Decode Ways
Problem Statement :
A message containing letters from A-Z can be encoded into numbers using the following mapping:
'A' -> "1"
'B' -> "2"
...
'Z' -> "26"
To decode an encoded message, all the digits must be grouped then mapped back into letters using the reverse of the mapping above (there may be multiple ways). For example, "11106" can be mapped into:
"AAJF" with the grouping (1 1 10 6)
"KJF" with the grouping (11 10 6)
Note that the grouping (1 11 06) is invalid because "06" cannot be mapped into 'F' since "6" is different from "06".
Given a string s containing only digits, return the number of ways to decode it.
The test cases are generated so that the answer fits in a 32-bit integer.
Example 1:
Input: s = "12"
Output: 2
Explanation: "12" could be decoded as "AB" (1 2) or "L" (12).
Example 2:
Input: s = "226"
Output: 3
Explanation: "226" could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).
Example 3:
Input: s = "06"
Output: 0
Explanation: "06" cannot be mapped to "F" because of the leading zero ("6" is different from "06").
Constraints:
1 <= s.length <= 100
s contains only digits and may contain leading zero(s).
Solution :
Solution in C :
int charToint(char *s){
int ans = 0;
ans = (s[0] - '0') * 10 + s[1] - '0';
return ans;
}
int numDecodings(char * s){
int lenS = strlen(s);
if(s[0] == '0')
return 0;
if(lenS == 0 || lenS ==1)
return 1;
int a = 1;
int b;
int temp = charToint(s);
if(temp == 10 || temp == 20)
b = 1;
else if(temp%10 == 0)
return 0;
else if(temp > 26)
b = 1;
else
b = 2;
if(lenS == 2)
return b;
int c;
for(int i = 2; i < lenS; i++ ){
temp = charToint(&s[i-1]);
if(temp == 0)
return 0;
else if(temp > 0 && temp < 10) // 01-09
c = b;
else if(temp == 10 || temp== 20 ) //10, 20
c = a;
else if(temp%10 == 0) //30 40 50 ...90
return 0;
else if(temp > 26)
c = b;
else
c = a + b;
a = b;
b = c;
}
return c;
}
Solution in C++ :
class Solution {
public:
int numDecodings(string s) {
int n = s.length();
if (n == 0 || s[0] == '0') {
return 0;
}
vector<int> dp(2, 0);
dp[0] = 1;
dp[1] = 1;
for (int i = 1; i < n; ++i) {
int ways = 0;
if (s[i] != '0') {
ways += dp[1];
}
if (s[i - 1] == '1' || (s[i - 1] == '2' && s[i] <= '6')) {
ways += dp[0];
}
dp[0] = dp[1];
dp[1] = ways;
}
return dp[1];
}
};
Solution in Java :
class Solution {
public int numDecodings(String s) {
int n = s.length();
if (n == 0) {
return 0;
}
// Initialize the DP array
int[] dp = new int[n + 1];
dp[n] = 1; // There is one way to decode an empty string
// Fill in the DP array from right to left
for (int i = n - 1; i >= 0; i--) {
// Check for leading zero
if (s.charAt(i) == '0') {
dp[i] = 0;
} else {
// Decode single digit
dp[i] += dp[i + 1];
// Decode two digits if possible
if (i + 1 < n && Integer.parseInt(s.substring(i, i + 2)) <= 26) {
dp[i] += dp[i + 2];
}
}
}
return dp[0];
}
}
Solution in Python :
class Solution(object):
def numDecodings(self, s):
n = len(s)
dp = [0] * (n + 1)
dp[n] = 1 # Base case: empty string is one valid decoding
for i in range(n - 1, -1, -1):
if s[i] == '0':
dp[i] = 0
else:
dp[i] = dp[i + 1]
if i + 1 < n and int(s[i:i+2]) <= 26:
dp[i] += dp[i + 2]
return dp[0]
View More Similar Problems
Inserting a Node Into a Sorted Doubly Linked List
Given a reference to the head of a doubly-linked list and an integer ,data , create a new DoublyLinkedListNode object having data value data and insert it at the proper location to maintain the sort. Example head refers to the list 1 <-> 2 <-> 4 - > NULL. data = 3 Return a reference to the new list: 1 <-> 2 <-> 4 - > NULL , Function Description Complete the sortedInsert function
View Solution →Reverse a doubly linked list
This challenge is part of a tutorial track by MyCodeSchool Given the pointer to the head node of a doubly linked list, reverse the order of the nodes in place. That is, change the next and prev pointers of the nodes so that the direction of the list is reversed. Return a reference to the head node of the reversed list. Note: The head node might be NULL to indicate that the list is empty.
View Solution →Tree: Preorder Traversal
Complete the preorder function in the editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's preorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the preOrder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the tree's
View Solution →Tree: Postorder Traversal
Complete the postorder function in the editor below. It received 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's postorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the postorder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the
View Solution →Tree: Inorder Traversal
In this challenge, you are required to implement inorder traversal of a tree. Complete the inorder function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values. Input Format Our hidden tester code passes the root node of a binary tree to your $inOrder* func
View Solution →Tree: Height of a Binary Tree
The height of a binary tree is the number of edges between the tree's root and its furthest leaf. For example, the following binary tree is of height : image Function Description Complete the getHeight or height function in the editor. It must return the height of a binary tree as an integer. getHeight or height has the following parameter(s): root: a reference to the root of a binary
View Solution →