Create 3234-count-the-number-of-substrings-with-dominant-ones.js

Author: everthis

3234-count-the-number-of-substrings-with-dominant-ones.js

@@ -0,0 +1,55 @@
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var numberOfSubstrings = function(s) {
+    const n = s.length;
+    let result = 0;
+
+    // Iterate through possible zero counts (1 to sqrt(n))
+    for (let k = 1; k <= Math.floor(Math.sqrt(n)); k++) {
+        const zeros = [];  // Array to store positions of zeros
+        let lastzero = -1; // Position of the zero before the first zero in our window
+        let ones = 0;      // Count of ones in our current window
+
+        // Scan through the string
+        for (let right = 0; right < n; right++) {
+            if (s[right] === '0') {
+                zeros.push(right);
+                // If we have more than k zeros, remove the leftmost one
+                while (zeros.length > k) {
+                    ones -= (zeros[0] - lastzero - 1);  // Subtract ones between lastzero and the removed zero
+                    lastzero = zeros.shift();
+                }
+            } else {
+                ones++;
+            }
+
+            // If we have exactly k zeros and at least k^2 ones
+            if (zeros.length === k && ones >= k ** 2) {
+                // Add the minimum of:
+                // 1. Number of ways to extend to the left (zeros[0] - lastzero)
+                // 2. Number of ways to extend to the right (ones - k^2 + 1)
+                result += Math.min(zeros[0] - lastzero, ones - k ** 2 + 1);
+            }
+        }
+    }
+
+    // Handle all-ones substrings
+    let i = 0;
+    while (i < n) {
+        if (s[i] === '0') {
+            i++;
+            continue;
+        }
+        let sz = 0;
+        while (i < n && s[i] === '1') {
+            sz++;
+            i++;
+        }
+        // Add number of all-ones substrings
+        result += (sz * (sz + 1)) / 2;
+    }
+
+    return result;
+};