Find the Running Median
Problem Statement :
The median of a set of integers is the midpoint value of the data set for which an equal number of integers are less than and greater than the value. To find the median, you must first sort your set of integers in non-decreasing order, then:
If your set contains an odd number of elements, the median is the middle element of the sorted sample. In the sorted set { 1, 2, 3 } , 2 is the median.
If your set contains an even number of elements, the median is the average of the two middle elements of the sorted sample. In the sorted set { 1, 2, 3 , 4}, 2 + 3 /2, 2.5 is the median.
Given an input stream of n integers, you must perform the following task for each ith integer:
Add the ith integer to a running list of integers.
Find the median of the updated list (i.e., for the first element through the ith element).
Print the list's updated median on a new line. The printed value must be a double-precision number scaled to 1 decimal place (i.e., 12.3 format).
Input Format
The first line contains a single integer, n, denoting the number of integers in the data stream.
Each line i of the n subsequent lines contains an integer, ai, to be added to your list.
Constraints
1 < = n <= 10 ^5
0 < = ai <= 10 ^5
Output Format
After each new integer is added to the list, print the list's updated median on a new line as a single double-precision number scaled to 1 decimal place (i.e., 12.3 format).
Solution :
Solution in C :
In C ++ :
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <iomanip>
using namespace std;
int main() {
size_t n;
cin >> n;
vector<size_t> numbers;
for (size_t i = 0; i < n; ++i){
size_t number;
cin >> number;
// keep numbers sorted
numbers.insert(lower_bound(numbers.begin(), numbers.end(), number), number);
double median;
if (i % 2 == 0){ // odd number of elements
median = numbers[i / 2];
}
else{ // even
median = (double(numbers[i / 2]) + numbers[i / 2 + 1]) / 2;
}
cout << fixed << setprecision(1) << median << "\n";
}
return 0;
}
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static class MinComparator implements Comparator<Integer> {
@Override
public int compare(Integer i1, Integer i2) {
if (i1 < i2) {
return -1;
} else if (i1 > i2) {
return 1;
} else {
return 0;
}
}
}
public static class MaxComparator implements Comparator<Integer> {
@Override
public int compare(Integer i1, Integer i2) {
if (i1 > i2) {
return -1;
} else if (i1 < i2) {
return 1;
} else {
return 0;
}
}
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
Queue<Integer> minHeap = new PriorityQueue<>(n/2+1, new MinComparator());
Queue<Integer> maxHeap = new PriorityQueue<>(n/2+1, new MaxComparator());
for (int i = 0; i < n; i++) {
int num = in.nextInt();
if (!maxHeap.isEmpty() && num > maxHeap.peek()) {
minHeap.offer(num);
} else {
maxHeap.offer(num);
}
if (minHeap.size() > maxHeap.size() + 1) {
maxHeap.offer(minHeap.poll());
} else if (maxHeap.size() > minHeap.size() + 1) {
minHeap.offer(maxHeap.poll());
}
double median = 0.0;
if (minHeap.size() == maxHeap.size()) {
median = 0.5 * (minHeap.peek() + maxHeap.peek());
} else {
median = (minHeap.size() > maxHeap.size()) ? minHeap.peek() : maxHeap.peek();
}
System.out.println(String.format("%1.1f", median));
}
}
}
In C :
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
int a[100000], n, i, j, t, k, start, end, mid;
scanf("%d", &n);
for(i=0; i < n; i++) {
scanf("%d", &t);
start = 0;
end = i;
mid = (end-start)/2;
while (start != mid && mid != end) {
if (a[mid] > t)
end = mid;
else if (a[mid] <= t)
start = mid;
mid = start + (end - start)/2;
}
for(j=mid; j < i; j++) {
if (a[j] > t) {
break;
}
}
//memcpy(&a[j+1], &a[j], (i-j)*sizeof(int));
for(k = i; k >= (j+1); k--)
a[k] = a[k-1];
a[j] = t;
//for(j = 0; j <= i; j++)
// printf("%d ", a[j]);
//printf("\n");
if(i%2 == 0) {
printf("%.1f\n", (float)a[i/2]);
} else {
printf("%.1f\n", (float)(a[i/2] + a[(i+1)/2])/2.0);
}
}
return 0;
}
In Python3 :
import heapq as hq
minheap = []
maxheap = []
N = int(input())
for i in range(0, N):
x = int(input())
if i % 2 == 0:
hq.heappush(maxheap, -1*x)
if len(minheap) == 0:
x = -1*maxheap[0]
print(float(x))
continue
elif -1*maxheap[0] > minheap[0]:
toMin = -1*hq.heappop(maxheap)
toMax = hq.heappop(minheap)
hq.heappush(minheap, toMin)
hq.heappush(maxheap, -1*toMax)
x = -1*maxheap[0]
print(float(x))
else:
toMin = -1*hq.heappushpop(maxheap, -1*x)
hq.heappush(minheap, toMin)
x = (-1*maxheap[0]+minheap[0])/2.0
print(x)
View More Similar Problems
Counting On a Tree
Taylor loves trees, and this new challenge has him stumped! Consider a tree, t, consisting of n nodes. Each node is numbered from 1 to n, and each node i has an integer, ci, attached to it. A query on tree t takes the form w x y z. To process a query, you must print the count of ordered pairs of integers ( i , j ) such that the following four conditions are all satisfied: the path from n
View Solution →Polynomial Division
Consider a sequence, c0, c1, . . . , cn-1 , and a polynomial of degree 1 defined as Q(x ) = a * x + b. You must perform q queries on the sequence, where each query is one of the following two types: 1 i x: Replace ci with x. 2 l r: Consider the polynomial and determine whether is divisible by over the field , where . In other words, check if there exists a polynomial with integer coefficie
View Solution →Costly Intervals
Given an array, your goal is to find, for each element, the largest subarray containing it whose cost is at least k. Specifically, let A = [A1, A2, . . . , An ] be an array of length n, and let be the subarray from index l to index r. Also, Let MAX( l, r ) be the largest number in Al. . . r. Let MIN( l, r ) be the smallest number in Al . . .r . Let OR( l , r ) be the bitwise OR of the
View Solution →The Strange Function
One of the most important skills a programmer needs to learn early on is the ability to pose a problem in an abstract way. This skill is important not just for researchers but also in applied fields like software engineering and web development. You are able to solve most of a problem, except for one last subproblem, which you have posed in an abstract way as follows: Given an array consisting
View Solution →Self-Driving Bus
Treeland is a country with n cities and n - 1 roads. There is exactly one path between any two cities. The ruler of Treeland wants to implement a self-driving bus system and asks tree-loving Alex to plan the bus routes. Alex decides that each route must contain a subset of connected cities; a subset of cities is connected if the following two conditions are true: There is a path between ever
View Solution →Unique Colors
You are given an unrooted tree of n nodes numbered from 1 to n . Each node i has a color, ci. Let d( i , j ) be the number of different colors in the path between node i and node j. For each node i, calculate the value of sum, defined as follows: Your task is to print the value of sumi for each node 1 <= i <= n. Input Format The first line contains a single integer, n, denoti
View Solution →