![]() ![]() Method 2 – In-Built Method – All permutations While calling the function, we obviously have to pass the array and indexes as 0 and length-1. (Refer to this)īelow is an output printing all permutation for an array. Note here that we use list(arr) to make sure we are doing a deep copy and not a shallow copy. ![]() We append all the permutation results into an array final. Notice that we keep passing smaller parts of the same array to the same function by modifying the index values and hence generate all possible permutations. At the end of that recursion, we will have all possible permutations Implementation in PythonĬonsider the following program, final = list() Hence, this process continues until we reach the last element and try it as the first element, in the first recursion depth. Now here, 3 is tried as the first element. In this recursion, this has to be the one and only first element, hence a permutation is printed and control goes back by one recursion depth. Then we call the array with the remaining element i.e. This recursion will take 2 as the first element. We take 1 as first element, then for the remaining part of the array, we call the same function. The idea is to take up every element in the array and place it at the beginning and for every such case, recursively do the same for a smaller instance of the same array. Method 1: generate all possible permutations in Python ![]() Prerequisites: Basics of loops and conditionals in Python. Hence if there is a repetition of elements in the array, the same permutation may occur twice. We consider numeric elements in an array here and do not consider repetition of the same elements. This post deals with methods to generate all possible permutations in Python, of a given set of elements. ![]()
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