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string permutation

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Mysterious Macaque answered on November 9, 2021 Popularity 9/10 Helpfulness 3/10

answer string permutation

related permutation in a string

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string permutation

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function permute(array, left = 0, right = array.length - 1) {
    if (left <= right) {

        console.log(array.toString());

        for (let i = left; i <= right; ++i) {

			// swap
            let temp = array[i];
            array[i] = array[left];
            array[left] = temp;

            permute(array, left + 1, right);

			// swap again when backtrack
            temp = array[i];
            array[i] = array[left];
            array[left] = temp;
        }
    }
}

const sequence = ["a", "b", "c"]
permute(sequence)

/*
	Output:
    a,b,c
    a,b,c
    a,b,c
    a,c,b
    b,a,c
    b,a,c
    b,c,a
    c,b,a
    c,b,a
    c,a,b

*/

Popularity 9/10 Helpfulness 3/10 Language javascript

Source: Grepper

Tags: javascript permutation string

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Contributed on Nov 13 2023

Mysterious Macaque

0 Answers Avg Quality 2/10

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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

string permutation

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// Permutation with combination
function combination(array = [], text = "") {
    if (array.length !== 0) {

        for (let i = 0; i < array.length; ++i) {
            const newArray = array.filter((element, index) => index !== i)
            
            text += array[i];

            if (array.length === 1) {
                console.log(text)
            }

            combination(newArray, text)
            text = text.slice(0, -1)
        }
    }
}


const array = [1, 2, 3]

combination(array)

/*
	[Output]
    123
    132
    213
    231
    312
    321
*/

Popularity 7/10 Helpfulness 1/10 Language javascript

Source: Grepper

Tags: javascript permutation per

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Contributed on Nov 13 2023

Mysterious Macaque

0 Answers Avg Quality 2/10

string permutation

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