solve

Author: sonjh1217

palindromic-substrings/sonjh1217.swift

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+class Solution {
+    //dynamic programming
+    // O(n^2) time / O(n^2) space
+    func countSubstrings1(_ s: String) -> Int {
+        var count = 0
+        var characters = Array(s)
+        var isPalindromes = Array(repeating: Array(repeating: false, count: characters.count), count: characters.count)
+        
+        for i in 0..<characters.count {
+            isPalindromes[i][i] = true
+            count += 1
+        }
+        
+        for right in 1..<characters.count {
+            for left in 0..<right {
+                if characters[left] == characters[right]
+                    && (isPalindromes[left+1][right-1] || right-left <= 2) {
+                    isPalindromes[left][right] = true
+                    count += 1
+                }
+            }
+        }
+        
+        return count
+    }
+    
+    //two pointers
+    // O(n^2) time / O(n) space
+    func countSubstrings2(_ s: String) -> Int {
+        var count = 0
+        var characters = Array(s)
+        var left: Int
+        var right: Int
+        
+        for i in 0..<characters.count{
+            left = i
+            right = i
+            
+            while left >= 0
+                    && right < characters.count
+                    && characters[left] == characters[right] {
+                count += 1
+                
+                left -= 1
+                right += 1
+            }
+            
+            left = i
+            right = i+1
+            
+            while left >= 0
+                    && right < characters.count
+                    && characters[left] == characters[right] {
+                count += 1
+                
+                left -= 1
+                right += 1
+            }
+        }
+        
+        return count
+    }
+    
+}
+