Complementary tutorial 2 – strings++

1. Complementary tutorial 2 – strings++ Prepared by: Valentin Kravtsov

2. char str[] = "abcdefghkl"; int i,str_len = strlen(str); for(i=0 ; i<str_len/2 ; i++){ swap(str+i , str+str_len – 1 – i ); } • Given a string, reverse it Output: lkhgfedcba

3. Given string of size n, rotate it m places to the right: No new arrays are allowed. O(nm) abcdefghk => fghkabcde (n=9, m=4). char tmp,str[] = "abcdefghk"; int i,j, m=4, n = strlen(str); for(i=0;i<m;i++){ tmp = str[n-1]; for(j=n-2;j>=0;j--) { str[j+1] = str[j]; } str[0] = tmp; } Output: fghkabcde

4. Given string of size n, rotate it m places to the right: (n=9, m=4). The efficient version. O(n) abcdefghk => fghkabcde The algorithm: • Identify the new first letter: “f” • Reverse both parts: edcbakhgf • Reverse the entire string: fghkabcde

5. The code: void reverse(char* str, int from, int to){ int i, half_len = (to-from+1)/2; for(i=0;i<half_len;i++){ swap(str+from+i, str+to-i); } } char str[] = "abcdefghk"; int i, m=4, n = strlen(str), break_point = n-m; reverse(str,0,break_point-1); reverse(str,break_point,n-1); reverse(str,0,n-1); Output: fghkabcde

6. Given string of with several words, reverse the order of words, (not characters), no additional arrays/strings are allowed. O(n) ima aba and bamba => bamba and aba ima The algorithm: • Reverse the whole string: abmab dna aba ami • Reverse the letters in each word: bamba and aba ima

7. Given string of with several words, reverse the order of words, (not characters), no additional arrays/strings are allowed. ima aba and bamba => bamba and aba ima char tmp,str[] = "ima aba and bamba"; int end,start=0, n = strlen(str); reverse(str,0,n-1); while( find_next_word(str,&start,&end) ){ reverse(str,start,end); start=end+1; }

8. Finds next word by updating start and end to appropriate values. Return true/false (1/0) if a new word is found. int find_next_word(char* str, int* start, int* end){ for( ; str[*start] && str[*start]==' '; (*start)++){} if( !str[*start] ) return 0; for(*end = *start ; str[*end+1] && str[*end+1] !=' ' ; (*end)++){} return 1; }

9. Compile/not compile questions

10. Compile/not compile questions

11. Compile/not compile questions

12. Compile/not compile questions

13. Given an array of characters of size n, and a new “order”, find if it is “newly ordered”.

14. Your very challenging homework… • Given an array of integers of size n, where one number appears more then n/2 times. Find this number, by scanning the array one time only.