![]() ![]() If you have any more approaches or you find an error/bug in the above solutions, please comment down below. What is the time complexity of iterative BFS? Is the time complexity for recursive and iterative approaches are same? If yes, then comment on their space complexities!Ĭomment down your iterative DFS approach for this problem! Why we only pushed those strings to result whose length is equal to the length of input digits? Is the number of digits in the input that maps to 4 letters,Ĭan we call the recursive solution as a backtracking solution?Ĭan you draw the recursion tree for some bigger input? Is the number of digits in the input that maps to 3 letters and String res = combination(digits, n, KEYS) Recurse(combination + letter, next_digits) For all the corresponding keypad characters of the input digit, we will be appending those characters to the front of the queue and then pushing them into the queue until the length of strings in the queue becomes equal to the length of the input digit string.Īdd a base case: If there are no more digits to check that means that the current combination is done. The dial pad mode is displayed above a soft key as for numeric, for upper case. ![]() If we will use the BFS approach, then we can take a queue of strings and will push an empty string to it. Use the dial pad to input numbers, letters, special characters, and a period. However, the answer would be the same in both the iterative approach. If we use the DFS approach then we can use stack and if we use BFS then we can use a queue. Here, if we look at the recursion tree, you can draw conclusions that to get all possible strings, we can make a DFS search or a BFS search. Generally, we can convert all-recursive solution to an iterative solution. Is there an iterative solution to this problem? In short, We keep adding each possible letter recursively and this will generate all the possible strings. In that case, the recursion tree will have further children until the result string length will be equal to the number of input digit What if we have more than 2 digits as input? The SPACE character is typically assigned to the. The green circles show the base case for the recursion when the length of the string is equal to the length of input digits. The 26 letters of the English alphabet are assigned to keys 2 through 9. ![]()
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