Given an array **A[]** consisting of distinct elements, the task is to obtain the largest possible modulus value that remains after repeatedly replacing adjacent elements by their modulus, starting from the first element, for any possible permutations of the given array.

(…(( A[1] mod A[2]) mod A[3]) …. ) mod A[N])

**Examples:**

Input:A[] = {7, 10, 12}Output:7Explanation:All possible values of the given expression across all permutations of the given array are as follows:

{7, 10, 12} = ((7 % 10) % 12) = 7

{10, 12 7} = ((10 % 12) % 7) = 3

{7, 12, 10} =((7 % 12) % 10) = 7

{10, 7, 12} = ((10 % 7) % 12) = 3

{12, 7, 10} = ((12 % 7) % 10) = 5

{12, 10, 7} = ((12 % 10) % 7) = 2

Therefore, the maximum possible value is 7.

Input:A[] = {20, 30}Output:20Explanation:

The maximum possible value from all the permutations of the given array is 20.

**Naive Approach:** The simplest approach to solve the problem is to generate all permutations of the given array and find the value of the given expression for all permutations. Finally, print the maximum value of the expression obtained. **Time Complexity:** O(N * N!) **Auxiliary Space: **O(N)

**Efficient Approach: **To optimize the above approach, the following observations need to be made:

- For any permutation
A, the value of the expression always lies in the range_{1}…..A_{N}[0, min(A._{2}…..A_{n})-1]- Considering
Kto be the smallest element in the array, the value of the expression will always be K for the permutations having K as the first element.- For all other permutations, the value of the expression will always be less than
K, as shown in the examples above. Therefore,Kis the maximum possible value of the expression for any permutation of the array.- Therefore, the maximum possible value will always be equal to the smallest element of the array.

Therefore, to solve the problem, simply traverse the array and find the minimum element present in the array and print it as the required answer.

Below is the implementation of the above approach:

## C++

`// C++ Program to implement` `// the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the minimum` `// of two numbers` `int` `min(` `int` `a, ` `int` `b)` `{` ` ` `return` `(a > b) ? b : a;` `}` `// Function to find the maximum value` `// possible of the given expression` `// from all permutations of the array` `int` `maximumModuloValue(` `int` `A[], ` `int` `n)` `{` ` ` `// Stores the minimum value` ` ` `// from the array` ` ` `int` `mn = INT_MAX;` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` `mn = min(A[i], mn);` ` ` `}` ` ` `// Return the answer` ` ` `return` `mn;` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `A[] = { 7, 10, 12 };` ` ` `int` `n = (` `sizeof` `(A) / (` `sizeof` `(A[0])));` ` ` `cout << maximumModuloValue(A, n)` ` ` `<< endl;` ` ` `return` `0;` `}` |

## Java

`// Java Program to implement` `// the above approach` `import` `java.io.*;` `class` `GFG{` ` ` `// Function to find the maximum value` ` ` `// possible of the given expression` ` ` `// from all permutations of the array` ` ` `static` `int` `maximumModuloValue(` `int` `A[], ` `int` `n)` ` ` `{` ` ` `// Stores the minimum value` ` ` `// from the array` ` ` `int` `mn = Integer.MAX_VALUE;` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++)` ` ` `{` ` ` `mn = Math.min(A[i], mn);` ` ` `}` ` ` `// Return the answer` ` ` `return` `mn;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `A[] = {` `7` `, ` `10` `, ` `12` `};` ` ` `int` `n = A.length;` ` ` `System.out.println(maximumModuloValue(A, n));` ` ` `}` `}` `// This code is contributed by AnkitRai01` |

## Python3

`# Python3 program to implement` `# the above approach` `import` `sys` `# Function to find the maximum value` `# possible of the given expression` `# from all permutations of the array` `def` `maximumModuloValue(A, n):` ` ` `# Stores the minimum value` ` ` `# from the array` ` ` `mn ` `=` `sys.maxsize` ` ` `for` `i ` `in` `range` `(n):` ` ` `mn ` `=` `min` `(A[i], mn)` ` ` `# Return the answer` ` ` `return` `mn` `# Driver Code` `# Given array arr[]` `A ` `=` `[ ` `7` `, ` `10` `, ` `12` `]` `n ` `=` `len` `(A)` `# Function call` `print` `(maximumModuloValue(A, n))` `# This code is contributed by Shivam Singh` |

## C#

`// C# Program to implement` `// the above approach` `using` `System;` `class` `GFG{` ` ` `// Function to find the maximum value` ` ` `// possible of the given expression` ` ` `// from all permutations of the array` ` ` `static` `int` `maximumModuloValue(` `int` `[]A,` ` ` `int` `n)` ` ` `{` ` ` `// Stores the minimum value` ` ` `// from the array` ` ` `int` `mn = ` `int` `.MaxValue;` ` ` `for` `(` `int` `i = 0; i < n; i++)` ` ` `{` ` ` `mn = Math.Min(A[i], mn);` ` ` `}` ` ` `// Return the answer` ` ` `return` `mn;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `Main(String[] args)` ` ` `{` ` ` `int` `[]A = {7, 10, 12};` ` ` `int` `n = A.Length;` ` ` `Console.WriteLine(maximumModuloValue(A, n));` ` ` `}` `}` `// This code is contributed by shikhasingrajput` |

## Javascript

`<script>` `// javascript Program to implement` `// the above approach` ` ` `// Function to find the maximum value` ` ` `// possible of the given expression` ` ` `// from all permutations of the array` ` ` `function` `maximumModuloValue(A , n) {` ` ` `// Stores the minimum value` ` ` `// from the array` ` ` `var` `mn = Number.MAX_VALUE;` ` ` `for` `(i = 0; i < n; i++) {` ` ` `mn = Math.min(A[i], mn);` ` ` `}` ` ` `// Return the answer` ` ` `return` `mn;` ` ` `}` ` ` `// Driver Code` ` ` ` ` `var` `A = [ 7, 10, 12 ];` ` ` `var` `n = A.length;` ` ` `document.write(maximumModuloValue(A, n));` `// This code contributed by umadevi9616` `</script>` |

**Output:**

7

**Time Complexity: **O(N)**Auxiliary Space: **O(1)

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the **Essential Maths for CP Course** at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**