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3721. Longest Balanced Subarray II
Description
You are given an integer array nums.
A subarray is called balanced if the number of distinct even numbers in the subarray is equal to the number of distinct odd numbers.
Return the length of the longest balanced subarray.
Example 1:
Input: nums = [2,5,4,3]
Output: 4
Explanation:
- The longest balanced subarray is
[2, 5, 4, 3]. - It has 2 distinct even numbers
[2, 4]and 2 distinct odd numbers[5, 3]. Thus, the answer is 4.
Example 2:
Input: nums = [3,2,2,5,4]
Output: 5
Explanation:
- The longest balanced subarray is
[3, 2, 2, 5, 4]. - It has 2 distinct even numbers
[2, 4]and 2 distinct odd numbers[3, 5]. Thus, the answer is 5.
Example 3:
Input: nums = [1,2,3,2]
Output: 3
Explanation:
- The longest balanced subarray is
[2, 3, 2]. - It has 1 distinct even number
[2]and 1 distinct odd number[3]. Thus, the answer is 3.
Constraints:
1 <= nums.length <= 1051 <= nums[i] <= 105
Solutions
Solution 1
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/** * * Idea: * - Treat each distinct odd number as +1, and each distinct even number as -1 * - Maintain prefix sums representing the balance * - When a number appears again, remove its previous contribution * - Use a segment tree to maintain min/max prefix sums with range add * - Binary search on the segment tree to find the earliest equal prefix sum */ class Solution { /** * Segment tree node */ static class Node { int l, r; // segment range int mn, mx; // minimum / maximum prefix sum int lazy; // lazy propagation (range add) } /** * Segment tree supporting: * - range add * - find the smallest index with a given prefix sum */ static class SegmentTree { Node[] tr; SegmentTree(int n) { tr = new Node[n << 2]; for (int i = 0; i < tr.length; i++) { tr[i] = new Node(); } build(1, 0, n); } // Build an empty tree with all prefix sums = 0 void build(int u, int l, int r) { tr[u].l = l; tr[u].r = r; tr[u].mn = tr[u].mx = 0; tr[u].lazy = 0; if (l == r) return; int mid = (l + r) >> 1; build(u << 1, l, mid); build(u << 1 | 1, mid + 1, r); } // Add value v to all prefix sums in [l, r] void modify(int u, int l, int r, int v) { if (tr[u].l >= l && tr[u].r <= r) { apply(u, v); return; } pushdown(u); int mid = (tr[u].l + tr[u].r) >> 1; if (l <= mid) modify(u << 1, l, r, v); if (r > mid) modify(u << 1 | 1, l, r, v); pushup(u); } // Binary search on the segment tree // Find the smallest index where prefix sum == target int query(int u, int target) { if (tr[u].l == tr[u].r) { return tr[u].l; } pushdown(u); int left = u << 1; int right = u << 1 | 1; if (tr[left].mn <= target && target <= tr[left].mx) { return query(left, target); } return query(right, target); } // Apply range add void apply(int u, int v) { tr[u].mn += v; tr[u].mx += v; tr[u].lazy += v; } // Update from children void pushup(int u) { tr[u].mn = Math.min(tr[u << 1].mn, tr[u << 1 | 1].mn); tr[u].mx = Math.max(tr[u << 1].mx, tr[u << 1 | 1].mx); } // Push lazy tag down void pushdown(int u) { if (tr[u].lazy != 0) { apply(u << 1, tr[u].lazy); apply(u << 1 | 1, tr[u].lazy); tr[u].lazy = 0; } } } public int longestBalanced(int[] nums) { int n = nums.length; SegmentTree st = new SegmentTree(n); // last[x] = last position where value x appeared Map<Integer, Integer> last = new HashMap<>(); int now = 0; // current prefix sum int ans = 0; // answer // Enumerate the right endpoint for (int i = 1; i <= n; i++) { int x = nums[i - 1]; int det = (x & 1) == 1 ? 1 : -1; // Remove previous contribution if x appeared before if (last.containsKey(x)) { st.modify(1, last.get(x), n, -det); now -= det; } // Add current contribution last.put(x, i); st.modify(1, i, n, det); now += det; // Find earliest position with the same prefix sum int pos = st.query(1, now); ans = Math.max(ans, i - pos); } return ans; } } -
/* * Segment tree node. * It maintains: * - [l, r] : the covered index range * - mn : minimum prefix sum in this range * - mx : maximum prefix sum in this range * - lazy : lazy propagation value (range add) */ class Node { public: int l = 0, r = 0; int mn = 0, mx = 0; int lazy = 0; }; /* * Segment Tree that supports: * 1. Range add * 2. Query the smallest index whose prefix sum equals a given value */ class SegmentTree { public: SegmentTree(int n) { tr.resize(n << 2); for (int i = 0; i < (int) tr.size(); ++i) { tr[i] = new Node(); } // Build the tree on range [0, n] build(1, 0, n); } /* * Add value v to all prefix sums in range [l, r] */ void modify(int u, int l, int r, int v) { if (tr[u]->l >= l && tr[u]->r <= r) { apply(u, v); return; } pushdown(u); int mid = (tr[u]->l + tr[u]->r) >> 1; if (l <= mid) modify(u << 1, l, r, v); if (r > mid) modify(u << 1 | 1, l, r, v); pushup(u); } /* * Binary search on the segment tree. * Find the smallest index pos such that prefix_sum[pos] == target. * * The key observation: * If target is within [mn, mx] of a segment, then such a position * must exist inside this segment. */ int query(int u, int target) { if (tr[u]->l == tr[u]->r) { return tr[u]->l; } pushdown(u); int lc = u << 1, rc = u << 1 | 1; if (tr[lc]->mn <= target && target <= tr[lc]->mx) { return query(lc, target); } return query(rc, target); } private: vector<Node*> tr; /* * Build an empty segment tree. * Initial prefix sums are all zero. */ void build(int u, int l, int r) { tr[u]->l = l; tr[u]->r = r; tr[u]->mn = tr[u]->mx = 0; tr[u]->lazy = 0; if (l == r) return; int mid = (l + r) >> 1; build(u << 1, l, mid); build(u << 1 | 1, mid + 1, r); } /* * Apply a range add to node u. */ void apply(int u, int v) { tr[u]->mn += v; tr[u]->mx += v; tr[u]->lazy += v; } /* * Push information from children to parent. */ void pushup(int u) { tr[u]->mn = min(tr[u << 1]->mn, tr[u << 1 | 1]->mn); tr[u]->mx = max(tr[u << 1]->mx, tr[u << 1 | 1]->mx); } /* * Push lazy tag down to children. */ void pushdown(int u) { if (tr[u]->lazy != 0) { apply(u << 1, tr[u]->lazy); apply(u << 1 | 1, tr[u]->lazy); tr[u]->lazy = 0; } } }; class Solution { public: int longestBalanced(vector<int>& nums) { int n = nums.size(); SegmentTree st(n); /* * last[x] = the most recent position where value x appeared */ unordered_map<int, int> last; int now = 0; // current prefix sum int ans = 0; // answer /* * Enumerate the right endpoint of the subarray */ for (int i = 1; i <= n; ++i) { int x = nums[i - 1]; /* * Contribution of x: * +1 if x is odd * -1 if x is even */ int det = (x & 1) ? 1 : -1; /* * If x appeared before, remove its previous contribution */ if (last.count(x)) { st.modify(1, last[x], n, -det); now -= det; } /* * Add current contribution of x */ last[x] = i; st.modify(1, i, n, det); now += det; /* * Find the smallest position pos such that * prefix_sum[pos] == now */ int pos = st.query(1, now); /* * Update the maximum balanced subarray length */ ans = max(ans, i - pos); } return ans; } }; -
class Solution: def longestBalanced(self, nums: List[int]) -> int: n = len(nums) class Node: __slots__ = ("l", "r", "mn", "mx", "lazy") def __init__(self): self.l = self.r = 0 self.mn = self.mx = 0 self.lazy = 0 tr = [Node() for _ in range((n + 1) * 4)] def build(u: int, l: int, r: int): tr[u].l, tr[u].r = l, r tr[u].mn = tr[u].mx = tr[u].lazy = 0 if l == r: return mid = (l + r) >> 1 build(u << 1, l, mid) build(u << 1 | 1, mid + 1, r) def apply(u: int, v: int): tr[u].mn += v tr[u].mx += v tr[u].lazy += v def pushdown(u: int): if tr[u].lazy: apply(u << 1, tr[u].lazy) apply(u << 1 | 1, tr[u].lazy) tr[u].lazy = 0 def pushup(u: int): tr[u].mn = min(tr[u << 1].mn, tr[u << 1 | 1].mn) tr[u].mx = max(tr[u << 1].mx, tr[u << 1 | 1].mx) def modify(u: int, l: int, r: int, v: int): if tr[u].l >= l and tr[u].r <= r: apply(u, v) return pushdown(u) mid = (tr[u].l + tr[u].r) >> 1 if l <= mid: modify(u << 1, l, r, v) if r > mid: modify(u << 1 | 1, l, r, v) pushup(u) def query(u: int, target: int) -> int: if tr[u].l == tr[u].r: return tr[u].l pushdown(u) if tr[u << 1].mn <= target <= tr[u << 1].mx: return query(u << 1, target) return query(u << 1 | 1, target) build(1, 0, n) last = {} now = ans = 0 for i, x in enumerate(nums, start=1): det = 1 if (x & 1) else -1 if x in last: modify(1, last[x], n, -det) now -= det last[x] = i modify(1, i, n, det) now += det pos = query(1, now) ans = max(ans, i - pos) return ans -
// Segment tree node type Node struct { l, r int // segment range mn, mx int // minimum / maximum prefix sum lazy int // lazy propagation (range add) } // Segment tree type SegmentTree struct { tr []Node } // Build a segment tree for range [0, n] func NewSegmentTree(n int) *SegmentTree { st := &SegmentTree{ tr: make([]Node, n<<2), } st.build(1, 0, n) return st } // Initialize all prefix sums to 0 func (st *SegmentTree) build(u, l, r int) { st.tr[u] = Node{l: l, r: r, mn: 0, mx: 0, lazy: 0} if l == r { return } mid := (l + r) >> 1 st.build(u<<1, l, mid) st.build(u<<1|1, mid+1, r) } // Add v to all prefix sums in [l, r] func (st *SegmentTree) modify(u, l, r, v int) { if st.tr[u].l >= l && st.tr[u].r <= r { st.apply(u, v) return } st.pushdown(u) mid := (st.tr[u].l + st.tr[u].r) >> 1 if l <= mid { st.modify(u<<1, l, r, v) } if r > mid { st.modify(u<<1|1, l, r, v) } st.pushup(u) } // Binary search on the segment tree // Find the smallest index where prefix sum == target func (st *SegmentTree) query(u, target int) int { if st.tr[u].l == st.tr[u].r { return st.tr[u].l } st.pushdown(u) left, right := u<<1, u<<1|1 if st.tr[left].mn <= target && target <= st.tr[left].mx { return st.query(left, target) } return st.query(right, target) } // Apply range addition func (st *SegmentTree) apply(u, v int) { st.tr[u].mn += v st.tr[u].mx += v st.tr[u].lazy += v } // Update from children func (st *SegmentTree) pushup(u int) { st.tr[u].mn = min(st.tr[u<<1].mn, st.tr[u<<1|1].mn) st.tr[u].mx = max(st.tr[u<<1].mx, st.tr[u<<1|1].mx) } // Push lazy value down func (st *SegmentTree) pushdown(u int) { if st.tr[u].lazy != 0 { v := st.tr[u].lazy st.apply(u<<1, v) st.apply(u<<1|1, v) st.tr[u].lazy = 0 } } // Main function func longestBalanced(nums []int) int { n := len(nums) st := NewSegmentTree(n) // last[x] = last position where value x appeared last := make(map[int]int) now := 0 // current prefix sum ans := 0 // answer // Enumerate right endpoint for i := 1; i <= n; i++ { x := nums[i-1] det := -1 if x&1 == 1 { det = 1 } // Remove previous contribution if x appeared before if pos, ok := last[x]; ok { st.modify(1, pos, n, -det) now -= det } // Add current contribution last[x] = i st.modify(1, i, n, det) now += det // Find earliest position with same prefix sum pos := st.query(1, now) ans = max(ans, i-pos) } return ans } -
function longestBalanced(nums: number[]): number { const n = nums.length; interface Node { l: number; r: number; mn: number; mx: number; lazy: number; } const tr: Node[] = Array.from({ length: (n + 1) * 4 }, () => ({ l: 0, r: 0, mn: 0, mx: 0, lazy: 0, })); function build(u: number, l: number, r: number) { tr[u].l = l; tr[u].r = r; if (l === r) return; const mid = (l + r) >> 1; build(u << 1, l, mid); build((u << 1) | 1, mid + 1, r); } function apply(u: number, v: number) { tr[u].mn += v; tr[u].mx += v; tr[u].lazy += v; } function pushdown(u: number) { if (tr[u].lazy !== 0) { apply(u << 1, tr[u].lazy); apply((u << 1) | 1, tr[u].lazy); tr[u].lazy = 0; } } function pushup(u: number) { tr[u].mn = Math.min(tr[u << 1].mn, tr[(u << 1) | 1].mn); tr[u].mx = Math.max(tr[u << 1].mx, tr[(u << 1) | 1].mx); } function modify(u: number, l: number, r: number, v: number) { if (tr[u].l >= l && tr[u].r <= r) { apply(u, v); return; } pushdown(u); const mid = (tr[u].l + tr[u].r) >> 1; if (l <= mid) modify(u << 1, l, r, v); if (r > mid) modify((u << 1) | 1, l, r, v); pushup(u); } function query(u: number, target: number): number { if (tr[u].l === tr[u].r) return tr[u].l; pushdown(u); if (tr[u << 1].mn <= target && target <= tr[u << 1].mx) { return query(u << 1, target); } return query((u << 1) | 1, target); } build(1, 0, n); const last = new Map<number, number>(); let now = 0, ans = 0; nums.forEach((x, idx) => { const i = idx + 1; const det = x & 1 ? 1 : -1; if (last.has(x)) { modify(1, last.get(x)!, n, -det); now -= det; } last.set(x, i); modify(1, i, n, det); now += det; const pos = query(1, now); ans = Math.max(ans, i - pos); }); return ans; }