add new solutions

solutions/18/q1888/solution.go

@@ -0,0 +1,40 @@
+// Package q1888 implements a solution for https://leetcode.com/problems/minimum-number-of-flips-to-make-the-binary-string-alternating/
+package q1888
+
+// Note: O(1) space is possible.
+
+var buf = make([][2]int, 1e5)
+
+func minFlips(s string) int {
+	backward := buf[:len(s)]
+
+	from0, from1 := 0, 0
+	var alt byte = 0 // alternating byte
+	for i := len(s) - 1; i >= 0; i-- {
+		if s[i]-'0' == alt {
+			from1++
+		} else {
+			from0++
+		}
+		alt = alt ^ 1
+		backward[i] = [2]int{from0, from1}
+	}
+
+	minCost := min(backward[0][0], backward[0][1])
+	from0, from1 = 0, 0
+	for i := range len(s) - 1 {
+		if s[i]-'0' == byte(i%2) {
+			from1++
+		} else {
+			from0++
+		}
+		minCost = min(
+			minCost,
+			from0+backward[i+1][1],
+			from1+backward[i+1][0],
+		)
+	}
+	return minCost
+}
+
+var _ = minFlips