Add js solution to day 13

05- May/13- Count Ways To Build Good Strings/13- Count Ways To Build Good Strings (Ahmed Gamal).js

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+// Author: Ahmed Gamal
+
+// You are given two integers low and high. Return the number of strings you can build such that:
+// The strings are unique and only contain the characters '0' and '1'.
+// The strings contain segments of zero zeros and one ones only. (can contain multiple successive segments of the same type).
+// The length of each string is between low and high, inclusive.
+
+// This is a simple Counting DP problem. We can use dp[i] to represent the number of good strings that start at index i.
+// We can then build the answer by starting at the end of the string and working backwards.
+// our base case is dp[i] = 1 for low <= i <= high, since we will have one good string if we end at index i.
+// The transition is dp[i] = dp[i] + dp[i + zero] + dp[i + one], since we can append a 0 or 1 to any good string that ends at index i + zero or i + one.
+// We can then return dp[0] as our answer.
+
+
+/**
+ * @param {number} low
+ * @param {number} high
+ * @param {number} zero
+ * @param {number} one
+ * @return {number}
+ */
+var countGoodStrings = function(low, high, zero, one) {
+    // MOD is the modulo we use to avoid overflow
+    const MOD = 1000000007;
+
+    // add two numbers and take the modulo
+    let add = (a, b) => (a + b) % MOD;
+
+    // dp[i] is the number of good strings that start at index i (initialized with some additional space to avoid index out of bounds)
+    let dp = Array(high << 1 | 1).fill(0);
+
+    // base case: dp[i] = 1 for low <= i <= high
+    for(let i = low; i <= high; i++) {
+        dp[i] = 1;
+    }  
+
+    // transition: dp[i] = dp[i] + dp[i + zero] + dp[i + one]
+    for(let i = high; ~i; --i) {
+        dp[i] = add(dp[i], add(dp[i + zero], dp[i + one]));
+    }
+
+    return dp[0];
+};