7

Hello how can I find the smallest and biggest number in delphi?

Suppose I have 10 different numbers stored in an array:

How can I find the biggest number and smallest numbers in my array?

menjaraz
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dnaz
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3 Answers3

6

Simply loop through the array in linear fashion. Keep a variable for the min value and one for the max values. Initialise both to the first value in the array. Then for each element, update the min or max value if that element is less than or greater than the min or max value respectively.

minval := a[0];
maxval := a[0];
for i := 1 to Count-1 do
begin
  if a[i]<minval then
    minval := a[i]
  else if a[i]>maxval then
    maxval := a[i];
end;

Obviously this code assumes Count>0.

Note that you could equally use the MinValue and MaxValue routines from the Math unit.

David Heffernan
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  • I think we were typing at the same time! I like your initialization better, but I prefer the more compact for..in structure, although it depends on which version of Delphi he had as to whether it is supported. – Richard A Aug 20 '11 at 12:15
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    +1 For `MinValue` and `MaxValue`. There is also [`MinIntValue`](http://docwiki.embarcadero.com/VCL/en/Math.MinIntValue) and [`MaxIntValue`](http://docwiki.embarcadero.com/VCL/en/Math.MaxIntValue). – NGLN Aug 20 '11 at 15:12
  • @ngln thanks for that, you are right of course, I habitually work with floating point data. – David Heffernan Aug 20 '11 at 16:18
  • but my numbers are in int64.and when I found minvalue there is a problem that is -112322654564545 is not a integer value.. – dnaz Aug 20 '11 at 21:17
4

Iterate through the array comparing to the previous found min and max.

Here is a code snippet. Following your clarification, I have edited the code to use Int64. 

Min := High(Int64);
Max := Low(Int64);
for ThisNumber in MyArray do
begin
  if ThisNumber < Min then
  begin
    Min := ThisNumber;
  end
  if ThisNumber > Max then
  begin
    Max := ThisNumber;
  end;
end;

It's interesting to note that MaxIntValue in Math.pas is implemented as:

function MaxIntValue(const Data: array of Integer): Integer;
var
  I: Integer;
begin
  Result := Data[Low(Data)];
  for I := Low(Data) + 1 to High(Data) do
    if Result < Data[I] then
      Result := Data[I];
end;

This implementation, similar to David's answer, uses the first array value as the initial value. This assumes that the array has at least one element. Note also that the loop can then start at Low(Data) + 1 and save one unnecessary comparison. For the data you have described, with 100 elements in each array, you would get a 1% speed improvement, at best.

If performance doesn't matter then MinIntValue and MaxIntValue will be more concise. If you roll your own, then you are only iterating through the array once instead of twice.

Richard A
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-1

Create a function that takes an array of numbers and return both the minimum and maximum numbers, in that order.

// Examples
// minMax([1, 2, 3, 4, 5]) ➞ [1, 5]

// minMax([2334454, 5]) ➞ [5, 2334454]

// minMax([1]) ➞ [1, 1]

const minMax = (arr) => {
  let newMinMax = [];
  let min = Math.min(...arr);

  newMinMax.push(min);
  let max = Math.max(...arr);

  newMinMax.push(max);
  return newMinMax;
};

// console.log(minMax([1, 2, 3, 4, 5]));
// console.log(minMax([2334454, 5]));
// console.log(minMax([1]));

Used javascript build in functions for that .Math.min function requires distinct number but when we provide array it will give you a NaN to avoid that use [...arr]
spread operator of Math.min.apply(Math,arr) function.
Tuseef ashraf
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