C# find highest array value and index

C#ArraysIndexing

C# Problem Overview


So I have an unsorted numeric array int[] anArray = { 1, 5, 2, 7 }; and I need to get both the value and the index of the largest value in the array which would be 7 and 3, how would I do this?

C# Solutions


Solution 1 - C#

This is not the most glamorous way but works.

(must have using System.Linq;)

 int maxValue = anArray.Max();
 int maxIndex = anArray.ToList().IndexOf(maxValue);

Solution 2 - C#

int[] anArray = { 1, 5, 2, 7 };
// Finding max
int m = anArray.Max();

// Positioning max
int p = Array.IndexOf(anArray, m);

Solution 3 - C#

If the index is not sorted, you have to iterate through the array at least once to find the highest value. I'd use a simple for loop:

int? maxVal = null; //nullable so this works even if you have all super-low negatives
int index = -1;
for (int i = 0; i < anArray.Length; i++)
{
  int thisNum = anArray[i];
  if (!maxVal.HasValue || thisNum > maxVal.Value)
  {
    maxVal = thisNum;
    index = i;
  }
}

This is more verbose than something using LINQ or other one-line solutions, but it's probably a little faster. There's really no way to make this faster than O(N).

Solution 4 - C#

A succinct one-liner:

var (number, index) = anArray.Select((n, i) => (n, i)).Max();

Test case:

var anArray = new int[] { 1, 5, 7, 4, 2 };
var (number, index) = anArray.Select((n, i) => (n, i)).Max();
Console.WriteLine($"Maximum number = {number}, on index {index}.");
// Maximum number = 7, on index 2.

Features:

  • Uses Linq (not as optimized as vanilla, but the trade-off is less code).
  • Does not need to sort.
  • Computational complexity: O(n).
  • Space complexity: O(n).

Remarks:

  • Make sure the number (and not the index) is the first element in the tuple because tuple sorting is done by comparing tuple items from left to right.

Solution 5 - C#

The obligatory LINQ one[1]-liner:

var max = anArray.Select((value, index) => new {value, index})
                 .OrderByDescending(vi => vi.value)
                 .First();

(The sorting is probably a performance hit over the other solutions.)

[1]: For given values of "one".

Solution 6 - C#

Here are two approaches. You may want to add handling for when the array is empty.

public static void FindMax()
{
    // Advantages: 
    // * Functional approach
    // * Compact code
    // Cons: 
    // * We are indexing into the array twice at each step
    // * The Range and IEnumerable add a bit of overhead
    // * Many people will find this code harder to understand

    int[] array = { 1, 5, 2, 7 };

    int maxIndex = Enumerable.Range(0, array.Length).Aggregate((max, i) => array[max] > array[i] ? max : i);
    int maxInt = array[maxIndex];

    Console.WriteLine($"Maximum int {maxInt} is found at index {maxIndex}");
}

public static void FindMax2()
{
    // Advantages: 
    // * Near-optimal performance

    int[] array = { 1, 5, 2, 7 };
    int maxIndex = -1;
    int maxInt = Int32.MinValue;

    // Modern C# compilers optimize the case where we put array.Length in the condition
    for (int i = 0; i < array.Length; i++)
    {
        int value = array[i];
        if (value > maxInt)
        {
            maxInt = value;
            maxIndex = i;
        }
    }

    Console.WriteLine($"Maximum int {maxInt} is found at index {maxIndex}");
}

Solution 7 - C#

int[] numbers = new int[7]{45,67,23,45,19,85,64}; 
int smallest = numbers[0]; 
for (int index = 0; index < numbers.Length; index++) 
{ 
 if (numbers[index] < smallest) smallest = numbers[index]; 
} 
Console.WriteLine(smallest);

Solution 8 - C#

 public static class ArrayExtensions
{
    public static int MaxIndexOf<T>(this T[] input)
    {
        var max = input.Max();
        int index = Array.IndexOf(input, max);
        return index;
    }
}

This works for all variable types...

var array = new int[]{1, 2, 4, 10, 0, 2};
var index = array.MaxIndexOf();


var array = new double[]{1.0, 2.0, 4.0, 10.0, 0.0, 2.0};
var index = array.MaxIndexOf();

Solution 9 - C#

anArray.Select((n, i) => new { Value = n, Index = i })
	.Where(s => s.Value == anArray.Max());

Solution 10 - C#

Output for bellow code:

00:00:00.3279270 - max1 00:00:00.2615935 - max2 00:00:00.6010360 - max3 (arr.Max())

With 100000000 ints in array not very big difference but still...

class Program
    {
        static void Main(string[] args)
        {
            int[] arr = new int[100000000];

            Random randNum = new Random();
            for (int i = 0; i < arr.Length; i++)
            {
                arr[i] = randNum.Next(-100000000, 100000000);
            }
            Stopwatch stopwatch1 = new Stopwatch();
            Stopwatch stopwatch2 = new Stopwatch();
            Stopwatch stopwatch3 = new Stopwatch();
            stopwatch1.Start();

            var max = GetMaxFullIterate(arr);

            Debug.WriteLine( stopwatch1.Elapsed.ToString());
         

            stopwatch2.Start();
            var max2 = GetMaxPartialIterate(arr);
          
            Debug.WriteLine( stopwatch2.Elapsed.ToString());

            stopwatch3.Start();
            var max3 = arr.Max();
            Debug.WriteLine(stopwatch3.Elapsed.ToString());

        }



 private static int GetMaxPartialIterate(int[] arr)
        {
            var max = arr[0];
            var idx = 0;
            for (int i = arr.Length / 2; i < arr.Length; i++)
            {
                if (arr[i] > max)
                {
                    max = arr[i];
                }

                if (arr[idx] > max)
                {
                    max = arr[idx];
                }
                idx++;
            }
            return max;
        }


        private static int GetMaxFullIterate(int[] arr)
        {
            var max = arr[0];
            for (int i = 0; i < arr.Length; i++)
            {
                if (arr[i] > max)
                {
                    max = arr[i];
                }
            }
            return max;
        }

Solution 11 - C#

public static void Main()
{
	int a,b=0;
	int []arr={1, 2, 2, 3, 3, 4, 5, 6, 5, 7, 7, 7, 100, 8, 1};
	
	for(int i=arr.Length-1 ; i>-1 ; i--)
		{
			a = arr[i];
			
			if(a > b)
			{
				b=a;	
			}
		}
	Console.WriteLine(b);
}

Solution 12 - C#

int[] Data= { 1, 212, 333,2,12,3311,122,23 };
int large = Data.Max();
Console.WriteLine(large);

Solution 13 - C#

Consider following:

    /// <summary>
    /// Returns max value
    /// </summary>
    /// <param name="arr">array to search in</param>
    /// <param name="index">index of the max value</param>
    /// <returns>max value</returns>
    public static int MaxAt(int[] arr, out int index)
    {
        index = -1;
        int max = Int32.MinValue;

        for (int i = 0; i < arr.Length; i++)
        {
            if (arr[i] > max)
            { 
                max = arr[i];
                index = i;
            }
        }

        return max;
    }

Usage:

int m, at;
m = MaxAt(new int[]{1,2,7,3,4,5,6}, out at);
Console.WriteLine("Max: {0}, found at: {1}", m, at);

Solution 14 - C#

Here is a LINQ solution which is O(n) with decent constant factors:

int[] anArray = { 1, 5, 2, 7, 1 };

int index = 0;
int maxIndex = 0;

var max = anArray.Aggregate(
    (oldMax, element) => {
        ++index;
        if (element <= oldMax)
            return oldMax;
        maxIndex = index;
        return element;
    }
);

Console.WriteLine("max = {0}, maxIndex = {1}", max, maxIndex);

But you should really write an explicit for lop if you care about performance.

Solution 15 - C#

Just another perspective using DataTable. Declare a DataTable with 2 columns called index and val. Add an AutoIncrement option and both AutoIncrementSeed and AutoIncrementStep values 1 to the index column. Then use a foreach loop and insert each array item into the datatable as a row. Then by using Select method, select the row having the maximum value.

Code

int[] anArray = { 1, 5, 2, 7 };
DataTable dt = new DataTable();
dt.Columns.AddRange(new DataColumn[2] { new DataColumn("index"), new DataColumn("val")});
dt.Columns["index"].AutoIncrement = true;
dt.Columns["index"].AutoIncrementSeed = 1;
dt.Columns["index"].AutoIncrementStep = 1;
foreach(int i in anArray)
    dt.Rows.Add(null, i);
        
DataRow[] dr = dt.Select("[val] = MAX([val])");
Console.WriteLine("Max Value = {0}, Index = {1}", dr[0][1], dr[0][0]);

Output

Max Value = 7, Index = 4

Find a demo here

Solution 16 - C#

If you know max index accessing the max value is immediate. So all you need is max index.

int max=0;

for(int i = 1; i < arr.Length; i++)
    if (arr[i] > arr[max]) max = i;
        

Solution 17 - C#

This is a C# Version. It's based on the idea of sort the array.

public int solution(int[] A)
 {
    // write your code in C# 6.0 with .NET 4.5 (Mono)
    Array.Sort(A);
    var max = A.Max();
    if(max < 0)
        return 1;
    else
        for (int i = 1; i < max; i++)
        {
            if(!A.Contains(i)) {
                return i;
            }
        }
    return max + 1;
}

Solution 18 - C#

This can be done with a bodiless for loop, if we're heading towards golf ;)

//a is the array


int mi = a.Length - 1;
for (int i=-1; ++i<a.Length-1; mi=a[mi]<a[i]?i:mi) ;

The check of ++i<a.Length-1 omits checking the last index. We don't mind this if we set it up as if the max index is the last index to start with.. When the loop runs for the other elements it will finish and one or the other thing is true:

  • we found a new max value and hence a new max index mi
  • the last index was the max value all along, so we didn't find a new mi, and we stuck with the initial mi

The real work is done by the post-loop modifiers:

  • is the max value (a[mi] i.e. array indexed by mi) we found so far, less than the current item?
    • yes, then store a new mi by remembering i,
    • no then store the existing mi (no-op)

At the end of the operation you have the index at which the max is to be found. Logically then the max value is a[mi]

I couldn't quite see how the "find max and index of max" really needed to track the max value too, given that if you have an array, and you know the index of the max value, the actual value of the max value is a trivial case of using the index to index the array..

Solution 19 - C#

Another answer in this long list, but I think it's worth it, because it provides some benefits that most (or all?) other answers don't:

  1. The method below loops only once through the collection, therefore the order is O(N).
  2. The method finds ALL indices of the maximum values.
  3. The method can be used to find the indices of any comparison: min, max, equals, not equals, etc.
  4. The method can look into objects via a LINQ selector.

Method:

///-------------------------------------------------------------------
/// <summary>
/// Get the indices of all values that meet the condition that is defined by the comparer.
/// </summary>
/// <typeparam name="TSource">The type of the values in the source collection.</typeparam>
/// <typeparam name="TCompare">The type of the values that are compared.</typeparam>
/// <param name="i_collection">The collection of values that is analysed.</param>
/// <param name="i_selector">The selector to retrieve the compare-values from the source-values.</param>
/// <param name="i_comparer">The comparer that is used to compare the values of the collection.</param>
/// <returns>The indices of all values that meet the condition that is defined by the comparer.</returns>
/// Create <see cref="IComparer{T}"/> from comparison function:
///   Comparer{T}.Create ( comparison )
/// Comparison examples:
/// - max:      (a, b) => a.CompareTo (b)
/// - min:      (a, b) => -(a.CompareTo (b))
/// - == x:     (a, b) => a == 4 ? 0 : -1
/// - != x:     (a, b) => a != 4 ? 0 : -1
///-------------------------------------------------------------------
public static IEnumerable<int> GetIndices<TSource, TCompare> (this IEnumerable<TSource> i_collection,
                                                                   Func<TSource, TCompare> i_selector,
                                                                   IComparer<TCompare> i_comparer)
{
  if (i_collection == null)
    throw new ArgumentNullException (nameof (i_collection));
  if (!i_collection.Any ())
    return new int[0];

  int index = 0;
  var indices = new List<int> ();
  TCompare reference = i_selector (i_collection.First ());

  foreach (var value in i_collection)
  {
    var compare = i_selector (value);
    int result = i_comparer.Compare (compare, reference);
    if (result > 0)
    {
      reference = compare;
      indices.Clear ();
      indices.Add (index);
    }
    else if (result == 0)
      indices.Add (index);

    index++;
  }

  return indices;
}

If you don't need the selector, then change the method to

public static IEnumerable<int> GetIndices<TCompare> (this IEnumerable<TCompare> i_collection,
                                                          IComparer<TCompare> i_comparer)

and remove all occurences of i_selector.

Proof of concept:

//########## test #1: int array ##########
int[] test = { 1, 5, 4, 9, 2, 7, 4, 6, 5, 9, 4 };

// get indices of maximum:
var indices = test.GetIndices (t => t, Comparer<int>.Create ((a, b) => a.CompareTo (b)));
// indices: { 3, 9 }

// get indices of all '4':
indices = test.GetIndices (t => t, Comparer<int>.Create ((a, b) => a == 4 ? 0 : -1));
// indices: { 2, 6, 10 }

// get indices of all except '4':
indices = test.GetIndices (t => t, Comparer<int>.Create ((a, b) => a != 4 ? 0 : -1));
// indices: { 0, 1, 3, 4, 5, 7, 8, 9 }

// get indices of all '15':
indices = test.GetIndices (t => t, Comparer<int>.Create ((a, b) => a == 15 ? 0 : -1));
// indices: { }

//########## test #2: named tuple array ##########
var datas = new (object anything, double score)[]
{
  (999,               0.1),
  (new object (),     0.42),
  ("hello",           0.3),
  (new Exception (),  0.16),
  ("abcde",           0.42)
};
// get indices of highest score:
indices = datas.GetIndices (data => data.score, Comparer<double>.Create ((a, b) => a.CompareTo (b)));
// indices: { 1, 4 }

Enjoy! :-)

Solution 20 - C#

Old post, but this is super easy with Lists:

For Maximum:

 List<int> lst = new List<int>(YourArray);
 int Max = lst.OrderByDescending(x => x).First();

For Minimum:

List<int> lst = new List<int>(YourArray);
int Max = lst.OrderBy(x => x).First();

Of course you can substitute "int" data type with any numeric variable type (float, decimal, etc).

This is very high performance BTW and beats any other method (IMHO)

Solution 21 - C#

Finds the biggest and the smallest number in the array:

int[] arr = new int[] {35,28,20,89,63,45,12};
int big = 0;
int little = 0;

for (int i = 0; i < arr.Length; i++)
{
    Console.WriteLine(arr[i]);

    if (arr[i] > arr[0])
    {
        big = arr[i];
    }
    else
    {
        little = arr[i];

    }
}

Console.WriteLine("most big number inside of array is " + big);
Console.WriteLine("most little number inside of array is " + little);

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