Check for Majority in a Sorted Array

Given a sorted array arr[] of size n, find if there is a majority element in the array or not. An element is called a majority element if it appears more than n/2 times in the array.

Examples:

Input: arr[] = [1, 2, 3, 3, 3, 3, 10]
Output: true
Explanation: 3 is majority in the array

Input: arr[] = [1, 1, 2, 4, 4, 4, 6, 6]
Output: false
Explanation: There is no majority in the array.

[Naive Approach] Linearly Count Middle - O(n) Time and O(1) Space

The idea is based on the property that if there is a majority in a sorted array, then it must be present at the mid index. So, we first take the middle element arr[n/2] and count how many times it appears in the array. If its count is greater than n/2, then it is a majority element; otherwise, it is not.

true

[Expected Approach] Binary Search - O(log n) Time and O(1) Space

We first take the middle element x = arr[n/2] and use Binary Search to find its first occurrence. Let the first occurrence index be i. If x is a majority element, then the element at index i + n/2 must also be equal to x. Otherwise, x cannot appear more than n/2 times.

• Find the middle element x = arr[n/2]
• Apply binary search to find the first occurrence of x
• If x is not found, return false
• Let i be the first occurrence index
• Check if: i + n/2 < n && arr[i + n/2] == x
• If true, return true, otherwise return false

Consider: arr[] = [1, 2, 3, 3, 3, 3, 10]

Middle element: x = arr[7/2] = arr[3] = 3

Step 1: Find First Occurrence of x

Initial range: low = 0, high = 3

Iteration 1

• mid = (0 + 3) / 2 = 1
• arr[mid] = 2
• Since arr[mid] < x, move right: low = mid + 1 = 2

Iteration 2

• mid = (2 + 3) / 2 = 2
• arr[mid] = 3
• Check first occurrence: arr[mid - 1] = arr[1] = 2 < 3
• So, first occurrence is found at index: i = 2

Step 2: Check Majority Condition

• i + n/2 = 2 + 3 = 5
• arr[5] = 3
• Since arr[i + n/2] == x, the middle element appears more than n/2 times.

Final Result: true

true