What is the best way to loop through an array and find an array element closest to a key?

Question

Here is my code so far and I am stuck. The closest value to 7 in terms of absolute value is 5. How can I check each element of the array to see if it is the closest element and then return that value. I know this is probably easier than I think but I am a beginner at programming.

#include <iostream>
#include <cmath>
using namespace std;

const int MAX = 25;

int searchArray(int [], int); \\prototype

int main(){

    int numArray[MAX] ={1,4,5,10,11,12,13,14,15,16,17,18,19,20,35,26,43,15,48,69,32,45,57,98,100};

searchArray(numArray, 7);
system("pause");
return 0;
}


int searchArray(int a[], int b){

int distance = 0;

for(int i=0;i<MAX;i++){
    if(i==b)
        return i;
    else if(i<b || i > b)
        abs(distance) == i-b;
    return distance;


}

}

and why are you comparing i to be and not a[i] , looks like you need to read up on your homework. — Markus Mikkolainen
– Markus Mikkolainen, Commented Jan 8, 2013 at 20:33
Ah you are right all of them should be a[i]. Like i said I am a beginner. I used abs(distance) == a[i] - b to try and find the absolute value of how far away an element is from the key 7. — Nick
– Nick, Commented Jan 8, 2013 at 20:35
Nick, please refrain from "thank you" comments. That's what voting is for. Only comment if you actually have something constructive to say. — pmr
– pmr, Commented Jan 8, 2013 at 20:44

Sarfaraz Nawaz · Accepted Answer · 2013-01-08 20:55:56Z

6

You can use std::min_element as:

#include <iostream>
#include <algorithm>
#include <cstdlib>

int main()
{
   int numArray[] ={1,4,5,10,11,12,13,14,15,16,17,18,19,20,35,26,43,15,48,69,32,45,57,98,100};

   //find nearest element to key            
   int key = 7;
   auto cmp = [&](int a, int b) { return std::abs(key-a) < std::abs(key-b); };
   int nearest_value = *std::min_element(std::begin(numArray),std::end(numArray),cmp);

   std::cout << nearest_value << std::endl;
}

Output (demo):

edited Jan 8, 2013 at 20:55

answered Jan 8, 2013 at 20:35

Sarfaraz Nawaz

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Comments

sampson-chen · Accepted Answer · 2013-01-08 20:44:37Z

3

Write your search function to the following:

int searchArray(int a[], int b){

    int min_dist = 0; // keep track of mininum distance seen so far
    int min_index = 0; // keep track of the index of the element 
                       // that has the min index

    for( int i = 0; i < a.Length; i++ ){ // a.Length is the size of the array
        if( a[i] == b ) { // if we find the exact one, stop
            return i;
        } else { // otherwise, keep looking for the closest
            if ( abs(a[i] - b) < min_dist ) { // if this one is closer, update
                min_dist = abs(a[i] - b);
                min_index = i;
            }
        }
    }
    return min_index; // when we finish return the index of the closest
}

edited Jan 8, 2013 at 20:44

answered Jan 8, 2013 at 20:37

sampson-chen

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1 Comment

sampson-chen Over a year ago

@Nick - see the edited version =) fixing a few other minor mistakes

K-ballo · Accepted Answer · 2013-01-08 20:48:03Z

3

You can use standard algorithms to do that for you:

struct closer_to
{
    int target_;

    explicit closer_to( int target ) : target_( target ){}

    bool operator ()( int left, int right ) const
    {
        return std::abs( target_ - left ) < std::abs( target_ - right );
    }
};

int* iter =
    std::min_element(
        numArray + 0, numArray + MAX
      , closer_to( 7 )
    );

min_element will return an iterator (pointer in this case) to the first element for which there is no other element in the Container that is closer_to( 7 ).

With C++11 and lambdas, it would look like this:

int* iter = 
    std::min_element(
        numArray + 0, numArray + MAX
      , []( int left, int right ) { return std::abs( target_ - left ) < std::abs( target_ - right ); }
    );

edited Jan 8, 2013 at 20:48

answered Jan 8, 2013 at 20:37

K-ballo

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Comments

imreal · Accepted Answer · 2013-01-08 20:38:19Z

1

You have to compare your number to every number on your array and keep track of the smallest distance so far.

int searchArray(int a[], int b){
    int minDistance = -1;

    for(int i=0;i<MAX;i++){
        if(minDistance == -1 || abs(minDistance - b) > abs(a[i] - b))
            minDistance = a[i];
    }
    return minDistance;
}

answered Jan 8, 2013 at 20:38

imreal

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