I have two different arrays that are widely different in length, arrayA, arrayB
for example:arrayA.Length=1000 and arrayB.Length=100,000
Create a boolean function that returns true if each item in arrayA are also present in arrayB, at least the same number of times as in arrayA.
Please review my code below and help me with siuggestions to improve this code below
private static bool IsMinRequiredFrequencyMatch(int[] arrA,int[] arrB)
{
bool result = false;
Array.Sort(arrA);
//create dictionary from shorter array
var dict = new Dictionary<int, int>();
for (int i = 0; i < arrA.Length; i++)
{
var ch = arrA[i];
if (!dict.ContainsKey(ch))
{
dict[ch] = 1; //add char count
}
else
{
dict[ch]++; //increase count
}
}
//loop through Dictionary
foreach (var kvp in dict)
{
var chr = kvp.Key; //target item to find
var expectedcount = kvp.Value; //expected count of item in arrB
int actualCount = 0; //actual count of item in arrB
//loop through arrayB
for (int i = 0; i < arrB.Length; i++)
{
if (arrB[i] == chr)
{
actualCount++;
}
if (actualCount >= expectedcount)
{
result = true;
break;
}
}
if (actualCount < expectedcount)
{
result = false;
break;
}
}
return result;
}
1 Answer 1
Sure, you can always write your own method for everything but if you are lazy you could just use LINQ with ToLookup
+ All
+ Count
var arr1 = new[] { 1, 2, 3, 4, 3, 2 };
var arr2 = new[] { 6, 7, 3, 4, 1, 2, 2, 3 };
var lookup1 = arr1.ToLookup(x => x);
var lookup2 = arr2.ToLookup(x => x);
var isFullyContained = lookup1.All(x => lookup2[x.Key].Count() == x.Count());
false - means arr1 is fully contained in arr2
The good thing about lookup is that you don't need the ContainsKey
method because if the key does not exist it'll return an empty collection.
Review
You create a dictionary only for the first array but you need one for the second array too or otherwise you need to loop over it each time which is unnecessary. Just count the items and compare their counts.
chr
,i
,kvp
,ch
, anddict
are horrible names that make reader think too much about unnecessary stuff. \$\endgroup\$