Check for Majority Element in a sorted array
Last Updated :
21 Dec, 2023
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Given an array arr of N elements, A majority element in an array arr of size N is an element that appears more than N/2 times in the array. The task is 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 the element at index i + n/2. If the element is present at i+n/2 then return 1 else return 0.
C++
/* 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 |
C
/* C Program to check for majority element in a sorted array */# include <stdio.h># include <stdbool.h>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 program to check above function */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)) printf("%d appears more than %d times in arr[]", x, n/2); else printf("%d does not appear more than %d times in arr[]", x, n/2); return 0;} |
Java
/* Program to check for majority element in a sorted array */import java.io.*;class Majority { static boolean isMajority(int arr[], int n, int x) { int i, last_index = 0; /* get last index according to n (even or odd) */ last_index = (n%2==0)? n/2: n/2+1; /* 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 true; } return false; } /* Driver function to check for above functions*/ public static void main (String[] args) { int arr[] = {1, 2, 3, 4, 4, 4, 4}; int n = arr.length; int x = 4; if (isMajority(arr, n, x)==true) System.out.println(x+" appears more than "+ n/2+" times in arr[]"); else System.out.println(x+" does not appear more than "+ n/2+" times in arr[]"); }}/*This article is contributed by Devesh Agrawal*/ |
Python3
'''Python3 Program to check for majority element in a sorted array'''def isMajority(arr, n, x): # get last index according to n (even or odd) */ last_index = (n//2 + 1) if n % 2 != 0 else (n//2) # search for first occurrence of x in arr[]*/ for i in range(last_index): # check if x is present and is present more than n / 2 times */ if arr[i] == x and arr[i + n//2] == x: return 1# Driver program to check above function */arr = [1, 2, 3, 4, 4, 4, 4]n = len(arr)x = 4if (isMajority(arr, n, x)): print ("% d appears more than % d times in arr[]" %(x, n//2))else: print ("% d does not appear more than % d times in arr[]" %(x, n//2))# This code is contributed by shreyanshi_arun. |
C#
// C# Program to check for majority// element in a sorted array using System;class GFG { static bool isMajority(int[] arr, int n, int x) { int i, last_index = 0; // Get last index according to // n (even or odd) last_index = (n % 2 == 0) ? n / 2 : n / 2 + 1; // 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 true; } return false; } // Driver code public static void Main() { int[] arr = { 1, 2, 3, 4, 4, 4, 4 }; int n = arr.Length; int x = 4; if (isMajority(arr, n, x) == true) Console.Write(x + " appears more than " + n / 2 + " times in arr[]"); else Console.Write(x + " does not appear more than " + n / 2 + " times in arr[]"); }}// This code is contributed by Sam007 |
Javascript
<script> // Javascript Program to check for majority // element in a sorted array function isMajority(arr, n, x) { let i, last_index = 0; // Get last index according to // n (even or odd) last_index = (n % 2 == 0) ? parseInt(n / 2, 10) : parseInt(n / 2, 10) + 1; // 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 + parseInt(n / 2, 10)] == x) return true; } return false; } let arr = [ 1, 2, 3, 4, 4, 4, 4 ]; let n = arr.length; let x = 4; 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> |
PHP
<?php// PHP Program to check for // majority element in a // sorted array // function returns majority// element in a sorted array function isMajority($arr, $n, $x){ $i; // get last index according // to n (even or odd) $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 $arr = array(1, 2, 3, 4, 4, 4, 4); $n = sizeof($arr); $x = 4; if (isMajority($arr, $n, $x)) echo $x, " appears more than " , floor($n / 2), " times in arr[]"; else echo $x, "does not appear more than " , floor($n / 2), "times in arr[]"; // This code is contributed by Ajit?> |
Output
4 appears more than 3 times in arr[]
Time Complexity: O(n)
Auxiliary Space: O(1)
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.
C++
// 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 |
C
/* C Program to check for majority element in a sorted array */# include <stdio.h># include <stdbool.h>/* If x is present in arr[low...high] then returns the index offirst 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/2times 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 offirst 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 program to check above functions */int 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)) printf("%d appears more than %d times in arr[]", x, n/2); else printf("%d does not appear more than %d times in arr[]", x, n/2); return 0;} |
Java
/* Java Program to check for majority element in a sorted array */import java.io.*;class Majority { /* If x is present in arr[low...high] then returns the index of first occurrence of x, otherwise returns -1 */ static 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; } /* This function returns true if the x is present more than n/2 times in arr[] of size n */ static boolean 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; } /*Driver function to check for above functions*/ public static void main (String[] args) { int arr[] = {1, 2, 3, 3, 3, 3, 10}; int n = arr.length; int x = 3; if (isMajority(arr, n, x)==true) System.out.println(x + " appears more than "+ n/2 + " times in arr[]"); else System.out.println(x + " does not appear more than " + n/2 + " times in arr[]"); }}/*This code is contributed by Devesh Agrawal*/ |
Python3
'''Python3 Program to check for majority element in a sorted array'''# This function returns true if the x is present more than n / 2# times in arr[] of size n */def isMajority(arr, n, x): # Find the index of first occurrence of x in arr[] */ 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)) and 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 */def _binarySearch(arr, low, high, x): if high >= low: 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 or x > arr[mid-1]) and (arr[mid] == x): return mid elif x > arr[mid]: return _binarySearch(arr, (mid + 1), high, x) else: return _binarySearch(arr, low, (mid -1), x) return -1# Driver program to check above functions */arr = [1, 2, 3, 3, 3, 3, 10]n = len(arr)x = 3if (isMajority(arr, n, x)): print ("% d appears more than % d times in arr[]" % (x, n//2))else: print ("% d does not appear more than % d times in arr[]" % (x, n//2))# This code is contributed by shreyanshi_arun. |
C#
// C# Program to check for majority// element in a sorted array */using System;class GFG { // If x is present in arr[low...high] // then returns the index of first // occurrence of x, otherwise returns -1 static 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; } // This function returns true if the x is // present more than n/2 times in arr[] // of size n static 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; } //Driver code public static void Main() { int[] arr = { 1, 2, 3, 3, 3, 3, 10 }; int n = arr.Length; int x = 3; if (isMajority(arr, n, x) == true) Console.Write(x + " appears more than " + n / 2 + " times in arr[]"); else Console.Write(x + " does not appear more than " + n / 2 + " times in arr[]"); }}// This code is contributed by Sam007 |
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> |
Output
3 appears more than 3 times in arr[]
Time Complexity: O(log n)
Auxiliary Space: O(1)
Algorithmic Paradigm: Divide and Conquer
