I'm solving this question as a assignment of the school.But the two of my test cases are coming out wrong when i submit the code? I don't know what went wrong. I have checked various other test cases and corner cases and it all coming out right.
Here is my code:
public static boolean isPermutation(String input1, String input2) {
if(input1.length() != input2.length())
{
return false;
}
int index1 =0;
int index2 =0;
int count=0;
while(index2<input2.length())
{
while(index1<input1.length())
{
if( input1.charAt(index1)==input2.charAt(index2) )
{
index1=0;
count++;
break;
}
index1++;
}
index2++;
}
if(count==input1.length())
{
return true;
}
return false;
}
abcde
baedc
true
abc
cbd
false
A simpler solution would be to sort the characters in both strings and compare those character arrays.
String.toCharArray()
returns an array of characters from a StringArrays.sort(char \[\])
to sort a character arrayArrays.equals(char \[\], char \[\])
to compare the arraysExample
public static void main(String[] args) {
System.out.println(isPermutation("hello", "olleh"));
System.out.println(isPermutation("hell", "leh"));
System.out.println(isPermutation("world", "wdolr"));
}
private static boolean isPermutation(String a, String b) {
char [] aArray = a.toCharArray();
char [] bArray = b.toCharArray();
Arrays.sort(aArray);
Arrays.sort(bArray);
return Arrays.equals(aArray, bArray);
}
A more long-winded solution without sorting would to be check every character in A is also in B
private static boolean isPermutation(String a, String b) {
char[] aArray = a.toCharArray();
char[] bArray = b.toCharArray();
if (a.length() != b.length()) {
return false;
}
int found = 0;
for (int i = 0; i < aArray.length; i++) {
char eachA = aArray[i];
// check each character in A is found in B
for (int k = 0; k < bArray.length; k++) {
if (eachA == bArray[k]) {
found++;
bArray[k] = '\uFFFF'; // clear so we don't find again
break;
}
}
}
return found == a.length();
}