{"id":"majority-check-single-bisect","text":"`is_majority_element` uses exactly one `bisect_left` call to achieve O(log n) majority checking — it avoids a second binary search by checking `nums[first + n//2] == target` directly.","truth_value":"IN","source":"entries/2026/06/06/check-if-a-number-is-majority-element-in-a-sorted-array-solution.md","source_url":"","source_hash":"","justifications":[],"dependents":[],"metadata":{},"created_at":"","updated_at":"","reviewed_at":"","verified_at":"","retracted_at":"","explanation":{"steps":[{"node":"majority-check-single-bisect","truth_value":"IN","reason":"premise"}]}}