Add solution #730

Author: JoshCrozier

README.md

@@ -553,6 +553,7 @@
 726|[Number of Atoms](./0726-number-of-atoms.js)|Hard|
 728|[Self Dividing Numbers](./0728-self-dividing-numbers.js)|Easy|
 729|[My Calendar I](./0729-my-calendar-i.js)|Medium|
+730|[Count Different Palindromic Subsequences](./0730-count-different-palindromic-subsequences.js)|Hard|
 733|[Flood Fill](./0733-flood-fill.js)|Easy|
 735|[Asteroid Collision](./0735-asteroid-collision.js)|Medium|
 739|[Daily Temperatures](./0739-daily-temperatures.js)|Medium|

solutions/0730-count-different-palindromic-subsequences.js

@@ -0,0 +1,59 @@
+/**
+ * 730. Count Different Palindromic Subsequences
+ * https://leetcode.com/problems/count-different-palindromic-subsequences/
+ * Difficulty: Hard
+ *
+ * Given a string s, return the number of different non-empty palindromic subsequences in s.
+ * Since the answer may be very large, return it modulo 109 + 7.
+ *
+ * A subsequence of a string is obtained by deleting zero or more characters from the string.
+ *
+ * A sequence is palindromic if it is equal to the sequence reversed.
+ *
+ * Two sequences a1, a2, ... and b1, b2, ... are different if there is some i for which ai != bi.
+ */
+
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var countPalindromicSubsequences = function(s) {
+  const MOD = 1e9 + 7;
+  const dp = new Array(s.length).fill().map(() => {
+    return new Array(s.length).fill(0);
+  });
+
+  for (let i = 0; i < s.length; i++) {
+    dp[i][i] = 1;
+  }
+
+  for (let len = 2; len <= s.length; len++) {
+    for (let start = 0; start + len <= s.length; start++) {
+      const end = start + len - 1;
+      const charStart = s[start];
+      const charEnd = s[end];
+
+      if (charStart !== charEnd) {
+        dp[start][end] = (dp[start + 1][end] + dp[start][end - 1] - dp[start + 1][end - 1]) % MOD;
+      } else {
+        let left = start + 1;
+        let right = end - 1;
+
+        while (left <= right && s[left] !== charStart) left++;
+        while (left <= right && s[right] !== charStart) right--;
+
+        if (left > right) {
+          dp[start][end] = (2 * dp[start + 1][end - 1] + 2) % MOD;
+        } else if (left === right) {
+          dp[start][end] = (2 * dp[start + 1][end - 1] + 1) % MOD;
+        } else {
+          dp[start][end] = (2 * dp[start + 1][end - 1] - dp[left + 1][right - 1]) % MOD;
+        }
+      }
+
+      if (dp[start][end] < 0) dp[start][end] += MOD;
+    }
+  }
+
+  return dp[0][s.length - 1];
+};