// Package q3714 implements a solution for https://leetcode.com/problems/longest-balanced-substring-ii/
package q3714

func longestBalanced(s string) int {
	longest := 0
	single := 0
	diff3 := make(map[[2]int]int, len(s))
	diffAB, diffAC := 0, 0

	diff2AB, diff2AC, diff2BC := map[int]int{}, map[int]int{}, map[int]int{}
	nDiff2AB, nDiff2AC, nDiff2BC := 0, 0, 0
	lastA, lastB, lastC := -1, -1, -1

	for i := range len(s) {
		// Single
		if i > 0 && s[i] == s[i-1] {
			single++
		} else {
			single = 1
		}
		longest = max(longest, single)

		// Dual
		var process2pair = func(diffIdx map[int]int, offset, i, diff int) int {
			if diff == 0 {
				return i - offset
			}
			if prev, ok := diffIdx[diff]; !ok {
				diffIdx[diff] = i
				return 0
			} else {
				return i - prev
			}
		}
		switch s[i] {
		case 'a':
			lastA = i
			nDiff2BC, diff2BC = 0, map[int]int{}
			nDiff2AB++
			nDiff2AC++
			t1 := process2pair(diff2AB, lastC, i, nDiff2AB)
			t2 := process2pair(diff2AC, lastB, i, nDiff2AC)
			longest = max(longest, t1, t2)

		case 'b':
			lastB = i
			nDiff2AC, diff2AC = 0, map[int]int{}
			nDiff2AB--
			nDiff2BC++
			t1 := process2pair(diff2AB, lastC, i, nDiff2AB)
			t2 := process2pair(diff2BC, lastA, i, nDiff2BC)
			longest = max(longest, t1, t2)

		case 'c':
			lastC = i
			nDiff2AB, diff2AB = 0, map[int]int{}
			nDiff2BC--
			nDiff2AC--
			t1 := process2pair(diff2BC, lastA, i, nDiff2BC)
			t2 := process2pair(diff2AC, lastB, i, nDiff2AC)
			longest = max(longest, t1, t2)
		}

		// Triple
		switch s[i] {
		case 'a':
			diffAB++
			diffAC++
		case 'b':
			diffAB--
		case 'c':
			diffAC--
		}
		key := [2]int{diffAB, diffAC}
		if key == [2]int{0, 0} {
			longest = i + 1
			continue
		}
		if prev, ok := diff3[key]; !ok {
			diff3[key] = i
		} else {
			longest = max(longest, i-prev)
		}
	}

	return longest
}

var _ = longestBalanced