all the character should``                ``# be vowel``                ``dp[i][j] ``=` `power(``5``, i, MOD)``            ``else``:``                ` `                ``# dp[i][j] relation with dp[i-1][j-1]``                ``dp[i][j] ``=` `dp[i ``-` `1``][j ``-` `1``] ``*` `5` `            ``dp[i][j] ``%``=` `MOD` `            ``# Adding dp[i][j] in the sum``            ``sum` `+``=` `dp[i][j]``            ``sum` `%``=` `MOD` `    ``return` `sum``    ` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``# Input``    ``N ``=` `3``    ``K ``=` `3` `    ``# Function Call``    ``print` `(kvowelwords(N, K))` `# This code is contributed by mohit kumar 29`

## C#

 `// C# program for the above approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG{` `// Power function to calculate``// long powers with mod``static` `int` `power(``int` `x, ``int` `y, ``int` `p)``{``    ``int` `res = 1;``    ``x = x % p;`` ` `    ``if` `(x == 0)``        ``return` `0;`` ` `    ``while` `(y > 0)``    ``{``        ``if` `((y & 1) != 0)``            ``res = (res * x) % p;``            ` `        ``y = y >> 1;``        ``x = (x * x) % p;``    ``}``    ``return` `res;``}`` ` `// Function for finding number of ways to``// create string with length N and atmost``// K contiguous vowels``static` `int` `kvowelwords(``int` `N, ``int` `K)``{``    ``int` `i, j;``    ``int` `MOD = 1000000007;`` ` `    ``// Array dp to store number of ways``    ``int``[,] dp = ``new` `int``[N + 1, K + 1];`` ` `    ``int` `sum = 1;``    ``for``(i = 1; i <= N; i++)``    ``{``        ` `        ``// dp[i] = (dp[i-1, 0]+dp[i-1, 1]..dp[i-1][k])*21``        ``dp[i, 0] = sum * 21;``        ``dp[i, 0] %= MOD;`` ` `        ``// Now setting sum to be dp[i]``        ``sum = dp[i, 0];`` ` `        ``for``(j = 1; j <= K; j++)``        ``{``            ` `            ``// If j>i, no ways are possible to create``            ``// a string with length i and vowel j``            ``if` `(j > i)``                ``dp[i, j] = 0;``                ` `            ``else` `if` `(j == i)``            ``{``                ` `                ``// If j = i all the character should``                ``// be vowel``                ``dp[i, j] = power(5, i, MOD);``            ``}``            ``else``            ``{``                ` `                ``// dp[i][j] relation with dp[i-1][j-1]``                ``dp[i, j] = dp[i - 1, j - 1] * 5;``            ``}`` ` `            ``dp[i, j] %= MOD;`` ` `            ``// Adding dp[i][j] in the sum``            ``sum += dp[i, j];``            ``sum %= MOD;``        ``}``    ``}``    ``return` `sum;``}` `// Driver Code``public` `static` `void` `Main()``{``    ` `    ``// Input``    ``int` `N = 3;``    ``int` `K = 3;`` ` `    ``// Function Call``    ``Console.Write(kvowelwords(N, K));``}``}` `// This code is contributed by code_hunt`

## Javascript

 ``
Output:
`17576`

Time Complexity: O(N×K)
Auxiliary Space: O(N×K)

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