daily-2026-02-12-3713-longest-balanced-substring-i

problems/3713-longest-balanced-substring-i/analysis.md

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+# 3713. Longest Balanced Substring I
+
+[LeetCode Link](https://leetcode.com/problems/longest-balanced-substring-i/)
+
+Difficulty: Medium
+Topics: Hash Table, String, Counting, Enumeration
+Acceptance Rate: 57.9%
+
+## Hints
+
+### Hint 1
+
+Think about what "balanced" means in terms of character frequencies. What data structure naturally tracks how often each character appears? Consider how you might check the balanced condition efficiently.
+
+### Hint 2
+
+Since the string length is at most 1000, an O(n^2) approach is feasible. Try enumerating substrings by fixing a left endpoint and expanding rightward, maintaining a frequency map as you go. At each step, check whether all frequencies are equal.
+
+### Hint 3
+
+The key insight for checking "balanced" efficiently: you don't need to compare every pair of frequencies. Instead, track the maximum frequency and the number of distinct characters. A substring is balanced if and only if `max_frequency * distinct_count == substring_length`. This lets you check in O(1) per expansion.
+
+## Approach
+
+1. Use two nested loops to enumerate all substrings. The outer loop fixes the left endpoint `i`, and the inner loop expands the right endpoint `j`.
+2. Maintain a frequency array of size 26 (one per lowercase letter) that tracks counts for the current window `s[i..j]`.
+3. As you extend `j`, increment the count for `s[j]`. Also maintain:
+   - `distinct`: the number of characters with a non-zero count.
+   - `maxFreq`: the maximum frequency among all characters.
+4. A substring is balanced when every distinct character has the same frequency. This is equivalent to checking `maxFreq * distinct == j - i + 1` (the total character count equals the number of distinct characters times the max frequency — which is only possible if all frequencies are equal to `maxFreq`).
+5. Whenever the substring is balanced, update the answer with the current length `j - i + 1`.
+6. When moving the left pointer (outer loop increments `i`), reset the frequency array, `distinct`, and `maxFreq`.
+
+### Why the O(1) balanced check works
+
+If there are `d` distinct characters and the maximum frequency is `f`, then the total length is at least `d * f` (each distinct character appears at most `f` times). Equality `d * f == length` forces every character to appear exactly `f` times, which is the definition of balanced.
+
+## Complexity Analysis
+
+Time Complexity: O(n^2) — two nested loops over the string, with O(1) work per step.
+Space Complexity: O(1) — a fixed-size frequency array of 26 elements.
+
+## Edge Cases
+
+- **Single character string**: `"a"` — the entire string is balanced (length 1).
+- **All same characters**: `"aaaa"` — any substring is balanced since there's only one distinct character. Answer is the full length.
+- **All distinct characters**: `"abcde"` — the full string is balanced (each character appears once). Answer is the full length.
+- **No balanced substring longer than 1**: This cannot happen since every single character is trivially balanced, so the answer is always at least 1.
+- **Multiple balanced substrings of the same max length**: Return the length, not the substring itself, so ties don't matter.

problems/3713-longest-balanced-substring-i/solution.go

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+package main
+
+// Enumerate all substrings with two nested loops, maintaining a frequency array.
+// A substring is balanced when maxFreq * distinct == length, checked in O(1).
+// Overall time: O(n^2), space: O(1).
+func longestBalancedSubstring(s string) int {
+	n := len(s)
+	ans := 1 // every single character is balanced
+
+	for i := 0; i < n; i++ {
+		var freq [26]int
+		distinct := 0
+		maxFreq := 0
+
+		for j := i; j < n; j++ {
+			c := s[j] - 'a'
+			freq[c]++
+			if freq[c] == 1 {
+				distinct++
+			}
+			if freq[c] > maxFreq {
+				maxFreq = freq[c]
+			}
+
+			length := j - i + 1
+			if maxFreq*distinct == length && length > ans {
+				ans = length
+			}
+		}
+	}
+
+	return ans
+}

problems/3713-longest-balanced-substring-i/solution_test.go

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+package main
+
+import "testing"
+
+func TestLongestBalancedSubstring(t *testing.T) {
+	tests := []struct {
+		name     string
+		s        string
+		expected int
+	}{
+		{"example 1: abbac", "abbac", 4},
+		{"example 2: zzabccy", "zzabccy", 4},
+		{"example 3: aba", "aba", 2},
+		{"edge case: single character", "a", 1},
+		{"edge case: all same characters", "aaaa", 4},
+		{"edge case: all distinct characters", "abcde", 5},
+		{"edge case: two characters alternating", "ababab", 6},
+		{"edge case: long balanced at end", "xyzaabbcc", 6},
+	}
+
+	for _, tt := range tests {
+		t.Run(tt.name, func(t *testing.T) {
+			result := longestBalancedSubstring(tt.s)
+			if result != tt.expected {
+				t.Errorf("longestBalancedSubstring(%q) = %d, want %d", tt.s, result, tt.expected)
+			}
+		})
+	}
+}