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checkpermutation in o(1) #3

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6007708
checkpermutation in o(1)
rishavranjan943 Oct 2, 2023
d2f3308
duplicate in o(1) space and o(n^2) time
rishavranjan943 Oct 2, 2023
963aa24
o(n) time
rishavranjan943 Oct 2, 2023
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16 changes: 6 additions & 10 deletions 1-Array-And-String/CheckPermutation.cpp
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Original file line number Diff line number Diff line change
Expand Up @@ -18,18 +18,14 @@ bool checkPermutation( string string1, string string2 ) {
if( string1.length() != string2.length() ) {
return false;
}
int charMap[128] = {0};
for( int i=0; i < string1.length(); i++ ) {
charMap[ string1[ i ] ]++;
int xorValue=0;
for(char ch : string1){
xorValue^=ch;
}

for( int i=0; i < string2.length(); i++ ) {
if( charMap[ string2[ i ] ] == 0 ) {
return false;
}
charMap[ string2[ i ] ]--;
for(char ch : string2){
xorValue^=ch;
}
return true;
return xorValue==0;
}

bool checkPermutationUsingSorting( string input1, string input2 ) {
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11 changes: 5 additions & 6 deletions 1-Array-And-String/IsUnique.cpp
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Expand Up @@ -20,12 +20,11 @@ bool isDupUsingHashTable( string input ) {
return true;
}

bool map[ 128 ] = { false };
for( int i=0; i < input.length(); i++ ) {
if( map[ input[ i ] ] == true )
return true;
else
map[ input[ i ] ]= true;
for(int i=0;i<input.length();i++){
for(int j=i+1;j<input.length();j++){
if(input[i]==input[j])
return true;
}
}
return false;
}
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36 changes: 13 additions & 23 deletions 1-Array-And-String/PalindromePermutation.cpp
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Original file line number Diff line number Diff line change
Expand Up @@ -11,33 +11,23 @@
using namespace std;

bool canFormPalindrome( string input ) {
int charFreqCount[ 26 ] = {0};
int charFreqCount[26] = {0};

int charCount = 0;
for( int i = 0; i < input.length(); i++ ) {
if( input[ i ] == ' ' ) {
continue;
for (char c : input) {
if (isspace(c)) continue;
if (isalpha(c)) {
charFreqCount[tolower(c) - 'a']++;
}
}
if( 'a' <= input[ i ] && input[ i ] <= 'z' ) {
charFreqCount[ input[ i ] - 'a' ]++;
charCount++;
continue;

int oddCount = 0;
for (int i = 0; i < 26; i++) {
if (charFreqCount[i] % 2 != 0) {
oddCount++;
}
}
if( 'A' <= input[ i ] && input[ i ] <= 'Z' ) {
charFreqCount[ input[ i ] - 'A' ]++;
charCount++;
}
}

int oddCount = 0;
for( int i=0; i < 26; i++ ) {
if( charFreqCount[ i ] % 2 != 0 )
oddCount++;
}

if( oddCount == 0 && charCount % 2 == 0 ) {
// when char count is even but oddCount is zero.
return true;
return oddCount <= 1;
}

if( charCount % 2 != 0 && oddCount == 1 ) {
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