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i was done it , without using Arrays.sort() ---> but Time Compelxity is O(n^2) in worst case scenario :(
char ch1[] = a.toLowerCase().toCharArray(); char ch2[] = b.toLowerCase().toCharArray();
for(int c = 0; c < ch1.length-1; c++){ for(int cc = 0; cc < ch1.length-c-1; cc++){ if(ch1[cc] > ch1[cc+1]){ //swap char temp = ch1[cc]; ch1[cc] = ch1[cc+1]; ch1[cc+1] = temp; } } } String ch1_str = new String(ch1); // for(int c = 0; c < ch2.length-1; c++){ for(int cc = 0; cc < ch2.length-c-1; cc++){ if(ch2[cc] > ch2[cc+1]){ //swap char temp = ch2[cc]; ch2[cc] = ch2[cc+1]; ch2[cc+1] = temp; } } } String ch2_str = new String(ch2); if(ch1_str.equals(ch2_str)){ return true; } return false; }
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Java Anagrams
You are viewing a single comment's thread. Return to all comments →
i was done it , without using Arrays.sort() ---> but Time Compelxity is O(n^2) in worst case scenario :(
char ch1[] = a.toLowerCase().toCharArray(); char ch2[] = b.toLowerCase().toCharArray();