Searching for an integer in arraylist in java

Question

public class FindNumber {

            static String findNumber(List<Integer> arr, int k) {
                String res = "YES";
    //Unable to identify problem with this part of the code
                for (int i = 0; i < arr.size(); i++) {
                    if (k == arr.get(i))
                        res = "YES";
                    else
                        res = "NO";

                }

                return res;

            }
}

Above code returns NO as the answer even if the integer is present in the list.

Using == to compare Integer objects only works if the values are between -128 and 127, all other values will need to be compared using the .equals() method instead — JonK
– JonK, Commented Jul 18, 2019 at 9:20
there are several things wrong with your code. If you want true or false, return a boolean (for instance) — Stultuske
– Stultuske, Commented Jul 18, 2019 at 9:24
@JonK there is no comparison between Integer objects. The comparison is between Integer object and int primitive. Integer object is auto unboxed in this case. — THe_strOX
– THe_strOX, Commented Jul 18, 2019 at 9:25
@THe_strOX Absolutely correct, and that's why it's a comment instead of an answer. It's quite an easy gotcha to get bitten by when you're first starting out, so I feel it's still worth mentioning. — JonK
– JonK, Commented Jul 18, 2019 at 9:28

Pavel Smirnov · Accepted Answer · 2019-07-18 09:33:20Z

1

You could just use arr.contains() to get a boolean value of whether Integer is on the list or not. An then you can translate this value to YES or NO (if you do really need it):

String yesNo = arr.contains(k) ? "YES" : "NO";

answered Jul 18, 2019 at 9:33

Pavel Smirnov

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1 Comment

Stultuske Over a year ago

or, change the returntype to boolean and say: return arr.contains(k);

Paolo Mossini · Accepted Answer · 2019-07-18 09:22:01Z

0

This will work:

static String findNumber(List<Integer> arr, int k) {
            String res = "YES";
            for (int i = 0; i < arr.size(); i++) {
                if (k == arr.get(i))
                    res = "YES";
                    break;
                else
                    res = "NO";

            }

            return res;

        }

Once you find the integer you have to stop the loop, and you can do this using a break

answered Jul 18, 2019 at 9:22

Paolo Mossini

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1 Comment

Stultuske Over a year ago

if you want to go for a fix like that, you should have just replaced the whole if else with if ( found ) return "YES";

Istiaque Hossain · Accepted Answer · 2019-07-18 09:34:55Z

0

try optimize your code....

way 1 (using for-each loop):

 static String findNumber(List<Integer> arr, int k) { 
        for (Integer integer : arr) {
            if (integer == k) {
                return "YES";
            }
        }
        return "NO"; 
    }

another way is (using ternary operator) :

static String findNumber(List<Integer> arr, int k) { 
    return arr.contains(k) ? "YES" : "NO";
}

edited Jul 18, 2019 at 9:34

answered Jul 18, 2019 at 9:28

Istiaque Hossain

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Comments

Michael · Accepted Answer · 2019-07-18 09:35:37Z

0

Using streams:

static String findNumber(List<Integer> arr, int k) {
    return arr.stream()
        .filter(e -> e == k)
        .findFirst()
        .map(e -> "YES")
        .orElse("NO");
}

answered Jul 18, 2019 at 9:35

Michael

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Comments

Evgeny Shagov · Accepted Answer · 2019-07-18 09:45:06Z

0

    public static String isListContainsNumber(List<Integer> nums, int n) {
           return nums.stream().anyMatch(el -> el.equals(n)) ? "YES" : "NO";
    }

answered Jul 18, 2019 at 9:45

Evgeny Shagov

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Comments

Sourish Mukherjee · Accepted Answer · 2019-07-18 09:52:39Z

0

The main problem with your code is that even if it finds an integer object in the ArrayList, after setting res = Yes, it still continues the iteration. Therefore there is a possibility that there might be other values inside the list of not the desired data type, thereby setting res back to No. The solution here is to use a jump statement such as break which would terminate the loop process as soon as an integer is encountered. Hope it helps!

answered Jul 18, 2019 at 9:52

Sourish Mukherjee

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1 Comment

NavsTrend Over a year ago

Thanks :) .. jump statement is something i missed. Adding "break" after return "YES" solved the issue.

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