Leetcode #1157: Online Majority Element In Subarray
In this guide, we solve Leetcode #1157 Online Majority Element In Subarray in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Design a data structure that efficiently finds the majority element of a given subarray. The majority element of a subarray is an element that occurs threshold times or more in the subarray.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Design, Binary Indexed Tree, Segment Tree, Array, Binary Search
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
Input
["MajorityChecker", "query", "query", "query"]
[[[1, 1, 2, 2, 1, 1]], [0, 5, 4], [0, 3, 3], [2, 3, 2]]
Output
[null, 1, -1, 2]
Explanation
MajorityChecker majorityChecker = new MajorityChecker([1, 1, 2, 2, 1, 1]);
majorityChecker.query(0, 5, 4); // return 1
majorityChecker.query(0, 3, 3); // return -1
majorityChecker.query(2, 3, 2); // return 2
Python Solution
class Node:
__slots__ = ("l", "r", "x", "cnt")
def __init__(self):
self.l = self.r = 0
self.x = self.cnt = 0
class SegmentTree:
def __init__(self, nums):
self.nums = nums
n = len(nums)
self.tr = [Node() for _ in range(n << 2)]
self.build(1, 1, n)
def build(self, u, l, r):
self.tr[u].l, self.tr[u].r = l, r
if l == r:
self.tr[u].x = self.nums[l - 1]
self.tr[u].cnt = 1
return
mid = (l + r) >> 1
self.build(u << 1, l, mid)
self.build(u << 1 | 1, mid + 1, r)
self.pushup(u)
def query(self, u, l, r):
if self.tr[u].l >= l and self.tr[u].r <= r:
return self.tr[u].x, self.tr[u].cnt
mid = (self.tr[u].l + self.tr[u].r) >> 1
if r <= mid:
return self.query(u << 1, l, r)
if l > mid:
return self.query(u << 1 | 1, l, r)
x1, cnt1 = self.query(u << 1, l, r)
x2, cnt2 = self.query(u << 1 | 1, l, r)
if x1 == x2:
return x1, cnt1 + cnt2
if cnt1 >= cnt2:
return x1, cnt1 - cnt2
else:
return x2, cnt2 - cnt1
def pushup(self, u):
if self.tr[u << 1].x == self.tr[u << 1 | 1].x:
self.tr[u].x = self.tr[u << 1].x
self.tr[u].cnt = self.tr[u << 1].cnt + self.tr[u << 1 | 1].cnt
elif self.tr[u << 1].cnt >= self.tr[u << 1 | 1].cnt:
self.tr[u].x = self.tr[u << 1].x
self.tr[u].cnt = self.tr[u << 1].cnt - self.tr[u << 1 | 1].cnt
else:
self.tr[u].x = self.tr[u << 1 | 1].x
self.tr[u].cnt = self.tr[u << 1 | 1].cnt - self.tr[u << 1].cnt
class MajorityChecker:
def __init__(self, arr: List[int]):
self.tree = SegmentTree(arr)
self.d = defaultdict(list)
for i, x in enumerate(arr):
self.d[x].append(i)
def query(self, left: int, right: int, threshold: int) -> int:
x, _ = self.tree.query(1, left + 1, right + 1)
l = bisect_left(self.d[x], left)
r = bisect_left(self.d[x], right + 1)
return x if r - l >= threshold else -1
# Your MajorityChecker object will be instantiated and called as such:
# obj = MajorityChecker(arr)
# param_1 = obj.query(left,right,threshold)
Complexity
The time complexity is O(log n) or O(n log n). The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.