How calculate array integers sum array with index belongs to an interval in Java?

user_15995870 0
Viewed 1234

I'm trying to calculate the sum of the integers of array whose index belongs to the interval [n1; n2]

n1 & n2 are int & 0 <= n1 <= n2 < array.length

int[] array = new int[] {0, 1, 2, 3, 4, 5, 3}; 
int zeroToOne = calc(array,0, 1); // Should return 1
int zeroToFive = calc(array,0, 5); // Should return 15
int doubleZero = calc(array,0, 0); // Should return 0
int zeroToSix = calc(array, 0,6); // Should return 18

So I've tried three methods

First method:

public static int calc(int[] array, int n1, int n2) {

    int sum = 0;
    for (int i = 0; i < array.length; i++){
        if (i <= n2 && n1 <= i) {
            sum += i;
        }
    }
    return sum;
}

}

I got:

1 15 0 21

Second method:

public static int calc(int[] array, int n1, int n2) {

    int sum = 0;
    for (int a: array){
        if (n1 <= a && a <= n2){
            sum += a;
        }
    }
    return sum;
}

I got:

1 18 0 18

Third method:

public static int calc(int[] array, int n1, int n2) {
      
    return Arrays.stream(array, n1, n2).sum();
}

And I got:

0 10 0 15

How can I solve this problem?

intervals indexing sum arrays java
5 Answers

A small change to your first function and it should work:

  public static int calc(int[] array, int n1, int n2) {

    int sum = 0;
    for (int i = n1; i <= n2; i++) {
      sum += array[i];
    }
    return sum;
  }

Note that we iterate over the elements from position n1 to n2, and summing them up.

user_132383100

The method you call:

Arrays.stream(array, n1, n2)

has the following definition in documentation:

stream(int[] array, int startInclusive, int endExclusive)

so you can see that the 3rd parameter is exclusive.

What you can do is just call it with the last index + 1. This should be fine:

return Arrays.stream(array, n1, n2+1).sum();
user_56062220

change first method to:

public static int calc(int[] array, int n1, int n2) {

    int sum = 0;
    for (int i = 0; i < array.length; i++){
        if (array[i] <= n2 && n1 <= array[i]) {
            sum += array[i];
        }
    }
    return sum;
}

and if you want to n1 & n2 are int & 0 <= n1 <= n2 < array.length,
you'd better check values of n1 and n2 and then if the condition was false do something like throw a exception

for do this:

public static int calc(int[] array, int n1, int n2) {
    if(!(n1>=0 && n1<=n2 && n1<=array.length && n2<=array.length)){
     // do something like throw a exception
    }
    int sum = 0;
    for (int i = 0; i < array.length; i++){
        if (array[i] <= n2 && n1 <= array[i]) {
            sum += array[i];
        }
    }
    return sum;
}
user_132899630

I have some proposition here!

This work for me in PHP, I have try to translate it to JAVA.

public static int calc(int[] array, int n1, int n2) {
    if(!(n1>=0 && n1<=n2 && n1<=array.length && n2<=array.length)){
     // Throw a exception here or return something
    }
    int sum = 0;
    for (int i = 0; i < array.length; i++){
        if (i >= n1 && i <= n2) {
            sum += array[i];
        }
    }
    return sum;
}

I translate that from my PHP code below.

function calc($array, $n1, $n2)
{
    if (!($n1 >= 0 && $n1 <= $n2 && $n1 <= count($array) && $n2 <=     count($array))) {
        throw new Exception('Invalid params.');
    }

    $sum = 0;
    for ($i = 0; $i < count($array); $i++) {
        if ($i >= $n1 && $i <= $n2) {
            $sum += $array[$i];
        }
    }
    return $sum;
}

//Example

$array = [0, 1, 2, 3, 4, 5, 3];

echo calc($array, 0, 1); // 1

echo calc($array, 0, 5); // 15
user_155589040 
var sum = 0

for (index , item) in array.enumerated() {
    if n1 <= index && index <= n2{
        print("index: \(index)")
        sum += item
    }
}
return sum
user_20164740
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