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all permutations of an array

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Pavya answered on May 1, 2020 Popularity 1/10 Helpfulness 1/10

answer all permutations of an array

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all permutations of an array

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using System;
using System.Collections.Generic;

public class GfG {

    // Function to swap elements in the array
    private static void Swap(int[] arr, int i, int j) {
        int temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
    }

    // Function to find the possible permutations.
    // Initial value of idx is 0.
    private static void Permutations(List<List<int>> res, int[] arr, int idx) {
        if (idx == arr.Length) {
            res.Add(new List<int>(arr));
            return;
        }

        for (int i = idx; i < arr.Length; i++) {
            Swap(arr, idx, i);
            Permutations(res, arr, idx + 1);
            Swap(arr, idx, i); // Backtracking
        }
    }

    // Function to get the permutations
    public static List<List<int>> Permute(int[] arr) {
        List<List<int>> res = new List<List<int>>();
        Permutations(res, arr, 0);
        return res;
    }

    // Driver code
    public static void Main() {
        int[] arr = {1, 2, 3};
        List<List<int>> res = Permute(arr);

        // Printing result
        foreach (List<int> perm in res) {
            Console.WriteLine(string.Join(" ", perm));
        }
    }
}

Popularity 1/10 Helpfulness 1/10 Language whatever

Source: www.geeksforgeeks.org

Tags: whatever

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Contributed on Sep 28 2024

Pavya

0 Answers Avg Quality 2/10

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how to find all permutations of an array with javascript

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function getArrayMutations(arr, perms = [], len = arr.length) {
  if (len === 1) perms.push(arr.slice(0))

  for (let i = 0; i < len; i++) {
    getArrayMutations(arr, perms, len - 1)

    len % 2 // parity dependent adjacent elements swap
      ? [arr[0], arr[len - 1]] = [arr[len - 1], arr[0]]
      : [arr[i], arr[len - 1]] = [arr[len - 1], arr[i]]
  }

  return perms
}

Popularity 10/10 Helpfulness 5/10 Language javascript

Source: Grepper

Tags: find

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Contributed on May 08 2020

m1ke510a

0 Answers Avg Quality 2/10

array permutation

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const permute = (input = [], permutation = []) => {
    if (input.length === 0) return [permutation]; // this will be one of the result

    // choose each number in a loop
    return input.reduce((allPermutations, current) => {
        // reduce the input by removing the current element
        // as we'll fix it by putting it in `permutation` array
        const rest = input.filter(n => n != current);
        return [
            ...allPermutations,
            // fixing our choice in the 2nd arg
            // by concatenationg current with permutation
            ...permute(rest, [...permutation, current])
        ];
    }, []);
}

Popularity 10/10 Helpfulness 3/10 Language javascript

Source: ankeetmaini.dev

Tags: javascript j

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Contributed on Dec 26 2021

Hilarious Hoopoe

0 Answers Avg Quality 2/10

find all permutations of a string

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void permute(string a, int l, int r)  
{  
    // Base case  
    if (l == r)  
        cout<<a<<endl;  
    else
    {  
        // Permutations made  
        for (int i = l; i <= r; i++)  
        {  
  
            // Swapping done  
            swap(a[l], a[i]);  
  
            // Recursion called  
            permute(a, l+1, r);  
  
            //backtrack  
            swap(a[l], a[i]);  
        }  
    }  
}

Popularity 10/10 Helpfulness 3/10 Language whatever

Source: Grepper

Tags: find

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Contributed on May 01 2020

Homeless Hornet

0 Answers Avg Quality 2/10

get all permutations of string

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# get all permutations of string
import itertools
for p in itertools.permutations('123'):
    print(p)					# ( ' 1 ', ' 2 ', ' 3 ') ( ' 1 ' , ' 3 ', ' 2 ' ) ( ' 2 ', ' 1 ', ' 3 ' )

Popularity 10/10 Helpfulness 2/10 Language python

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Contributed on Mar 21 2022

Impossible Impala

0 Answers Avg Quality 2/10

find permutations

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void find_permutations(vector<int> &array, size_t index, vector<int> current_perm, vector<vector<int>> &res){
    if(index == array.size()) 
        res.push_back(current_perm);
    else{
        for(size_t i = 0; i <= current_perm.size(); ++i){
            vector<int> new_perm(current_perm.begin(), current_perm.end());
            new_perm.insert(new_perm.begin()+i, array[index]);
            find_permutations(array, index+1, new_perm, res);
        }
    }
}

Popularity 10/10 Helpfulness 2/10 Language cpp

Source: practice.geeksforgeeks.org

Tags: c++

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Contributed on Aug 05 2021

pecador

0 Answers Avg Quality 2/10

golang convert array to set

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var empty struct{}  // variable with empty value
nums := []int{1,2,2,3,3,3,4}
hmap := make(map[int]struct{})
for i:=0; i<len(nums); i++ {
  hmap[nums[i]] = empty
}

Popularity 6/10 Helpfulness 1/10 Language go

Source: Grepper

Tags: go set

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Contributed on Sep 28 2024

OwO People

0 Answers Avg Quality 2/10

all permutations of an array

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