# Count of non-palindromic strings of length M using given N characters

Given two positive integers **N** and **M**, the task is to calculate the number of non-palindromic strings of length **M** using given **N** distinct characters. **Note:** Each distinct characters can be used more than once.

**Examples:**

Input:N = 3, M = 2Output:6Explanation:

Since only 3 characters are given, those 3 characters can be used to form3different strings. Out of these, only 3 strings are palindromic. Hence, the remaining 6 strings are palindromic.^{2}

Input:N = 26, M = 5Output:11863800

**Approach:**

Follow the steps below to solve the problem:

- Total number of strings of length
**M**using given**N**characters will be**N**.^{M} - For a string to be a palindrome, the first half and second half should be equal. For
**even**values of**M**, we need to select only**M/2**characters from the given N characters. For**odd**values, we need to select**M/2 + 1**characters from the given N characters. Since repetitions are allowed, the total number of palindromic strings of length**M**will be**N**.^{(M/2 + M%2)} - The required count of non-palindromic strings is given by the following equation:

N^{M - N(M/2 + M%2)}

Below is the implementation of the above approach:

## C++

`// C++ Program to count` `// non-palindromic strings` `// of length M using N` `// distinct characters` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Iterative Function to calculate` `// base^pow in O(log y)` `unsigned ` `long` `long` `power(` ` ` `unsigned ` `long` `long` `base,` ` ` `unsigned ` `long` `long` `pow` `)` `{` ` ` `unsigned ` `long` `long` `res = 1;` ` ` `while` `(` `pow` `> 0) {` ` ` `if` `(` `pow` `& 1)` ` ` `res = (res * base);` ` ` `base = (base * base);` ` ` `pow` `>>= 1;` ` ` `}` ` ` `return` `res;` `}` `// Function to return the` `// count of non palindromic strings` `unsigned ` `long` `long` `countNonPalindromicString(` ` ` `unsigned ` `long` `long` `n,` ` ` `unsigned ` `long` `long` `m)` `{` ` ` `// Count of strings using n` ` ` `// characters with` ` ` `// repetitions allowed` ` ` `unsigned ` `long` `long` `total` ` ` `= power(n, m);` ` ` ` ` `// Count of palindromic strings` ` ` `unsigned ` `long` `long` `palindrome` ` ` `= power(n, m / 2 + m % 2);` ` ` ` ` `// Count of non-palindromic strings` ` ` `unsigned ` `long` `long` `count` ` ` `= total - palindrome;` ` ` `return` `count;` `}` `int` `main()` `{` ` ` `int` `n = 3, m = 5;` ` ` `cout<< countNonPalindromicString(n, m);` ` ` `return` `0;` `}` |

## Java

`// Java program to count non-palindromic` `// strings of length M using N distinct` `// characters` `import` `java.util.*;` `class` `GFG{` `// Iterative Function to calculate` `// base^pow in O(log y)` `static` `long` `power(` `long` `base, ` `long` `pow)` `{` ` ` `long` `res = ` `1` `;` ` ` `while` `(pow > ` `0` `)` ` ` `{` ` ` `if` `((pow & ` `1` `) == ` `1` `)` ` ` `res = (res * base);` ` ` `base = (base * base);` ` ` `pow >>= ` `1` `;` ` ` `}` ` ` `return` `res;` `}` `// Function to return the` `// count of non palindromic strings` `static` `long` `countNonPalindromicString(` `long` `n,` ` ` `long` `m)` `{` ` ` ` ` `// Count of strings using n` ` ` `// characters with` ` ` `// repetitions allowed` ` ` `long` `total = power(n, m);` ` ` ` ` `// Count of palindromic strings` ` ` `long` `palindrome = power(n, m / ` `2` `+ m % ` `2` `);` ` ` ` ` `// Count of non-palindromic strings` ` ` `long` `count = total - palindrome;` ` ` `return` `count;` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `n = ` `3` `, m = ` `5` `;` ` ` ` ` `System.out.println(` ` ` `countNonPalindromicString(n, m));` `}` `}` `// This code is contributed by offbeat` |

## Python3

`# Python3 program to count non-palindromic strings` `# of length M using N distinct characters` `# Iterative Function to calculate` `# base^pow in O(log y)` `def` `power(base, pwr):` ` ` `res ` `=` `1` ` ` `while` `(pwr > ` `0` `):` ` ` `if` `(pwr & ` `1` `):` ` ` `res ` `=` `res ` `*` `base` ` ` `base ` `=` `base ` `*` `base` ` ` `pwr >>` `=` `1` ` ` `return` `res` `# Function to return the count` `# of non palindromic strings` `def` `countNonPalindromicString(n, m):` ` ` `# Count of strings using n` ` ` `# characters with` ` ` `# repetitions allowed` ` ` `total ` `=` `power(n, m)` ` ` `# Count of palindromic strings` ` ` `palindrome ` `=` `power(n, m ` `/` `/` `2` `+` `m ` `%` `2` `)` ` ` `# Count of non-palindromic strings` ` ` `count ` `=` `total ` `-` `palindrome` ` ` `return` `count` `# Driver code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `n ` `=` `3` ` ` `m ` `=` `5` ` ` ` ` `print` `(countNonPalindromicString(n, m))` `# This code is contributed by Shivam Singh` |

## C#

`// C# program to count non-palindromic` `// strings of length M using N distinct` `// characters` `using` `System;` `class` `GFG{` `// Iterative Function to calculate` `// base^pow in O(log y)` `static` `long` `power(` `long` `Base, ` `long` `pow)` `{` ` ` `long` `res = 1;` ` ` `while` `(pow > 0)` ` ` `{` ` ` `if` `((pow & 1) == 1)` ` ` `res = (res * Base);` ` ` ` ` `Base = (Base * Base);` ` ` `pow >>= 1;` ` ` `}` ` ` `return` `res;` `}` `// Function to return the` `// count of non palindromic strings` `static` `long` `countNonPalindromicString(` `long` `n,` ` ` `long` `m)` `{` ` ` ` ` `// Count of strings using n` ` ` `// characters with` ` ` `// repetitions allowed` ` ` `long` `total = power(n, m);` ` ` ` ` `// Count of palindromic strings` ` ` `long` `palindrome = power(n, m / 2 + m % 2);` ` ` ` ` `// Count of non-palindromic strings` ` ` `long` `count = total - palindrome;` ` ` `return` `count;` `}` `// Driver code` `public` `static` `void` `Main(String[] args)` `{` ` ` `int` `n = 3, m = 5;` ` ` ` ` `Console.WriteLine(` ` ` `countNonPalindromicString(n, m));` `}` `}` `// This code is contributed by PrinciRaj1992` |

## Javascript

`<script>` `// JavaScript program to count non-palindromic` `// strings of length M using N distinct` `// characters` `// Iterative Function to calculate` `// base^pow in O(log y)` `function` `power(base, pow)` `{` ` ` `let res = 1;` ` ` `while` `(pow > 0)` ` ` `{` ` ` `if` `((pow & 1) == 1)` ` ` `res = (res * base);` ` ` ` ` `base = (base * base);` ` ` `pow >>= 1;` ` ` `}` ` ` `return` `res;` `}` `// Function to return the` `// count of non palindromic strings` `function` `countNonPalindromicString(n, m)` `{` ` ` ` ` `// Count of strings using n` ` ` `// characters with` ` ` `// repetitions allowed` ` ` `let total = power(n, m);` ` ` ` ` `// Count of palindromic strings` ` ` `let palindrome = power(n, m / 2 + m % 2);` ` ` ` ` `// Count of non-palindromic strings` ` ` `let count = total - palindrome;` ` ` `return` `count;` `}` ` ` `// Driver Code` `let n = 3, m = 5;` `document.write(countNonPalindromicString(n, m));` `// This code is contributed by sanjoy_62` `</script>` |

**Output:**

216

**Time Complexity:** O(log(N))