CODEWITHSUNDEEP
java
user13027438
user13027438

Reputation: 25

Compare 2 int arrays for matching numbers - java

Hi may someone suggest the fastest way to compare 2 arrays in java returning a string of an intersection array or a 'nil' value if there are no matches

static int [] WinnerNumbers; static int [] ClientNumbers;

After executing my program, I end up with 2 filled arrays (WinnerNumbers and ClientNumbers) of which I need to compare and return a match or null.

I would like to do this via a function

public static String WinnerChecker(){

      prize = (......)

      return prize;
}

Upvotes: 0

Views: 152

Answers (3)

WJS
WJS

Reputation: 40057

I'd do it like this. It returns an empty String if no intersection (which is what I presumed you meant by nil).

        int[] a = { 1, 2, 4, 3 ,5, 9};
        int[] b = { 4, 5, 6, 7, 8, 9, 11, 12, 13 };

        String intersection = getIntersection(a, b);
        System.out.println(intersection);

        public static String getIntersection(int[] a, int[] b) {
            return Arrays.stream(a)
                .flatMap(i -> Arrays.stream(b).filter(k -> i == k))
                .mapToObj(String::valueOf)
                .collect(Collectors.joining(""));
        }

Prints

459

If there are duplicates in either array and they are not wanted in the output, then add the method .distinct() right before the mapToObj process.

If one wants to return the result as an array, the following works.

    public static int[] getIntersection(int[] a, int[] b) {
        return Arrays.stream(a)
                .flatMap(i -> Arrays.stream(b).filter(k -> i == k))
                .toArray();

    }

Another way which may be more efficient than those above is using sets.

    public static String getIntersection(int[] a, int[] b) {
        Set<Integer> seta =
                Arrays.stream(a).boxed().collect(Collectors.toSet());
        Set<Integer> setb =
                Arrays.stream(b).boxed().collect(Collectors.toSet());

        seta.retainAll(setb);
        return seta.stream().map(String::valueOf)
                .collect(Collectors.joining(""));
    }

The actual set could be returned instead if that is desired.

Upvotes: 1

chaitanya dalvi
chaitanya dalvi

Reputation: 1669

Hope below code useful for you.

 String prize="";
    int[] winnerNumbers={ 23,84,32,98,11};
    int[] clientNumbers ={ 82, 70,81,98,41,32};
    for(int i=0; i< winnerNumbers.length; i++){
        for(int j=0; j< clientNumbers.length; j++){

            if(winnerNumbers[i] == clientNumbers[j]){
                prize = prize + winnerNumbers[i] +" ";
            }
        }
    }
    System.out.println(prize);

Here linear search operation perform to locate number. You can Java 8 stream class or other searching algorithm as per you array size. For more searching algorithm refer below link. https://www.geeksforgeeks.org/check-if-a-value-is-present-in-an-array-in-java/

Upvotes: 0

shakhawat
shakhawat

Reputation: 2725

Create a set with the winner array. For each number of the client check if the number exists in the set.

  • Set contains is O(1)
  • Iterate over client array O(N)

Overall complexity will be O(N)

private static String findWinners(int[] winnerNumbers, int[] clientNumbers) {
    Set<Integer> winnerNumberSet = IntStream.of(winnerNumbers).boxed().collect(Collectors.toSet());

    Optional<String> output = IntStream.of(clientNumbers)
            .filter(winnerNumberSet::contains)
            .mapToObj(String::valueOf)
            .reduce((result, element) ->
                    result + ", " + element);

    return output.orElse(null);
}

Upvotes: 0

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