{"id":8569,"date":"2022-06-21T10:07:17","date_gmt":"2022-06-21T10:07:17","guid":{"rendered":"https:\/\/www.prepbytes.com\/blog\/?p=8569"},"modified":"2023-11-28T04:42:44","modified_gmt":"2023-11-28T04:42:44","slug":"heap-algorithm-for-generating-permutations","status":"publish","type":"post","link":"https:\/\/prepbytes.com\/blog\/heap-algorithm-for-generating-permutations\/","title":{"rendered":"Heap Algorithm for Generating Permutations"},"content":{"rendered":"

\"\"<\/p>\n

Heaps algorithms are employed for generating all conceivable permutations of n-decimals within a number. This algorithm minimizes the need for extensive rearrangements, essentially constructing each permutation from its predecessor by swapping a single element while leaving the remaining elements undisturbed.<\/p>\n

How to use Heap algorithm for Generating Permutations?<\/h2>\n

The Heap’s algorithm is a well-known method for generating all permutations of a set. It efficiently produces all permutations of a given sequence in a specific order without recursion. The algorithm minimizes memory usage and executes in a way that it generates each permutation by only swapping elements in the sequence.<\/p>\n

For n numbers, it takes O(n!) time complexity as there are n! Permutations.<\/p>\n

For example:<\/em><\/strong>
\nLet n = 789
\nPossible permutations are:
\n789, 798, 879, 897, 987, 978<\/p>\n

So, here we have 6 total permutations i.e N! (3!(3 decimal digits) = 6)<\/p>\n

Algorithm:<\/h3>\n
    \n
  • Calculate all the possible permutations of the first n-1 digits adjoining the last element to each of these permutations.<\/li>\n
  • Iterate the digits, and check if n is an odd number then swap the first digit with the last digit and if n is an even number then swap the ith element with the last element.<\/li>\n
  • After repeating the above steps, Algorithm produces all the permutations of N.<\/li>\n<\/ul>\n

    Code Implementation<\/h3>\n\t\t\t\t\t\t