New Problem Solution - "1849. Splitting a String Into Descending Consecutive Values"

README.md

@@ -9,6 +9,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|1849|[Splitting a String Into Descending Consecutive Values](https://leetcode.com/problems/splitting-a-string-into-descending-consecutive-values/) | [C++](./algorithms/cpp/splittingAStringIntoDescendingConsecutiveValues/SplittingAStringIntoDescendingConsecutiveValues.cpp)|Medium|
 |1848|[Minimum Distance to the Target Element](https://leetcode.com/problems/minimum-distance-to-the-target-element/) | [C++](./algorithms/cpp/minimumDistanceToTheTargetElement/MinimumDistanceToTheTargetElement.cpp)|Easy|
 |1847|[Closest Room](https://leetcode.com/problems/closest-room/) | [C++](./algorithms/cpp/closestRoom/ClosestRoom.cpp)|Hard|
 |1846|[Maximum Element After Decreasing and Rearranging](https://leetcode.com/problems/maximum-element-after-decreasing-and-rearranging/) | [C++](./algorithms/cpp/maximumElementAfterDecreasingAndRearranging/MaximumElementAfterDecreasingAndRearranging.cpp)|Medium|

algorithms/cpp/splittingAStringIntoDescendingConsecutiveValues/SplittingAStringIntoDescendingConsecutiveValues.cpp

@@ -0,0 +1,100 @@
+// Source : https://leetcode.com/problems/splitting-a-string-into-descending-consecutive-values/
+// Author : Hao Chen
+// Date   : 2021-05-03
+
+/***************************************************************************************************** 
+ *
+ * You are given a string s that consists of only digits.
+ * 
+ * Check if we can split s into two or more non-empty substrings such that the numerical values of the 
+ * substrings are in descending order and the difference between numerical values of every two 
+ * adjacent substrings is equal to 1.
+ * 
+ * 	For example, the string s = "0090089" can be split into ["0090", "089"] with numerical 
+ * values [90,89]. The values are in descending order and adjacent values differ by 1, so this way is 
+ * valid.
+ * 	Another example, the string s = "001" can be split into ["0", "01"], ["00", "1"], or ["0", 
+ * "0", "1"]. However all the ways are invalid because they have numerical values [0,1], [0,1], and 
+ * [0,0,1] respectively, all of which are not in descending order.
+ * 
+ * Return true if it is possible to split s​​​​​​ as described above, or false otherwise.
+ * 
+ * A substring is a contiguous sequence of characters in a string.
+ * 
+ * Example 1:
+ * 
+ * Input: s = "1234"
+ * Output: false
+ * Explanation: There is no valid way to split s.
+ * 
+ * Example 2:
+ * 
+ * Input: s = "050043"
+ * Output: true
+ * Explanation: s can be split into ["05", "004", "3"] with numerical values [5,4,3].
+ * The values are in descending order with adjacent values differing by 1.
+ * 
+ * Example 3:
+ * 
+ * Input: s = "9080701"
+ * Output: false
+ * Explanation: There is no valid way to split s.
+ * 
+ * Example 4:
+ * 
+ * Input: s = "10009998"
+ * Output: true
+ * Explanation: s can be split into ["100", "099", "98"] with numerical values [100,99,98].
+ * The values are in descending order with adjacent values differing by 1.
+ * 
+ * Constraints:
+ * 
+ * 	1 <= s.length <= 20
+ * 	s only consists of digits.
+ ******************************************************************************************************/
+
+class Solution {
+private:
+    int pos;
+public:
+    bool getNum(string& s, long target) {
+        
+        long n = 0;
+        while(s[pos] == '0') pos++; 
+        while(pos < s.size()){
+            n = (n*10 + s[pos++] - '0') ;
+            if (n == target) return true;
+            if (n > target ) return false; // n is already greater than target
+        }
+        return target == n;
+    }
+    
+    long firstNum(string& s, int len) {
+        long n = 0;
+        while(s[pos] == '0') pos++; 
+        for(; pos< s.size() && len > 0; pos++, len--) {
+            n = (n*10 + s[pos] - '0') ;
+        }
+        return n;  
+    }
+    
+    bool splitString(string s) {
+        pos = 0;
+        long num;
+        for (int len=1; len<= s.size()/2+1; len++) {
+            pos = 0;
+            num = firstNum(s, len); 
+            if (pos >= s.size()) continue; // only have one number
+            
+            bool result = true;
+            while( pos < s.size() ) {
+                if (getNum(s, --num) == false){
+                    result = false;
+                    break;
+                }
+            }
+            if (result) return true;
+        }
+        return false;
+    }
+};