saikrishna chunchu
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4of 4 votes
AnswersGiven two strings, find that the if the letters in both the strings are same? i.e. can we be able to make string2 out of string1 by shuffling the words and vice versa.
- saikrishna chunchu in India| Report Duplicate | Flag | PURGE
Amazon Software Engineer / Developer Algorithm
Can you elaborate the question?
- saikrishna chunchu April 17, 2012Exmples:
String1 - dad, String2 - add
String1 - mad, String2 - dam
String1 - cat, String2 - act
In the above, shuffling the characters in String1 makes String2.
Complexity of the solution:
Lets take that, no collision in the Hashmap, in order to insert or search the data, complexity is O(1) - may not be the case always.
In order to insert a String into the Hasmap, Complexity is O(n)
In order to compare the second String with hasmap, Complexity is O(n).
Hence, total complexity is O(n)+O(n) ==> O(2n). We have not considered the complexity of the hasmap yet.
Space overhead is O(n).
Can we have a better solution??
How about the worst case complexity?
If i take, quick sort for sorting the strings, then O(n2) would be the complexity for each string.
With this solution, to sort two strings it can reach to O(n2) + O(n2).
Can we have better solution?
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- saikrishna chunchu April 18, 2012Here is the solution in JAVA public static void compositions(int num){ for(int i =1;i<num;i +=2){ int rem = num%i; int divident = num/i; if(rem%2 !=0) System.out.print(rem); if(rem ==0 || rem%2 !=0){ // Print the composition based on divident for(int j=divident;j>0;j--) System.out.print("+"+i); System.out.println(""); if(i>1){ // get the i in 1s String iIn1s = ""; for(int temp=i;temp>0;temp--) iIn1s += "+"+1; for(int j=1;j<divident;j++){ if(rem%2 !=0) System.out.print(rem); for(int k=j;k>0;k--) System.out.print(iIn1s); for(int l=(divident-j);l>0;l--) System.out.print("+"+i); System.out.println(""); } } } } }