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Sign upAdded Anagram.java #1646
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Added Anagram.java #1646
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strings/Anagram.java
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| Scanner scan=new Scanner(System.in); | ||
| String a=scan.nextLine(); | ||
| String b=scan.nextLine(); | ||
| boolean ana=true; |
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rbshealy
Oct 9, 2020
Contributor
Fix indentation
Fix indentation
strings/Anagram.java
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| for(char c : a.toCharArray()) | ||
| { | ||
| int i=(int)c; | ||
| u[i]++; | ||
| } | ||
| for(char c : b.toCharArray()) | ||
| { | ||
| int i=(int)c; | ||
| v[i]++; | ||
| } |
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rbshealy
Oct 9, 2020
Contributor
What is going on here?
What is going on here?
sj056
Oct 10, 2020
Author
- Here i have initialized 2 arrays u and v first for string a and second for b .
- Converted the strings to lowercase.
- Then alternatively, while converting the char of the string to int(we'll get the ASCII value of that character) then
incrementing that particular index of the string by 1(for eg : if char is 'a' then we'll increment the 97th(ASCII value of
a) index of the array).
- We'll follow these steps for both the strings (hence 2 for loops are used) .
- At the end we'll check if both the arrays are equal then the strings are anagram of each other else not.
- Here i have initialized 2 arrays u and v first for string a and second for b .
- Converted the strings to lowercase.
- Then alternatively, while converting the char of the string to int(we'll get the ASCII value of that character) then
incrementing that particular index of the string by 1(for eg : if char is 'a' then we'll increment the 97th(ASCII value of
a) index of the array). - We'll follow these steps for both the strings (hence 2 for loops are used) .
- At the end we'll check if both the arrays are equal then the strings are anagram of each other else not.
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Describe your change:
References
Checklist:
Fixes: #{$ISSUE_NO}.