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MajorityElement.cpp
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MajorityElement.cpp
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/*
Problem Introduction
===================
An element of a sequence of length n is called a majority element if it appears in the sequence strictly more
than n/2 times.
Task.
=====
The goal in this code problem is to check whether an input sequence contains a majority element.
Max Time : less than 1 sec.
Input Format.
=============
The 1st line contains an integer n, the next one contains a sequence of n non-negative
integers a0; a1; : : : ; an1.
*/
#include <iostream>
#define ans 3
typedef long long Long;
Long* MergeSort(Long* arr, Long num);
// This is an in-effective algorothm, as it takes around
// O(n^2) running time. A better algorithm would be one that
// uses less than quadratic time. Given after this one.
Long getMajority(Long *arr, Long num) {
Long count = 0;
for (Long i = 0; i < num; ++i) {
for (Long j = 0; j < num; ++j) {
if (arr[i] == arr[j])
++count;
}
if (count >= (num / 2))
return arr[i];
count = 0;
}
return -1;
}
//Optimized algorithm
//This algorithm sorts the array with O(nlogn) time and
// then finds the majority element in O(n) time.
Long findMajority(Long *arr, Long num) {
arr = MergeSort(arr, num); // Merge Sort with O(nlogn) time
Long count = 1, i = 0;
Long b4 = 0, element = 0;
while (i < num-1) { // O(n) time to find majority
if (arr[i] == arr[i + 1])
++count;
else {
if (count >= (num / 2))
return arr[i];
count = 1;
}
++i;
}
if (count >= (num / 2))
return arr[i];
return -1; // No Majority
}
int main() {
std::cout << "'Enter The Number Of Elements and the array will be randomly created, '-1' means no majority'\n\n";
std::cout << "Enter The Number Of Elements : ";
Long num = 0;
std::cin >> num;
Long *arr = new Long[num];
Long *tmp = new Long[num];
for (Long i = 0; i < num; ++i) {
arr[i] = rand() % 5;
tmp[i] = arr[i];
}
std::cout << "Fast Algorithm : Start!\n";
Long a = findMajority(arr, num);
std::cout << "Majority Element : " << a << "\n";
std::cout << "\nSlow Algorithm : Start!\n";
a = getMajority(tmp, num);
std::cout << "Majority Element : " << a << "\n";
}
Long* Merge(Long*arr1, Long*arr2, Long* arr, Long num);
Long* MergeSort(Long *arr, Long num) {
if (num < 2) {
return arr;
}
Long mid = num / 2;
Long* half1 = new Long[mid];
Long *half2 = new Long[num - mid];
half1 = MergeSort(arr, mid);
half2 = MergeSort(&arr[mid], num - mid);
return Merge(half1, half2, arr, num);
}
Long* Merge(Long*arr1, Long*arr2, Long* arr, Long num) {
Long mid = num / 2;
Long h1 = 0, h2 = 0;
Long* temp = new Long[num];
Long i = 0;
while (h1 < mid && h2 < (num - mid)) {
if (arr1[h1] < arr2[h2]) {
temp[i] = arr1[h1];
++h1;
}
else {
temp[i] = arr2[h2];
++h2;
}
++i;
}
while (h1 < mid) {
temp[i] = arr1[h1];
++i; ++h1;
}
while (h2 < (num - mid)) {
temp[i] = arr2[h2];
++h2; ++i;
}
for (Long i = 0; i < num; ++i) {
arr[i] = temp[i];
}
return arr;
}
//.
//