This logic is to find the number n in the array in which the range between n and n + 5 will include the most numbers in the array. I came up with a solution but it requires a nested loop, therefore it is kind of slow. Is there any way to improve its performance? Thanks in advance.
The array is guaranteed to sorted.
int[] myArray = new int[]{1,2,4,5,7,9,15,19};
int bestNumber = 0;
int MaxMatchFound = 0;
for (int o = 0; o < myArray.Length; o++)
{
int TempMatchFound = 0;
for (int i = 0; i < myArray.Length; i++)
{
if (myArray[i] >= myArray[o] && myArray[i] <= (myArray[o] + 5))
{
TempMatchFound++;
}
}
if (TempMatchFound > MaxMatchFound)
{
bestNumber = myArray[o];
MaxMatchFound = TempMatchFound;
}
}
return bestNumber;
Bucket the values, then loop over the values v and sum the associated counts for all values w that satisfy v <= w <= v + 5, then find the max count:
var buckets = myArray.GroupBy(n => n)
.ToDictionary(g => g.Key, g => g.Count());
var rangeCounts =
buckets.Keys
.Select(v =>
new {
Value = v,
Count = Enumerable.Range(0, 6)
.Sum(i => buckets.ContainsKey(v + i) ?
buckets[v + i] :
0
)
}
);
var bestRange = rangeCounts.MaxBy(x => x.Count);
Now, bestRange.Value is the starting point for your best range and bestRange.Count is the number of elements falling into the range [bestRange.Value, bestRange.Value + 5]. Here, I've used MaxBy.
Think this gets you linear performance. Building dictionary is linear, building rangeCounts is linear, MaxBy is linear. Even works on non-sorted.
Here you go: This runs in O(N) time, and O(1) memory. This forms the buckets described in other solutions, then discards them as you move through the array. The Queue is used to track which buckets are 'active' in the sense that they can be added to. The Dictionary will never have more than 6 entries in it, neither will the Queue.
int[] myArray = new int[]{1,2,4,5,7,9,15,19};
Dictionary<int, int> counts = new Dictionary<int, int>();
Queue<int> q = new Queue<int>();
int n = 0;
int currentMaxCount = 0;
for(int i = 0; i < myArray.Length; i++)
{
var currentNum = myArray[i];
if(counts.ContainsKey(currentNum))
{
counts[currentNum]++;
}
else
{
counts[currentNum] = 1;
q.Enqueue(currentNum);
}
for(int j = 1; j <= 5; j++)
{
if(counts.ContainsKey(currentNum - j))
counts[currentNum - j]++;
}
if(q.Peek() + 5 < currentNum)
{
if(counts[q.Peek()] > currentMaxCount)
{
currentMaxCount = counts[q.Peek()];
n = q.Peek();
}
counts.Remove(q.Dequeue());
}
}
while(q.Count > 0)
{
if(counts[q.Peek()] > currentMaxCount)
{
currentMaxCount = counts[q.Peek()];
n = q.Peek();
}
counts.Remove(q.Dequeue());
}
Console.WriteLine("There are {0} matches between {1} and {2}", currentMaxCount, n, n + 5);
Here's a solution that is O(n) and uses O(1) extra space regardless of the range.
It does a single pass through the array, always making 2N comparisons. I don't see any way to improve on this algorithm, although there are certainly micro optimizations that could squeeze a little more speed out of the implementation.
private int FindRange(int[] myArray)
{
const int range = 5;
int start = 0;
int maxMatchFound = 0;
int maxIndex = 0;
for (int i = 0; i < myArray.Length; ++i)
{
if (myArray[i] > myArray[start] + range)
{
int matchLength = i - start;
if (matchLength > maxMatchFound)
{
maxMatchFound = matchLength;
maxIndex = start;
}
// move forward until within range
do
{
++start;
} while (myArray[i] > myArray[start] + range);
}
}
// Final check, from myArray[start] to end of array
int len = myArray.Length - start;
if (len > maxMatchFound)
{
maxMatchFound = len;
maxIndex = start;
}
return maxIndex;
The idea here is that if a particular number a[x] falls within the range for a[i] then it will fall within the range for a[i+1], assuming that x > i. (So in your original array, the value at a[3] falls within the range of a[0], so it will also fall within the range of a[1] and a[2]).
So the index i is incremented until the value it references falls out of the range of a[start]. Then, start is incremented until a[i] is in range again. The two indexes move forward through the array in that alternating fashion.
Here's a one-line LINQ option. Not the best in terms of performance (it iterates multiple times). Still worth noting.
var result = myArray
.OrderByDescending(i => myArray.Count(i2 => i2 >= i && i2 <= i + 5))
.First();
Related
I am trying to count how many times a number appears in an array 1 (a1) and then trying to print out that number by storing it in array 2 (a2) just once and then try to print array 2. But first using for loop and a function, I will check that if a number already exist in array 2 then move to next index in array 1, unfortunateley this code is not working; can someone please help me in trying to fix it, I don't need some complex solutions like dictionaries or lists athe moment, although it might be helpful too. thanks, I am not an expert in programming and I try to practise it in my free time, so please help me.
I just want this code to be fixed for my understanding and knowledge
class Program
{
static void Main(string[] args)
{
int i, j;
int[] a1 = new int[10];
int[] a2 = new int[10];
int[] a3 = new int[10];
//takes an input
for (i = 0; i < a1.Length; i++)
{
a1[i] = Convert.ToInt32(Console.ReadLine());
}
for (i = 0; i < a1.Length; i++)
{
Cn(a1, a2); //calls in function
i++; //increments is if true
int count = 0;
for (j = 0; j < a1.Length; j++)
{
//if a number matches with a number in second array
if (a1[i] == a1[j])
{
//do count ++
count++;
// store that number into second array
a2[i] = a1[i];
}
}
//store the number of counts in third array
a3[i] = count;
}
for (i = 0; i < a2.Length; i++)
{
if (a2[i] != 0)
{
Console.WriteLine(a2[i]);
}
}
Console.ReadLine();
}
//function to check if element at current index of array 1 exists in array 2 if yes than break
public static void Cn (int[] aa1, int [] aa2)
{
int k, j;
for ( k = 0; k < aa1.Length; k++)
{
for (j = 0; j < aa2.Length; j++)
{
if (aa1[k] == aa2[j])
break;
}
}
}
}
You probably want to do a group by count:
int[] a1 = new int[10];
var rnd = new Random();
//takes an input
for (int i = 0; i < a1.Length; i++)
{
a1[i] = Convert.ToInt32(rnd.Next(0, 11)); // or Console.ReadLine()
}
var grouped = a1
.GroupBy(x => x)
.Select(g => new
{
Item = g.Key,
Count = g.Count()
}).ToList(); // ToList() is optional, materializes the IEnumerable
foreach (var item in grouped)
{
Console.WriteLine($"number: {item.Item}, count: {item.Count}");
}
This uses a Hash algorithm internally.
You can solve this without a Hash or Dictionary but it wouldn't be very efficient because you need to do lots of linear searches through the arrays.
The advantage of a Hash algorithm is that your lookups or groupings are much faster than if you loop over a complete array to find / increment an item.
I have five long integers p, q, s, m and x. An array numbers[] is created by the following formula.
numbers[0] = s;
for(int i=1; i<numbers.Length;i++){
numbers[i] = (p * numbers[i-1] + q) % m;
}
The first value of numbers (numbers[0]) is s.
What is the most efficient way to find index j where i < j and |numbers[j] - numbers[i]| <= x or |numbers[j] - numbers[i]| >= m-x.
For instance, in a case where p = 3, q= 7, s= 1, m= 29 en x= 1 the array will be:
numbers[0] = 1, numbers[1] = 10, numbers[2] = 8 and numbers[3] = 2.
In this case index j would be 3, because numbers[3] - numbers[0]<=x, because x is 1.
I thought about using something such as a variant of counting sort or radix sort but I can't get anything to work.
As i < j, then you need to grant that numbers has a length of at least 2.
You could do two nested loops, the outer one ranging from j = 1 to numbers.Length - 1 (granting the possible solution to be the smallest j) to i = 0 to i < j.
Then you compare both positions according your specs. If true, return j. If it finishes both loops, then there is no solution.
Edit: Code Sample
public int GetSmallestIndex(long[] numbers, long x, long m)
{
if (numbers.Length >= 2)
{
for (int j = 1; j < numbers.Length; j++)
{
for (int i = 0; i < j; i++)
{
long diff = Math.Abs(numbers[j] - numbers[i]);
if (diff <= x || diff >= m - x)
return j;
}
}
}
return -1; //If no solution is found, return -1 as convention
}
The only way to find out if something is more efficient is to benchmark it by using the StopWatch in System.Diagnostics.
using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.Linq;
namespace ConsoleApplication1
{
class Program
{
public static void Main()
{
var stopWatch = new Stopwatch();
stopWatch.Start();
const long x = 1;
var numbers = new long[] {3, 7, 1, 29};
var theSmallestIndex = SmallestIndex(x, numbers);
stopWatch.Stop();
Console.WriteLine("Elapsed Time: {0}", stopWatch.Elapsed);
Console.WriteLine("Smallest Index: {0}", theSmallestIndex);
Console.ReadKey();
}
public static long SmallestIndex(long x, long[] numbers)
{
var values = ValuesMinusTheValueOfPreviousIndex(x, numbers.ToList());
var smallest = values.Values.OrderBy(n => n).FirstOrDefault();
var result = values.Where(n => n.Value.Equals(smallest));
return result.FirstOrDefault().Key;
}
public static Dictionary<int, long> ValuesMinusTheValueOfPreviousIndex(long x, List<long> numbers)
{
var results = new Dictionary<int, long>();
foreach (var number in numbers)
{
var index = numbers.IndexOf(number);
var previousNumber = index > 0 ? numbers.ElementAt(index - 1) : 0;
var result = number - previousNumber;
results.Add(index, result);
}
return results;
}
}
}
EDIT: Added Math.Abs as you requested in the comments
long p = 3, q = 7, s = 1, m = 29, x = 1;
long[] numbers = new long[10];
numbers[0] = s;
for (int i = 1; i < numbers.Length; i++)
{
numbers[i] = (p * numbers[i - 1] + q) % m;
}
// Find the smallest index j in numbers, where i < j &&
// (numbers[j] - numbers[i] <= x || numbers[j] - numbers[i] >= m-x)
int smallestIndex = 0;
long comparison;
for (int j = 1; j < numbers.Length; j++)
{
for (int i = 0; i < j; i++)
{
comparison = Math.Abs(numbers[j] - numbers[i]);
if (comparison <= x || comparison >= m - x)
{
smallestIndex = j;
break;
}
}
if (smallestIndex != 0) break;
}
if (smallestIndex == 0)
{
// No result matches the conditions
}
else
{
// j is the smallest index matching the conditions
// Before using Abs, in the example j is 2, because 8 - 10 = -2, lesser than x
// Now using absolute values, In the example j is 3
}
I have this array of integers:-
int[] numbers = new int[] { 10, 20, 30, 40 };
I am trying to create an array which will have first element, last element, second element, second-last element and so on..
So, my resulting output will be:-
int[] result = {10,40,20,30};
This was my approach, in one loop start from first and go till the middle & in second loop start from last and get to the middle and select items accordingly, but I totally messed it up. Here is my attempted code:-
private static IEnumerable<int> OrderedArray(int[] numbers)
{
bool takeFirst = true;
if (takeFirst)
{
takeFirst = false;
for (int i = 0; i < numbers.Length / 2; i++)
{
yield return numbers[i];
}
}
else
{
takeFirst = true;
for (int j = numbers.Length; j < numbers.Length / 2; j--)
{
yield return numbers[j];
}
}
}
Need Help.
You might try this:
int[] result = numbers.Zip(numbers.Reverse(), (n1,n2) => new[] {n1, n2})
.SelectMany(x =>x)
.Take(numbers.Length)
.ToArray();
Explanation: This approach basically pairs up the elements of the original collection with the elements of its reverse ordered collection (using Zip). So you get a collection of pairs like [first, last], [second, second from last], etc.
It then flattens those collection of pairs into a single collection (using SelectMany). So the collection becomes [first, last, second, second from last,...].
Finally, we limit the number of elements to the length of the original array (n). Since we are iterating through twice as many elements (normal and reverse), it works out that iterating through n elements allow us to stop in the middle of the collection.
As a different approach, this is a modification on your existing method:
private static IEnumerable<int> OrderedArray(int[] numbers)
{
var count = (numbers.Length + 1) / 2;
for (int i = 0; i < count; i++)
{
yield return numbers[i];
int reverseIdx = numbers.Length - 1 - i;
if(i != reverseIdx)
yield return numbers[reverseIdx];
}
}
ok,
public static class Extensions
{
public static IEnumerable<T> EndToEnd<T>(this IReadOnlyList<T> source)
{
var length = source.Count;
var limit = length / 2;
for (var i = 0; i < limit; i++)
{
yield return source[i];
yield return source[length - i - 1];
}
if (length % 2 > 0)
{
yield return source[limit];
}
}
}
Which you could use like this,
var result = numbers.EndToEnd().ToArray();
more optimally,
public static class Extensions
{
public static IEnumerable<T> EndToEnd<T>(this IReadOnlyList<T> source)
{
var c = source.Count;
for (int i = 0, f = 0, l = c - 1; i < c; i++, f++, l--)
{
yield return source[f];
if (++i == c)
{
break;
}
yield return source[l];
}
}
}
no divide or modulus required.
With a simple for;
int len = numbers.Length;
int[] result = new int[len];
for (int i = 0, f = 0, l = len - 1; i < len; f++, l--)
{
result[i++] = numbers[f];
if (f != l)
result[i++] = numbers[l];
}
Based on Selman22's now deleted answer:
int[] numbers = new int[] { 10, 20, 30, 40 };
int[] result = numbers
.Select((x,idx) => idx % 2 == 0
? numbers[idx/2]
: numbers[numbers.Length - 1 -idx/2])
.ToArray();
result.Dump();
(The last line is LinqPad's way of outputting the results)
Or in less LINQy form as suggested by Jeppe Stig Nielsen
var result = new int[numbers.Length];
for (var idx = 0; idx < result.Length; idx++) {
result[idx] = idx % 2 == 0 ? numbers[idx/2] : numbers[numbers.Length - 1 -idx/2];
}
The principle is that you have two sequences, one for even elements (in the result) and one for odd. The even numbers count the first half of the array and the odds count the second half from the back.
The only modification to Selman's code is adding the /2 to the indexes to keep it counting one by one in the right half while the output index (which is what idx basically is in this case) counts on.
Came up with this
static void Main(string[] args)
{
List<int> numbers = new List<int>() { 10, 20, 30, 40, 50, 60, 70};
List<int> numbers2 = new List<int>();
int counter1 = 0;
int counter2 = numbers.Count - 1;
int remainder = numbers.Count % 2 == 0 ? 1: 0;
while (counter1-1 < counter2)
{
if (counter1 + counter2 % 2 == remainder)
{
numbers2.Add(numbers[counter1]);
counter1++;
}
else
{
numbers2.Add(numbers[counter2]);
counter2--;
}
}
string s = "";
for(int a = 0; a< numbers2.Count;a++)
s+=numbers2[a] + " ";
Console.Write(s);
Console.ReadLine();
}
This late answer steals a lot from the existing answers!
The idea is to allocate the entire result array at once (since its length is known). Then fill out all even-indexed members first, from one end of source. And finally fill out odd-numbered entries from the back end of source.
public static TElement[] EndToEnd<TElement>(this IReadOnlyList<TElement> source)
{
var count = source.Count;
var result = new TElement[count];
for (var i = 0; i < (count + 1) / 2; i++)
result[2 * i] = source[i];
for (var i = 1; i <= count / 2; i++)
result[2 * i - 1] = source[count - i];
return result;
}
Came up with this
public int[] OrderedArray(int[] numbers)
{
int[] final = new int[numbers.Length];
var limit=numbers.Length;
int last = numbers.Length - 1;
var finalCounter = 0;
for (int i = 0; finalCounter < numbers.Length; i++)
{
final[finalCounter] = numbers[i];
final[((finalCounter + 1) >= limit ? limit - 1 : (finalCounter + 1))] = numbers[last];
finalCounter += 2;
last--;
}
return final;
}
Suppose I have an array of integers:
int[] A = { 10, 3, 6, 8, 9, 4, 3 };
My goal is to find the largest difference between A[Q] and A[P] such that Q > P.
For example, if P = 2 and Q = 3, then
diff = A[Q] - A[P]
diff = 8 - 6
diff = 2
If P = 1 and Q = 4
diff = A[Q] - A[P]
diff = 9 - 3
diff = 6
Since 6 is the largest number between all the difference, that is the answer.
My solution is as follows (in C#) but it is inefficient.
public int solution(int[] A) {
int N = A.Length;
if (N < 1) return 0;
int difference;
int largest = 0;
for (int p = 0; p < N; p++)
{
for (int q = p + 1; q < N; q++)
{
difference = A[q] - A[p];
if (difference > largest)
{
largest = difference;
}
}
}
return largest;
}
How can I improve this so it will run at O(N)? Thanks!
Simply getting the max and min wont work. Minuend (Q) should come after the Subtrahend (P).
This question is based on the "Max-profit" problem in codility (http://codility.com/train/). My solution only scored 66%. It requires O(N) for a score of 100%.
The following code runs in O(n) and should conform to the specification (preliminary tests on codility were successful):
public int solution(int[] A)
{
int N = A.Length;
if (N < 1) return 0;
int max = 0;
int result = 0;
for(int i = N-1; i >= 0; --i)
{
if(A[i] > max)
max = A[i];
var tmpResult = max - A[i];
if(tmpResult > result)
result = tmpResult;
}
return result;
}
Update:
I submitted it as solution and it scores 100%.
Update 02/26/16:
The original task description on codility stated that "each element of array A is an integer within the range [0..1,000,000,000]."
If negative values would have been allowed as well, the code above wouldn't return the correct value. This could be fixed easily by changing the declaration of max to int max = int.MinValue;
Here is the O(n) Java implementation
public static int largestDifference(int[] data) {
int minElement=data[0], maxDifference=0;
for (int i = 1; i < data.length; i++) {
minElement = Math.min(minElement, data[i]);
maxDifference = Math.max(maxDifference, data[i] - minElement);
}
return maxDifference;
}
After some attempts, I end up with this:
int iMax = N - 1;
int min = int.MaxValue, max = int.MinValue;
for (int i = 0; i < iMax; i++) {
if (min > A[i]) min = A[i];
if (max < A[N - i - 1]){
iMax = N - i - 1;
max = A[iMax];
}
}
int largestDiff = max - min;
NOTE: I have just tested it with some cases. Please if you find any case in which it doesn't work, let me know in the comment. I'll try to improve it or remove the answer. Thanks!
int FirstIndex = -1;
int SecondIndex = -1;
int diff = 0;
for (int i = A.Length-1; i >=0; i--)
{
int FirstNo = A[i];
int tempDiff = 0;
for (int j = 0; j <i ; j++)
{
int SecondNo = A[j];
tempDiff = FirstNo - SecondNo;
if (tempDiff > diff)
{
diff = tempDiff;
FirstIndex = i;
SecondIndex = j;
}
}
}
MessageBox.Show("Diff: " + diff + " FirstIndex: " + (FirstIndex+1) + " SecondIndex: " + (SecondIndex+1));
PHP Solution
<?php
$a = [0,5,0,5,0];
$max_diff = -1;
$min_value = $a[0];
for($i = 0;$i<count($a)-1;$i++){
if($a[$i+1] > $a[$i]){
$diff = $a[$i+1] - $min_value;
if($diff > $max_diff){
$max_diff = $diff;
}
} else {
$min_value = $a[$i+1];
}
}
echo $max_diff;
?>
We can do it in a much simpler way by calculating biggest and smallest element of the array. I know that you're also looking for time complexity. But for anyone looking to understand and solve this problem in a simple and easy to understand way, then here is my code:
#include<stdio.h>
#define N 6
int main()
{
int num[N], i, big, small, pos = 0;
printf("Enter %d integer numbers\n", N);
for(i = 0; i < N; i++)
scanf("%d", &num[i]);
big = small = num[0];
for(i = 1; i < N; i++)
{
if(num[i] > big)
{
big = num[i];
pos = i;
}
}
for(i = 1; i < pos; i++)
{
if(num[i] < small)
small = num[i];
}
printf("The largest difference is %d, ", (big - small));
printf("and its between %d and %d.\n", big, small);
return 0;
}
Output:
Enter 6 integer numbers
7
9
5
6
13
2
The largest difference is 8, and its between 13 and 5.
Source: C Program To Find Largest Difference Between Two Elements of Array
C++ solution for MaxProfit of codility test task giving 100/100 https://app.codility.com/programmers/lessons/9-maximum_slice_problem/max_profit/
int Max(vector<int> &A)
{
if (A.size() == 1 || A.size() == 0)
return 0;
int min_price = A[0];
int max_val = 0;
for (int i = 1; i < A.size(); i++)
{
max_val = std::max(max_val, A[i] - min_price);
min_price = std::min(min_price, A[i]);
}
return max_val;
}
My 100% JavaScript solution with O(N) time complexity:
function solution(A) {
// each element of array A is an integer within the range [0..200,000]
let min = 200000;
// The function should return 0 if it was impossible to gain any profit.
let maxDiff = 0;
for (const a of A) {
min = Math.min(min, a);
// find the maximum positive difference (profit) between current global minimum and current value of a
maxDiff = Math.max(maxDiff, a - min);
}
return maxDiff;
}
function solution(A) {
var n = A.length;
var min = Infinity, max = -Infinity, maxNet=0;
// find smallest and largest in the array following each other
for(let i = 0; i < n; i++){
if(A[i]<min) { // if you are updating the min you cannot consider the old max
min = A[i];
max = -Infinity;
} else if(A[i]> max){
max = A[i];
}
if(max!=-Infinity && max-min>maxNet) maxNet = max-min;
}
return maxNet;
}
PHP solution for MaxProfit of codility test task giving 100/100 found at http://www.rationalplanet.com/php-related/maxprofit-demo-task-at-codility-com.html
function solution($A) {
$cnt = count($A);
if($cnt == 1 || $cnt == 0){
return 0;
}
$max_so_far = 0;
$max_ending_here = 0;
$min_price = $A[0];
for($i = 1; $i < $cnt; $i++){
$max_ending_here = max(0, $A[$i] - $min_price);
$min_price = min($min_price, $A[$i]);
$max_so_far = max($max_ending_here, $max_so_far);
}
return $max_so_far;
}
100% score JavaScript solution.
function solution(A) {
if (A.length < 2)
return 0;
// Init min price and max profit
var minPrice = A[0];
var maxProfit = 0;
for (var i = 1; i < A.length; i++) {
var profit = A[i] - minPrice;
maxProfit = Math.max(maxProfit, profit);
minPrice = Math.min(minPrice, A[i]);
}
return maxProfit;
}
Python solution
def max_diff_two(arr):
#keep tab of current diff and min value
min_value = arr[0]
#begin with something
maximum = arr[1] - arr[0]
new_min = min_value
for i,value in enumerate(arr):
if i == 0:
continue
if value < min_value and value < new_min:
new_min = value
current_maximum = value - min_value
new_maximum = value - new_min
if new_maximum > current_maximum:
if new_maximum > maximum:
maximum = new_maximum
min = new_min
else:
if current_maximum > maximum:
maximum = current_maximum
return maximum
100% for Javascript solution using a more elegant functional approach.
function solution(A) {
var result = A.reverse().reduce(function (prev, val) {
var max = (val > prev.max) ? val : prev.max
var diff = (max - val > prev.diff) ? max - val : prev.diff
return {max: max, diff: diff}
}, {max: 0, diff: 0})
return result.diff
}
I have an array with "n" numbers and I need to print all repeated numbers only one time
i made this code, but something is wrong
for (int i = 0; i < numbers.Length; i++)
{
for (int j = 1; j < numbers.Length; j++)
{
if (numbers[i] == numbers[j] && i!=j)
{
Console.WriteLine(numbers[i]);
break;
}
}
}
then if my array have the elements {2,3,1,5,2,3}
the program prints :
2
3
3
what can I do?
You can use:
using System.Linq;
…
foreach(var number in numbers.Distinct()) Console.WriteLine(number);
edit
I may have misunderstood the requirement. If you only want to output the numbers that appear more than once, then you can use:
foreach(var group in numbers.GroupBy(n => n).Where(g => g.Count() > 1))
Console.WriteLine(group.Key);
var query = numbers.GroupBy(x => x)
.Where(g => g.Skip(1).Any())
.Select(g => g.Key);
foreach (int n in query)
{
Console.WriteLine(n);
}
Or, alternatively...
var dict = new Dictionary<int, int>();
foreach (int n in numbers)
{
int count;
dict.TryGetValue(n, out count);
if (count == 1)
{
Console.WriteLine(n);
}
dict[n] = count + 1;
}
The problem in your code: you get 3 repeated because when i is 1 (looking at the first 3) there's another 3 in the list at the end, and when i is 5 (looking at the last 3) there's another three in the list near the beginning.
Instead you should look at only those numbers which come after your current position - change to int j = i; so that you only look at the positions after your current position, and you won't get repeated results.
for (int i = 0; i < numbers.Length; i++)
{
for (int j = i; j < numbers.Length; j++)
{
if (numbers[i] == numbers[j] && i!=j)
{
Console.WriteLine(numbers[i]);
break;
}
}
}
Having said that, your algorithm is not as efficient as using a built in algorithm. Try GroupBy
var duplicates = numbers.GroupBy(n => n)
.Where(group => group.Count() > 1);
foreach (var group in duplicates)
{
Console.WriteLine("{0} appears {1} times", group.Key, group.Count());
}
One way to get distinct numbers is
var uniqueNumbers = numbers.Distinct().ToArray()
and then iterate over uniqueNumbers like you have in your snippet with numbers.
You can add the number to a HashSet as you loop and then only print if not in hash set.
That allows you to do it in one pass.
The algorithm above is n^2 which should be avoided.
// deletes an integer if it appears double
#include <iostream.h>
#include <conio.h>
int main ()
{
int count=0;
int ar[10]={1,2,3,3,3,4,5,6,7,7};
for (int i=0; i<10; i++)
{
if (ar[i]==ar[i+1])
count++;
else
cout << ar[i];
}
getch();
return 0;
}