Question: Write a function to find if a given integer x appears more than n/2 times in a sorted array of n integers.
Basically, we need to write a function say isMajority() that takes an array (arr[] ), array’s size (n) and a number to be searched (x) as parameters and returns true if x is a majority element (present more than n/2 times).
Examples:
Input: arr[] = {1, 2, 3, 3, 3, 3, 10}, x = 3
Output: True (x appears more than n/2 times in the given array)
Input: arr[] = {1, 1, 2, 4, 4, 4, 6, 6}, x = 4
Output: False (x doesn't appear more than n/2 times in the given array)
Input: arr[] = {1, 1, 1, 2, 2}, x = 1
Output: True (x appears more than n/2 times in the given array)
METHOD 1 (Using Linear Search)
Linearly search for the first occurrence of the element, once you find it (let at index i), check element at index i + n/2. If element is present at i+n/2 then return 1 else return 0.
Output:
4 appears more than 3 times in arr[]
Time Complexity: O(n)
Auxiliary Space: O(1)
As constant extra space is used.
METHOD 2 (Using Binary Search)
Use binary search methodology to find the first occurrence of the given number. The criteria for binary search is important here.
Javascript
<script> // Javascript Program to check for majority // element in a sorted array */ // If x is present in arr[low...high] // then returns the index of first // occurrence of x, otherwise returns -1 function _binarySearch(arr, low, high, x) { if (high >= low) { let mid = parseInt((low + high) / 2, 10); //low + (high - low)/2; // Check if arr[mid] is the first // occurrence of x. arr[mid] is // first occurrence if x is one of // the following is true: // (i) mid == 0 and arr[mid] == x // (ii) arr[mid-1] < x and arr[mid] == x if ((mid == 0 || x > arr[mid - 1]) && (arr[mid] == x)) return mid; else if (x > arr[mid]) return _binarySearch(arr, (mid + 1), high, x); else return _binarySearch(arr, low, (mid - 1), x); } return -1; } // This function returns true if the x is // present more than n/2 times in arr[] // of size n function isMajority(arr, n, x) { // Find the index of first occurrence // of x in arr[] let i = _binarySearch(arr, 0, n - 1, x); // If element is not present at all, // return false if (i == -1) return false; // check if the element is present // more than n/2 times if (((i + parseInt(n / 2, 10)) <= (n - 1)) && arr[i + parseInt(n / 2, 10)] == x) return true; else return false; } let arr = [ 1, 2, 3, 3, 3, 3, 10 ]; let n = arr.length; let x = 3; if (isMajority(arr, n, x) == true) document.write(x + " appears more than " + parseInt(n / 2, 10) + " times in arr[]"); else document.write(x + " does not appear more than " + parseInt(n / 2, 10) + " times in arr[]"); </script> |
3 appears more than 3 times in arr[]
Time complexity: O(1)
Auxiliary Space: O(1)
Please refer complete article on Check for Majority Element in a sorted array for more details!
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