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

Source: Grepper

Tags: get

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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

print all permutations of a string

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def permute(s):
    if len(s) == 0:
        return []
    elif len(s) == 1:
        return [s]
    else:
        permutations = []
        for i in range(len(s)):
            char = s[i]
            remaining_chars = s[:i] + s[i+1:]
            for p in permute(remaining_chars):
                permutations.append(char + p)
        return permutations

string = "abc"  # Replace with the desired input string
print(permute(string))

Popularity 8/10 Helpfulness 2/10 Language python

Source: Grepper

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Contributed on Sep 12 2023

YOU.com

0 Answers Avg Quality 2/10

how to print all permutations of a string

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void permutation(string s)
{
    sort(s.begin(),s.end());
	do{
		cout << s << " ";
	}
    while(next_permutation(s.begin(),s.end()); // std::next_permutation
    
    cout << endl;
}

Popularity 10/10 Helpfulness 2/10 Language whatever

Source: Grepper

Tags: string st

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Contributed on Sep 29 2020

Your Friend

0 Answers Avg Quality 2/10

permutation in a string

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import collections
class Solution:
    def checkInclusion(self, s1, s2):
        """
        :type s1: str
        :type s2: str
        :rtype: bool
        """
        '''
        Apprach 3:
        Just changing the collections.Counter to collections.defaultdict
        improved the speed
        94.36%, Nice. :)
        
        '''
        ctr1 = collections.defaultdict(int)
        ctr2 = collections.defaultdict(int)
        for x in s1: ctr1[x] += 1
        for x in s2[:len(s1)]: ctr2[x] += 1
            
        i = 0; j = len(s1)
        
        while j < len(s2):
            if ctr2 == ctr1: return True
            ctr2[s2[i]] -= 1
            if ctr2[s2[i]] < 1: ctr2.pop(s2[i]); 
            ctr2[s2[j]] = ctr2.get(s2[j], 0) + 1
            i += 1; j += 1
            
        return ctr2 == ctr1
        
        '''
        Approach 2:
        Use the same counter but much more effeciently
        63% beated, okay not bad..
        two pointers i(front) j(last), delete and add accordingly and check for the counts
        
        # Code Below
        '''
        ctr1 = collections.Counter(s1)
        ctr2 = collections.Counter(s2[:len(s1)])
            
        i = 0; j = len(s1)
        
        while j < len(s2):
            if ctr2 == ctr1: return True
            ctr2[s2[i]] -= 1
            if ctr2[s2[i]] < 1: ctr2.pop(s2[i]); 
            ctr2[s2[j]] = ctr2.get(s2[j], 0) + 1
            i += 1; j += 1
            
        return ctr2 == ctr1
    
        
        '''
        1. Approach 1
        Things i Missed: 
            - Had the len(s2) - len(s1) + 1: forgot to add the 1 in the if condition 
        Ultra easy solution
        Slice and check for the equality
        5% beat, yuck !! 
        # Code below
       '''
        ctr1 = collections.Counter(s1)
        i = 0
        while i < len(s2) - len(s1) + 1:
            if s2[i] in ctr1:
                ctr2 = collections.Counter(s2[i: i+len(s1)])
                if ctr1 == ctr2: return True
            i += 1
        return False

Popularity 10/10 Helpfulness 1/10 Language whatever

Source: leetcode.com

Tags: permutation perm

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Contributed on Nov 09 2021

Upset Unicorn

0 Answers Avg Quality 2/10

finding permutations of a string

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