C++ Program to Check for Majority Element in a sorted array
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.
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/* C++ Program to check for majority element in a sorted array */#include<bits/stdc++.h>using namespace std;bool isMajority(int arr[], int n, int x){ int i; /* get last index according to n (even or odd) */ int last_index = n % 2 ? (n / 2 + 1): (n / 2); /* search for first occurrence of x in arr[]*/ for (i = 0; i < last_index; i++) { /* check if x is present and is present more than n/2 times */ if (arr[i] == x && arr[i + n / 2] == x) return 1; } return 0;}/* Driver code */int main(){ int arr[] ={1, 2, 3, 4, 4, 4, 4}; int n = sizeof(arr)/sizeof(arr[0]); int x = 4; if (isMajority(arr, n, x)) cout << x <<" appears more than "<< n/2 << " times in arr[]"<< endl; else cout <<x <<" does not appear more than" << n/2 <<" times in arr[]" << endl;return 0;}// This code is contributed by shivanisinghss2110 |
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.
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// C++ program to check for majority// element in a sorted array #include<bits/stdc++.h>using namespace std;// If x is present in arr[low...high] // then returns the index of first// occurrence of x, otherwise returns -1 int _binarySearch(int arr[], int low, int high, int x);// This function returns true if the x// is present more than n/2 times in// arr[] of size n bool isMajority(int arr[], int n, int x){ // Find the index of first occurrence // of x in arr[] int 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 + n / 2) <= (n - 1)) && arr[i + n / 2] == x) return true; else return false;}// If x is present in arr[low...high] then // returns the index of first occurrence // of x, otherwise returns -1 int _binarySearch(int arr[], int low, int high, int x){ if (high >= low) { int mid = (low + high)/2; /*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;}// Driver codeint main(){ int arr[] = { 1, 2, 3, 3, 3, 3, 10 }; int n = sizeof(arr) / sizeof(arr[0]); int x = 3; if (isMajority(arr, n, x)) cout << x << " appears more than " << n / 2 << " times in arr[]" << endl; else cout << x << " does not appear more than" << n / 2 << " times in arr[]" << endl; return 0;}// This code is contributed by shivanisinghss2110 |
Output:
3 appears more than 3 times in arr[]
Time Complexity: O(Logn)
Auxiliary Space: O(1)
As constant extra space is used.
Algorithmic Paradigm: Divide and Conquer
METHOD 3: If it is already given that the array is sorted and there exists a majority element, checking if a particular element is as easy as checking if the middle element of the array is the number we are checking against.
Since a majority element occurs more than n/2 times in an array, it will always be the middle element. We can use this logic to check if the given number is the majority element.
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#include <iostream>using namespace std;bool isMajorityElement(int arr[], int n, int key){ if (arr[n / 2] == key) return true; else return false;}int main(){ int arr[] = { 1, 2, 3, 3, 3, 3, 10 }; int n = sizeof(arr) / sizeof(arr[0]); int x = 3; if (isMajorityElement(arr, n, x)) cout << x << " appears more than " << n / 2 << " times in arr[]" << endl; else cout << x << " does not appear more than" << n / 2 << " times in arr[]" << endl; return 0;}// This code is contributed by shivanisinghss2110 |
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!
