Created
January 3, 2021 08:33
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Write Java code for this problem and analyze its complexity: Given an array of distinct integers and a target sum, find two numbers in the array that add up to the target sum.
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| import java.util.LinkedHashSet; | |
| import java.util.Set; | |
| public class Test { | |
| public static void main(String[] args) { | |
| int[] arr = {4, 5, 7, -5 - 3, 2, 8, 5, -1, 3, 8}; | |
| int[] distinctIntArray = makeDistinct(arr); | |
| getPairsCount(distinctIntArray, 9); | |
| } | |
| public static void getPairsCount(int[] arr, int targetSum) { | |
| for (int i = 0; i < arr.length; i++) { | |
| for (int j = i + 1; j < arr.length; j++) { | |
| if ((arr[i] + arr[j]) == targetSum) { | |
| System.out.println(String.format("Sum of [%d] and [%d] yield [%d] which is the target sum!", arr[i], arr[j], targetSum)); | |
| } | |
| } | |
| } | |
| } | |
| public static int[] makeDistinct(int[] ints){ | |
| Set<Integer> setString = new LinkedHashSet<>(); | |
| for(int i=0;i<ints.length;i++){ | |
| setString.add(ints[i]); | |
| } | |
| int[] distinctArray = setString.stream().mapToInt(Number::intValue).toArray(); | |
| return distinctArray; | |
| } | |
| } |
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